Descripción: solutions for mathematics for economic analysis by Sydsaeter and Hammond...
Answers to even-numbered problems in Mathematics for Economic Analysis Knut Sydsæter Peter Hammond
Preface Mathematics for Economic Analysis, Prentice Hall, 1995 has been out for a long time, and over the years we have had many request for supplying solutions to the even-numbered problems. (Answers to the odd-numbered problems are given in the main text.) This manual provides answers to all the even-numbered problems in the text. These answers are taken from the old Instructors Manual which is no longer available. (In fact, some of the old answers have been extended.) For many of the more interesting and/or difficult problems, detailed solutions are provided. For some of the simpler problems, only the final answer is presented. Appendix A in the main text reviews elementary algebra. This manual includes a Test I, designed for the students themselves to see if they need to review particular sections of Appendix A. Many students using our text will probably have some background in calculus. The accompanying Test II is designed to give information to the students about what they actually know about single variable calculus, and about what needs to be studied more closely, perhaps in Chapters 6 to 9 of the text. Oslo and Coventry, November 2010 Knut Sydsæter and Peter Hammond Contact addresses:
[email protected] [email protected]
Version 1.0 07122010 1113 © Knut Sydsæter and Peter Hammond 2010
CHAPTER 1
INTRODUCTION
1
Chapter 1 Introduction 1.3 2. (a) Put p/100 = x. Then the given expression becomes a + ax − (a + ax)x = a(1 − x 2 ), as required. (b) $2000 · 1.05 · 0.95 = $1995. (c) The result is precisely the formula in (a). (d) With the notation used in the answer to (a), a − ax + (a − ax)x = a(1 − x 2 ), which is the same expression as in (a). 4. (a) F = 32 yields C = 0; C = 100 yields F = 212. (b) F = 95 C + 32 (c) F = 40 for C ≈ 4.4, F = 80 for C ≈ 26.7. The assertion is meaningless. R(1 − p) − S(1 − q) pS − qR , y= (p − q)(1 − p − q) (p − q)(1 − p − q) (Use, for example, (A.39) in Section A.9.)
6. x =
if (p − q)(1 − p − q) = 0.
1.4 2. (a) Correct. (b) Incorrect. (c) Incorrect. (d) Correct. (e) Incorrect. (f) Incorrect. (“Usually” the sum of two irrationals is irrational, but not always. For example, π and −π are both irrationals, but π + (−π) = 0, which is rational.) 4. (a) y ≤ 3 − 43 x
(c) y ≤ (m − px)/q
(b) y > 23 z
6. |2 · 0 − 3| = | − 3| = 3, |2 ·
1 2
− 3| = | − 2| = 2, |2 ·
7 2
− 3| = |4| = 4
8. (a) 3 − 2x = 5 or 3 − 2x = −5, so −2x = 2 or √ −8. Hence √ x = −1 or x = 4. (b) −2 ≤ x ≤ 2 (c) 1 ≤ x ≤ 3 √(d) −1/4 ≤ x ≤ 1 (e) x > 2 or x < − 2 √ (f) 1 ≤ x 2 ≤ 3, and so 1 ≤ x ≤ 3 or − 3 ≤ x ≤ −1
1.5 2. (a) ⇒ right, ⇐ wrong (b) ⇒ wrong, ⇐ right (c) (d) ⇒ and ⇐ both right (e) ⇒ wrong (0 · 5 = 0 · 4, but 5 = 4), ⇐ right
⇒ right, ⇐ wrong (f) ⇒ right, ⇐ wrong
4. x = 2. (x = −1, 0 or 1 make the equation meaningless. Multiplying each term by the common denominator x(x − 1)(x + 1) yields (x + 1)3 + (x − 1)3 − 2x(3x + 1) = 0. Expanding and simplifying, 2x 3 − 6x 2 + 4x = 0, or 2x(x 2 − 3x + 2) = 0, or 2x(x − 1)(x − 2) = 0. Hence, x = 2 is the only solution.) √ √ 6. (a) No solutions. (Squaring each side yields x−4 = x+5−18 x + 5+81, which reduces to x + 5 = 5, with solution x = 20. But x√= 20 does not satisfy the given equation.) (b) Just as in part (a) we find that x must be a solution of x + 5 = 5, which has the solution x = 20. Inserting this value of x into the given equation we find that x = 20 is the solution. √ √ (1) (2) 8. (a) x + x + 4 = 2 ⇒ x + 4 = 2−x ⇒ x +4 = 4−4x +x 2 ⇒ x 2 −5x = 0 ⇒ x −5 = 0 ⇐ x = 5. Here implication (1) is incorrect (x 2 − 5x = 0 ⇒ x − 5 = 0 or x = 0.) Implication (2) is correct, but it breaks the chain of implications. (b) x = 0. (By modifying the argument we see that the given equation implies x = 5 or x = 0. But only x = 0 is a solution.) 10. No. (To study law it seems that you need a course in logic.) © Knut Sydsæter and Peter Hammond 2010
2
CHAPTER 1
INTRODUCTION
1.6 2. Logically the two statements are equivalent, but the second statement is still an expressive poetic reinforcement.
1.7 2. (a) No, not necessarily.
(b) Yes (if anybody is both a painter and a poet).
4. F ∩ B ∩ C is the set of all female biology students in the university choir; M ∩ F the female mathematics students; (M ∩ B) \ C \ T the students who study both mathematics and biology but neither play tennis nor belong to the university choir. 6. (b) and (c) are true, the others are wrong. (Counter example for (a), (d), and (f): A = {1, 2}, B = {1}, C = {1, 3}. As for (e), note in particular that A ∪ B = A ∪ C = A whenever B and C are subsets of A, even if B = C.) 8. 50 − 35 = 15 liked only coffee, 40 − 35 = 5 liked only tea, 35 liked both, and 10 did not like either. In all there were 15 + 5 + 35 + 10 = 65 who responded. 10. (a) Consider Fig. 1.7.10(a). Both sets consist of the elements in (1) and in (3). (b) Consider Fig. 1.7.10(b): A B consists of (3), (4), (2), and (5). Then (A B) C consists of (4), (5), (6), and (7), which is described verbally in the problem.
B
A (5)
(1) (4) (7)
(2)
(3) A
(1)
(2)
(3)
B (8)
(6) C
Figure 1.7.10(a)
Figure 1.7.10(b)
12. (a) Consider Fig. 1.7.10(b), and let ni denote the number of people in the set marked (i), for i = 1, 2, . . . , 8. The responses reported imply that: n1 +n3 +n4 +n7 = 420; n1 +n2 +n5 +n7 = 316; n2 +n3 +n6 +n7 = 160; n1 + n7 = 116; n3 + n7 = 100; n2 + n7 = 30; n7 = 16. From these equations we easily find n1 = 100, n2 = 14, n3 = 84, n4 = 220, n5 = 186, n6 = 46, n7 = 16. (i) n3 + n4 = 304 had read A but not B; (ii) n6 = 46; (iii) 1000 − (n1 + n2 + n3 + n4 + n5 + n6 + n7 ) = 334. (b) (i) n(A \ B) (ii) n(C \ (A ∪ B)) (iii) n( \ (A ∪ B ∪ C)) = n() − n(A ∪ B ∪ C)) (c) The equality is n1 + n2 + n3 + n4 + n5 + n6 + n7 = (n1 + n3 + n4 + n7 ) + (n1 + n2 + n5 + n7 ) + (n2 + n3 + n6 + n7 ) − (n1 + n7 ) − (n3 + n7 ) − (n2 + n7 ) + n7 , which is easily verified. The last equality is a special case of n( \ D) = n() − n(D): The number of persons who are in , but not in D, is the number of persons in all of minus the number of those who are in D. © Knut Sydsæter and Peter Hammond 2010
CHAPTER 2
FUNCTIONS OF ONE VARIABLE: INTRODUCTION
3
Chapter 2 Functions of One Variable: Introduction 2.2 2. F (0) = F (−3) = 10, F (a + h) − F (a) = 10 − 10 = 0 √ √ √ √ 2/3, f ( π ) = π /(1 + π ), 4. (a) f (−1/10) = −10/101, f (0) = 0, f (1/ 2) = f (2) = 2/5 (b) f (−x) = −x/(1 + (−x)2 ) = −x/(1 + x 2 ) = −f (x) and moreover, f (1/x) = (1/x)/[1 + (1/x)2 ] = (1/x) · x 2 /[1 + (1/x)2 ] · x 2 = x/(1 + x 2 ) = f (x) 6. F (0) = 2, F (−3) =
√ √ 19, F (t + 1) = t 2 + 3
8. (a) b(0) = 0, b(50) = 100/11, b(100) = 200 (b) b(50 + h) − b(50) is the additional cost of removing h% more than 50% of the impurities. 10. (a) No. f (2 + 1) = f (3) = 2 · 32 = 18, whereas f (2) + f (1) = 2 ·√22 + 2 · 12 = 8 + 2 = 10. (b)Yes. f (2+1) = f (2)+f (1) = −9 (c) No. f (2+1) = f (3) = 3 ≈ 1.73, whereas f (2)+f (1) = √ 2 + 1 ≈ 2.41. 12. See Figs. 2.2.12(a) and 2.2.12(b).
1
x·1
1·1
1 1
x
x2
1·x
x
1
Figure 2.2.12(a). Area = (x + 1)2 = x 2 + 2x + 1
x
x Figure 2.2.12(b) Area = x 2 +1
14. (a) (−∞, 2) ∪ (2, ∞) (b) f (8) = 5 3x + 6 (c) f (x) = = 3 ⇐⇒ 3x + 6 = 3(x − 2) ⇐⇒ 6 = −6, which is absurd. x−2 16. (a) is invalid: |2 + (−2)| = 0, whereas |2| + | − 2| = 2 + 2 = 4. (b), called the triangle inequality, |x + y| ≤ |x| + |y|, is valid. In fact, if x and y have the same sign, then |x + y| = |x| + |y|. If x and y have opposite signs, then we see that |x + y| < |x| + |y|. (See Problem 12.4.8 for a generalization.) (c) is valid: |xy| = |x| · |y|. (Look at the 4 different sign combinations of x and y: If x is positive and y is negative, then xy is negative, |xy| = −xy and |x|·|y| = (−x)·y = −xy, and so on.) (d) is easily seen to be valid. (e) is not valid (put x = 1). (f) is easily seen (Note that −4)2 = −(−2) = 2.) to be valid. (g) is valid: | − 2x| = 2|x|. (h) is valid because |x| − |y| ≤ |x − y| ⇔ x 2 − 2|x||y| + y 2 ≤ x 2 − 2xy + y 2 ⇔ −2|x||y| ≤ −2xy ⇔ |x||y| ≥ x · y, which is true because (c) is valid. © Knut Sydsæter and Peter Hammond 2010
4
CHAPTER 2
FUNCTIONS OF ONE VARIABLE: INTRODUCTION
2.3 2. See Figs. 2.3.2(a)—2.3.2(f). y
y
y 4 3 2
1
1
1
x
1
x
1
Figure 2.3.2(a)
1
Figure 2.3.2(b)
y
2
x
3
Figure 2.3.2(c)
y
y
1
1
2 1 2
1
1
2
x
x
1
x
1
1
Figure 2.3.2(d)
Figure 2.3.2(e)
Figure 2.3.2(f)
4. See Figs. 2.3.4(a)–(c). y
y
y
2
2
2
1
1
1
−3 −2 −1 −1
1
2
−1
−3 −2 −1 −1
−2
−2
3x
−2
Figure 2.3.4(a)
1
2
3
4
5
x
Figure 2.3.4(b)
1
2
3x
Figure 2.3.4(c)
y y y A = (3, 2)
2 (2, 4) 2
x
1
2 P 2
5
B = (5, −4)
x
Figure 2.3.6
© Knut Sydsæter and Peter Hammond 2010
Figure 2.3.10
1 Figure 2.3.12
x
CHAPTER 2
FUNCTIONS OF ONE VARIABLE: INTRODUCTION
5
6. (5 − 2)2 + (y − 4)2 = 13, or y 2 − 8y + 12 √ = 0, with solutions y = 2 and y = 6. Geometric explanation: The circle with center at (2, 4) and radius 13 intersects the line x = 5 at two points. If the radius were 2, the circle would not intersect the line x = 5. See Fig. 2.3.6. 8. (a) (x − 2)2 + (y − 3)2 = 16
(b) (x − 2)2 + (y − 5)2 = 13
10. (x − 3)2 + (y − 2)2 = (x − 5)2 + (y + 4)2 , which reduces to x − 3y = 7. See Fig. 2.3.10 12. (x, y) = (1, 0), (0, 1), (1/4, 1/4) are all solutions. See Fig. 2.3.12. 14. The condition is: p + r x 2 + y 2 = p + s (x − a)2 + y 2 . Cancelling p, then squaring each side and rearranging yields (r 2 − s 2 )(x 2 + y 2 ) = s 2 (a 2 − 2ax). If r = s, then the markets are separated by the straight line x = 21 a. Otherwise, completing squares yields the equation [x + as 2 /(r 2 − s 2 )]2 + y 2 = [ars/(r 2 − s 2 )]2 . This is a circle with center at − as 2 /(r 2 − s 2 ), 0 and radius |ars/(r 2 − s 2 )|.
2.4 2. (a) f (−5) = 0, f (−3) = −3, f (−2) = 0, f (0) = 2, f (3) = 4, f (4) = 0 (b) Df = [−5, 4], Vf = [−3, 4] 4. See Figs. 2.4.4(a)–(d). y
y
y
y
1 1 1
1
1 1
x
3
Figure 2.4.4(a)
x
1
Figure 2.4.4(b)
x
1
x
1
Figure 2.4.4(c)
Figure 2.4.4.4(d)
2.5 2. 0.78
4. See Figs. 2.5.4(a)–(c). y
y
y
1
4 3
−1
2
−2 −3
1 1
2
Figure 2.5.4(a)
3
4
x
4 1 2
3 4
5 6 7
8
10 x
3 2 1
−4 −5
−1
Figure 2.5.4(b)
Figure 2.5.4(c)
1
2
3
4
5 x
6. (a) Assume F = aC + b. Then 32 = a · 0 + b and 212 = a · 100 + b. Therefore a = 180/100 = 9/5 and b = 32, so F = (9/5)C + 32. (b) If X = (9/5)X + 32, then X = −40. © Knut Sydsæter and Peter Hammond 2010
6
CHAPTER 3
POLYNOMIALS, POWERS, AND EXPONENTIALS
8. L: y = 3x − 2; M: y = − 43 x + 45 . P : (13/15, 3/5). N : y = − 43 x − 47 . See Fig. 2.5.8. 10. (a) 75 − 3P e = 20 + 2P e , and so P e = 11.
(b) P e = 90
12. C = 45 y + 100. (The general equation is C = ay + b. Here 900 = a · 1000 + b, and a = 80/100 = 4/5, so b = 100.) 1 1 − x0 + h x0 . Multiplying both numerator and denominator by the product (x0 + h)x0 , then 14. The slope is x0 + h − x 0 simplifying, yields the given expression. 16. See Figs. 2.5.16(a)–(c). y
L
1
y
y
y
1
1
1
x
1
x
1
x
1
1
x
M N Figure 2.5.8
Figure 2.5.16(a)
Figure 2.5.16(b)
Figure 2.5.16(c)
Chapter 3 Polynomials, Powers, and Exponentials 3.1 2. (a)
−4
−3
−2
−1
0
1
2
−2.5
0
1.5
2
1.5
0
−2.5
x f (x)
(b) See Fig. 3.1.2(b). (c) f (x) = (−1/2)(x + 1)2 + 2. Maximum point (−1, 2). (d) x = −3 and x = 1 (e) f (x) > 0 in the interval (−3, 1), while f (x) < 0 for x < −3 and for x > 1. y 3 2 f (x) = − 21 x 2 − x + 1 −4 −3 −2−1 −1 −2 −3 −4 Figure 3.1.2(b)
© Knut Sydsæter and Peter Hammond 2010
1 2 3 x
3 2
CHAPTER 3
POLYNOMIALS, POWERS, AND EXPONENTIALS
7
4. (a) x(x + 4). Zeros 0 and −4. (b) Factorization not possible. No zeros. √ √ (c) −3(x − x1 )(x − x2 ), where the zeros are x1 = 5 + √15 and x2 = 5 − √ 15. (d) 9(x − x1 )(x − x2 ), where the zeros are x1 = 1/3 + 5 and x2 = 1/3 − 5. (e) −(x + 300)(x − 100), where the zeros are −300 and 100. (f) (x + 200)(x − 100), where the zeros are −200 and 100. 6. (a) If the other side is y, then 2x + 2y = L, and so y = L/2 − x. The area is then A(x) = x(L/2 − x) = Lx/2 − x 2 . The square with side L/4 gives the largest area. (b) Yes; the radius is L/2π so the area is L2 /4π > L2 /16. 8. (a) x = ± 1, x = ± 2 (b) (i) x 2 = 9 (x 2 < 0 is impossible), so x = ± 3 (ii) x 3 = 1 or x 3 = 8, so x = 1 or x = 2 10. y = 2x 2 + x − 6. ((1, −3) belongs to the graph if −3 = a + b + c, (0, −6) belongs to the graph if −6 = c, and (3, 15) belongs to the graph if 15 = 9a + 3b + c. It follows that a = 2, b = 1, and c = −6.) 12. (a) (i) 289 ≤ 290 (ii) 361 ≤ 377 (b) If B 2 − 4AC were > 0, then according to formula [3.2] the equation f (x) = 0 would have two distinct solutions, which is impossible when f (x) ≥ 0 for all x. We find that A = a12 + a22 + · · · + an2 , B = 2(a1 b1 + a2 b2 + · · · + an bn ), and C = b12 + b22 + · · · + bn2 , so the conclusion follows.
3.3 2. (a) Integer roots must divide 6. Thus ±1, ±2, ±3, and ±6 are the only possible integer solutions. We find that −2, −1, 1, 3 all are roots, and since there can be no more than 4 roots in a polynomial equation of degree 4, we have found them all. (b) The same possible integer solutions. Only −6 and 1 are integer solutions. (The third root is −1/2.) (c) Neither 1 nor −1 satisfies the equation, so there are no integer roots. (d) First multiply the equation by 4 to have integer coefficients. Then ±1, ±2, and ±4 are seen to be the only possible integer solutions. In fact, 1, 2, −2 are all solutions. 4. (a) The answer is 2x 2 + 2x + 4 + 3/(x − 1), because + 2x − 1) ÷ (x − 1) = 2x 2 + 2x + 4 (2x 3 3 2 2x − 2x 2x 2 + 2x − 1 2x 2 − 2x 4x − 1 4x − 4 3 (b) The answer is
x2
remainder
+ 1, because (x 4 + x 3 + x 2 + x) ÷ (x 2 + x) = x 2 + 1 x4 + x3 x2 + x x2 + x 0
© Knut Sydsæter and Peter Hammond 2010
no remainder
8
CHAPTER 3
POLYNOMIALS, POWERS, AND EXPONENTIALS
(c) The answer is 3x 5 + 6x 3 − 3x 2 + 12x − 12 + (28x 2 − 36x + 13)/(x 3 − 2x + 1): x2
+ 1) ÷ (x 3 − 2x + 1) = 3x 5 + 6x 3 − 3x 2 + 12x − 12
x2
+ 1
− 3x 5 + 12x 4 − 6x 3 + x 2 + 6x 3 − 3x 2 − 3x 5
+ 1
(3x 8 3x 8 − 6x 6 + 3x 5 6x 6 − 3x 5 + − 12x 4 + 6x 3 6x 6
12x 4 − 12x 3 + 4x 2 + 1 − 24x 2 + 12x 12x 4 − 12x 3 + 28x 2 − 12x + 1 + 24x − 12 − 12x 3 28x 2 − 36x + 13
remainder
(d) The answer is x 3 − 4x 2 + 3x + 1 − 4x/(x 2 + x + 1), because (x 5 − 3x 4 x5 + x4 + x3
+ 1) ÷ (x 2 + x + 1) = x 3 − 4x 2 + 3x + 1
− 4x 4 − x 3 − 4x 4 − 4x 3 − 4x 2
+1
3x 3 + 4x 2 +1 3 2 3x + 3x + 3x x 2 − 3x + 1 x2 + x + 1 − 4x
remainder
6. (a) p(x) = x(x 2 + x − 12) = x(x − 3)(x + 4), because x 2 + x − 12 = 0 for x = 3 and x = −4. (b) ±1, ±2, ±4, ±8 are the only possible integer zeros. By trial and error we find that q(2) = q(−4) = 0, so 2(x − 2)(x + 4) = 2x 2 + 4x − 16 is a factor for q(x). By polynomial division we find that q(x) ÷ (2x 2 + 4x − 16) = x − 1/2, so q(x) = 2(x − 2)(x + 4)(x − 1/2).
3.4 2. (a) 2.511886 (b) 0.000098 (c) 0.530094 √ 16 100 50 (Find first the number that, when raised to the power of 100, yields 50. Then this 4. 500.16 = number is raised to the power of 16.) 1 6. (a) 23 = 8, so x = 3/2 (b) 81 = 3−4 , so 3x + 1 = −4, and then x = −5/3. 2 2 (c) x − 2x + 2 = 2, so x − 2x = 0, implying that x = 0 or x = 2.
8. (a) Multiply by 5K 1/2 to obtain K 1/2 = 15L1/3 . Squaring each side, K = 225L2/3 . (b) abx0b−1 = p, so x0b−1 = p/ab. Now raise each side to the power 1/(b − 1). −1/ρ b (c) x = − 2a . (Multiply each side by (ax + b)2/3 .) (d) b = λ1/ρ c−ρ − (1 − λ)a −ρ 10. K = 57 315.86 for Y = 100, L = 6, t = 10. © Knut Sydsæter and Peter Hammond 2010
CHAPTER 3
POLYNOMIALS, POWERS, AND EXPONENTIALS
9
3.5 ∗
(b) The doubling time t ∗ is given by the equation (1.034)t = 2, and we
2. (a) P = 1.22 · 1.034t million find t ∗ ≈ 20.7 (years). 4. (a) 210 = 1024
(b) f (t) = 2t
(c) f (20) = (210 )2 = (1024)2 > 10002 = 1 million.
6. The graph is shown in Fig. 3.5.6. x 2
x2
−2
−1
0
1
2
16
2
1
2
16
6 5
y
4 y 6
3 2 1
y
4 −2 −1 −1
1 2
y = x 2 2x
3 x 2
−2 −3 −4
1 1
x
Figure 3.5.6
−5 −4 −3 −2 −1
Figure 3.5.10
1
2 x
Figure 3.5.12
∗
8. If the initial time is t, the doubling time t ∗ is given by the equation Aa t+t = 2Aa t , which implies ∗ ∗ Aa t a t = 2Aa t , so a t = 2, independent of t. 10. (a) The graph is shown in Fig. 3.5.10.
x
−3
−2
−1
0
1
2
3
4
1 − 2−x
−7
−3
−1
0
1/2
3/4
7/8
15/16
(b) 1 − 2−x gets close to 1 as x becomes very large, and becomes very large negative as x becomes large negative. (Formally, limx→∞ f (x) = 1, limx→−∞ f (x) = −∞. See Section 6.1.) 12. The graph is drawn in Fig. 3.5.12. x
−10
−5
−4
−3
−2
−1
0
1
2
x 2 2x
0.1
0.8
1.0
1.1
1.0
0.5
0
2.0
16
14. I (t) = I0 (1/2)t/8 remains after t days. © Knut Sydsæter and Peter Hammond 2010
10
CHAPTER 4
SINGLE VARIABLE DIFFERENTIATION
3.6 2. The function in (b) is one-to-one and has an inverse: the rule mapping each youngest child alive today to his/her mother. (Though the youngest child of a mother with several children will have been different at different dates.) The function in (d) is one-to-one and has an inverse: the rule mapping the surface area to the volume. The function in (e) is one-to-one and has an inverse: the rule that maps (u, v) to (u − 3, v). The other functions are many-to-one, in general, and so have no inverses.
Chapter 4 Single Variable Differentiation 4.2 2. f (x) = 6x + 2, f (0) = 2, f (−2) = −10, f (3) = 20. The tangent equation is y = 2x − 1. 4.
f (x + h) − f (x) 1/(x + h) − 1/x x − (x + h) −h −1 1 = = = = → − 2 as h → 0, h h hx(x + h) hx(x + h) x(x + h) x which proves the implication.
6. (a) f (x + h) − f (x) = a(x + h)2 + b(x + h) + c − (ax 2 + bx + c) = 2ahx + bh + ah2 , so [f (x + h) − f (x)]/ h = 2ax + b + ah → 2ax + b as h → 0. Thus f (x) = 2ax + b. (b) f (x) = 0 for x = −b/2a. The tangent is parallel to the x-axis at the minimum/maximum point. √ √ 8. (a) Use [A.12] in Section A.3. (b) and (c) [f (x + h) − f (x)]/ h = x + h − x)/ h. Multiplying both √ √ numerator and denominator by x + h + x , then using the identity in (a), yields the result. Letting √ √ h → 0, the formula follows. (d) x = x 1/2 and 1/ x = x −1/2 . 10. (a) [f (x + h) − f (x)]/ h = [(x + h)1/3 − x 1/3 ]/ h. Then use the hint. (b) Follows from (a) by letting h → 0.
4.3 2. I is the fixed cost, whereas k is the marginal cost, and also the incremental cost of producing one additional unit. 4. T (y) = t, so the marginal tax rate is constant.
4.4 2. (a) The following table seems to indicate that the limit is 9. x
0.9
0.99
0.999
1
1.001
1.01
1.1
x 2 +7x−8 x−1
8.9
8.99
8.999
∗
9.001
9.01
9.1
*not defined (b) x 2 + 7x − 8 = (x − 1)(x + 8), so for x = 1,
4. (a) 22 + 3 · 2 − 5 = 5 6. (a) 4
(b) 5
(c) 6
(b) 1/5
(d) 2a + 2
© Knut Sydsæter and Peter Hammond 2010
(c) 1
x 2 + 7x − 8 = x + 8 → 9 as x → 1. x−1
(d) −2
(e) 2a + 2
(e) 3x 2
(f) 4a + 4
(f) h2 8. (a) 1/6
(b) 1/27
(c) n
CHAPTER 4
SINGLE VARIABLE DIFFERENTIATION
11
4.5 2. (a) 2g (x) (b) (−1/6)g (x) (c) (1/3)g (x) 6. Add an arbitrary constant to these answers:
(b) A(b + 1)y b (c) (−5/2)A−7/2 x a+1 (b) x 2 + 3x (c) a+1
4. (a) 8π r (a) (1/3)x 3
4.6 2. (a) (6/5)x − 14x 6 − (1/2)x −1/2
(b) 4x(3x 4 − x 2 − 1)
(c) 10x 9 + 5x 4 + 4x 3 − x −2
3 −x 4 + 5x 2 + 18x + 2 4. (a) √ √ (b) (c) −2(1+2x)x −3 (x 2 + 2)2 (x + 3)2 2 x( x + 1)2 2(x 3 − x 2 + x + 1) (f) 3x 3 (x + 1)2 √ √ √ √ 6. (a) x = 2 (b) x = − √ 3, x = 0, x = 3 (c) x = − 2, x = 2 √ (d) x = 0, x = −1/2 − 5/2, x = −1/2 + 5/2 8. (a)
ad − bc (ct + d)2
(b) a(n + 1/2)t n−1/2 + nbt n−1
(c)
(d)
4x (x 2 + 1)2
(e)
−2x 2 + 2 (x 2 − x + 1)2
−(2at + b) (at 2 + bt + c)2
10. Differentiating f (x) · f (x) = x gives f (x) · f (x) + f (x) · f (x) = 1, so 2f (x) · f (x) = 1. Hence, 1 1 f (x) = = √ . 2f (x) 2 x 12. The Newton quotient of F is f (x + h)/g(x + h) − f (x)/g(x) F (x + h) − F (x) = h h Multiplying the numerator and denominator by g(x)g(x+h) yields the expression for the Newton quotient given in the hint. Letting h → 0 yields the desired result.
4.7 2. dy/dx = ax a−1 − ax −a−1 and d 2 y/dx 2 = a(a − 1)x a−2 + a(a + 1)x −a−2 4. g (t) = (t 2 − 2t)/(t − 1)2 , so g (t) = [(2t − 2)(t − 1)2 − (t 2 − 2t)2(t − 1)]/(t − 1)4 = 2 for t = 2. 6. For n = 1, the formula is true. Suppose it is true for n = k, so that y = x k ⇒ y (k) = k! . Then (d k+1 /dx k+1 )x k+1 = (d k /dx k )(d/dx)(x k+1 ) = (d k /dx k )(k +1)x k = (k +1)(d k /dx k )x k = (k +1)·k!, by the induction hypothesis. This is equal to (k + 1)!, so the given formula is also true for n = k + 1. By induction, the formula is true for all n.
© Knut Sydsæter and Peter Hammond 2010
12
CHAPTER 5
MORE ON DIFFERENTIATION
Chapter 5 More on Differentiation 5.1 2. (a) 3(2x + 1)2 · 2 = 6(2x + 1)2 (b) −5(1 − x)4 (c) 2(x 2 − 2x + 2)(2x − 2) (d) (x + 1)4 (4x − 1)/x 2 (e) −21(3x − 4)−8 (f) −2(4x + 3)(2x 2 + 3x − 4)−3 −b(a + 1) at + b a 4. (a) −6at (at 2 + 1)−4 (b) na(at + b)n−1 (c) nt 2 nt 6. dx/dt = Ar(Ap + B)r−1 (2at + b) 8. Putting y = 1/(g(x))n implies that y = [0·(g(x))n −1·n(g(x))n−1 g (x)]/(g(x))2n = −n(g(x))−n−1 g (x).
k+1 = 10. For a = 1, the formula is true. Suppose the formula is true for a = k. Consider y = g(x)
k g(x) · g(x). The product rule for differentiation and the induction hypothesis yield
k−1
k
k y = k g(x) · g (x) · g(x) + g(x) · g (x) = (k + 1) g(x) · g (x), which is the given formula for a = k + 1. By induction, the formula is true for all natural numbers a.
5.2 2. (a) dY /dt = (dY /dV )(dV /dt) = (−3)5(V + 1)4 t 2 = −15(t 3 /3 + 1)4 t 2 (b) dK/dt = (dK/dL)(dL/dt) = AaLa−1 b = Aab(bt + c)a−1 4. dY /dt = (dY /dK) · (dK/dt) = Y (K(t0 ))K (t0 ) √ 1 dx a √ ap − c = b − u, with u = ap − c. Then = − √ u = − √ . (The restriction dp 2 ap − c 2 u should be p > c/a, for x to be differentiable.) 8. (a) b(t) is the total fuel consumption after t hours. (b) b (t) = B s(t) s (t). So the rate of fuel consumption per hour is equal to the rate per kilometer multiplied by the speed in kph.
6. x = b −
10. f (f (x)) = f (3x + 7) = 3(3x + 7) + 7 = 9x + 28. Then 9x + 28 = 100 for x = 8. 14. p (x) = 2(x − a)q(x) + (x − a)2 q (x), so p (a) = 0. 12. dC/dx = q(25 − 21 x)−1/2 16. F (x) = f x n g(x) nx n−1 g(x) + x n g (x)
5.3 2. By implicit differentiation, 4x + 6y + 6xy + 2yy = 0. So y = − at the point (1, 2).
2x + 3y . In particular, y = −8/5 3x + y
4. (a) 2x + 2yy = 0, and solve for y to get y = −x/y. √ √ √ (b) 1/2 x + y /2 y = 0, and solve for y to get y = − y/x. (c) 4x 3 − 4y 3 y = 2xy 3 + x 2 3y 2 y , and so y = 2x(2x 2 − y 3 )/y 2 (3x 2 + 4y). 6. (a) No relationship.
(b) f (x0 ) = g (x0 )
8. Differentiating y n = x m w.r.t. x yields ny n−1 y = mx m−1 , so y = mx m−1 /ny n−1 = mx m−1 /n(x m/n )n−1 = (m/n)x m−1−(m/n)(n−1) = (m/n)x (m/n)−1 . © Knut Sydsæter and Peter Hammond 2010
CHAPTER 5
MORE ON DIFFERENTIATION
13
5.4 2. f (x) ≈ f (0) + f (0)x =
1 9
−
10 27 x
4. F (1) = A, F (K) = αAK α−1 , so F (1) = αA. Then F (K) ≈ F (1) + F (1)(K − 1)= A + αA(K − 1) = A(1 + αA(K − 1)). 6. (a) (i) y = 0.61, dy = 0.6 (ii) y = 0.0601, dy = 0.06 (b) (i) y = 0.011494, dy = 0.011111 (ii) y = 0.001124, dy = 0.001111 (c) (i) y = 0.012461, dy = 0.0125 (ii) y = 0.002498, dy = 0.0025 8. g(0) = A − 1 and g (μ) = Aa/(1 + b) (1 + μ)[a/(1+b)]−1 , so g (0) = Aa/(1 + b). Hence, g(μ) ≈ aA g(0) + g (0)μ = A − 1 + μ 1+b
5.5 2. Follows from formula [5.9] with f = U , a = y, x = y + M − s. 4. We find x(0) ˙ = 2[x(0)]2 = 2 · 1 = 2. Differentiating the expression for x(t) ˙ yields x(t) ¨ = x(t) + t x(t) ˙ + 2 = 4[x(t)]x(t), ˙ and so x(0) ¨ = x(0)+4[x(0)]x(0) ˙ = 1+4·1·2 = 9. Hence, x(t) ≈ x(0)+ x(0)t ˙ + 21 x(0)t ¨ 1 + 2t + 29 t 2 . (px p−1 − qx q−1 )(x p + x q ) − (x p − x q )(px p−1 + qx q−1 ) 2(p − q)x p+q−1 = , so h (1) = (x p + x q )2 (x p + x q )2 1 1 2 (p − q). Since h(1) = 0, h(x) ≈ h(1) + h (1)(x − 1) = 2 (p − q)(x − 1).
6. h (x) =
5.6 2. ElK T = 1.06. A 1% increase in expenditure on road building leads to an increase in the traffic volume of approx. 1.06 %. 4. (a) Elx f (x) = 0
(b) Elx f (x) = x/(x + 1) (c) Elx f (x) = −20x 2 /(1 − x 2 ) f (x)Elx f (x) 6. Elx Af (x) = Elx f (x), Elx A + f (x) = A + f (x) 1 + 2x 2x 2 30x 3 (c) 3 (d) Elx 5x 2 = 2, so Elx Elx 5x 2 = 0 (e) 1+x x +1 1 + x2 x−1 x 5x 5 x Elx x x 5 Elx x 5 (f) Elx = Elx (x − 1) − Elx (x 5 + 1) = − 5 = − 5 (Some answers 5 x +1 x−1 x +1 x−1 x +1 are easier if the results of Problem 6 are used as well.)
8. (a) −3
(b)
10. (a) 6Elx y = 5, so Elx y = 5/6 (b) Using rules (c) and (b) in Problem 7, we obtain Elx y − Elx x = Elx (x + 1)a + Elx (y − 1)b . Here Elx x = 1, whereas Elx (x + 1)a = [x/(x + 1)a ]a(x + 1)a−1 = ax/(x + 1) Elx (y − 1)b = [x/(y − 1)b ]b(y − 1)b−1 y = [by/(y − 1)]Elx y Thus, Elx y − 1 = ax/(x + 1) + [by/(y − 1)]Elx y. Solving for Elx y yields (ax + x + 1)(y − 1) Elx y = (y − by − 1)(x + 1) © Knut Sydsæter and Peter Hammond 2010
14
CHAPTER 6
LIMITS, CONTINUITY, AND SERIES
Chapter 6 Limits, Continuity, and Series 6.1 x−3 (1/x) − (3/x 2 ) 2. (a) 2 = −→ x +1 1 + (1/x 2 ) x→∞
(b)
2 + 3x = x−1
√ 3 + 2/x −→ 3 1 − 1/x x→∞
(c) a 2
4. limx→∞ fi (x) = ∞ for i = 1, 2, 3; limx→∞ f4 (x) = 0. Then: (a) ∞ (b) 0 (c) −∞ (d) 1 (e) 0 (f) ∞ (g) 1 (h) ∞ 6. y = Ax + A(b − c) + d is an asymptote as x → ∞.
6.2 2. (a) None of the six functions is continuous at a. (b) Only in (i) does f have a limit: limx→a f (x) = A (c) (i) limx→a − f (x) = limx→a + f (x) = limx→a f (x) = A, (ii) limx→a − f (x) = f (a), limx→a + f (x) = A; (iii) limx→a − f (x) = limx→a + f (x) = ∞; (iv) limx→a − f (x) = −∞, limx→a + f (x) = f (a); (v) limx→a − f (x) = ∞, limx→a + f (x) = f (a); (vi) limx→a − f (x) = A, limx→a + f (x) = B. (d) Only the function in (ii) is left-continuous at x = a. The functions in (iv), (v), and (vi) are right-continuous at x = a. (e) (v) limx→∞ f (x) = A; (vi) limx→∞ f (x) does not exist. 4. (a) Continuous for all x. (b) Continuous for all√ x = 1. (c) Continuous √ for all x < 2. (d) Continuous for all x. (e) Continuous for all x where x = 3 − 1 and x = − 3 − 1. (f) Continuous for all x where x ≤ −1 or x > 1. (g) Continuous for all x > 0. (h) Continuous for all x = 0. (i) Continuous for all x > 0. 6. See Fig. 6.2.6; y is discontinuous at x = a. In the case where y is the nearest point on the ground, it would be a continuous function. y
a
x
Figure 6.2.6
8. Because limx→0− f (x) = −2 and limx→2+ f (x) = 3, we need f (0) = −2 and f (2) = 3. Because f is linear on [0, 2], put f (x) = (5/2)x − 2 for x ∈ [0, 2].
6.3 2. f (0+ ) = 1, f (0− ) = −1. At x = 0, f is continuous, but not differentiable. © Knut Sydsæter and Peter Hammond 2010
CHAPTER 6
4. The tax function is
⎧ ⎨ 0.15x t (x) = 0.28x − 2 632.50 ⎩ 0.31x − 4 111.50
LIMITS, CONTINUITY, AND SERIES
15
for x ∈ [0, 20 250] for x ∈ (20 250, 49 300] for x ∈ (49 300, ∞)
In particular, t (22 000) = 3527.50 and t (50 000) = 11388.50.
6.4 2. (a) Converges to 5.
(b) Diverges (to ∞).
(c) √
3n 2n2 − 1
=
√ 3 3 2 −→ √ = 2 2 − 1/n2 n→∞ 2 3
6.5 2. (a) Geometric series with quotient 1/8. Its sum is 8/(1 − 1/8) = 64/7. (b) Geometric with quotient −3. It diverges. (c) Geometric, with sum 21/3 /(1 − 2−1/3 ). (d) Not geometric. (One can show that the series is convergent with sum ln 2.) 4. Geometric series with quotient (1 + p/100)−1 . Its sum is b/[1 − (1 + p/100)−1 ] = b(1 + 100/p). 6. The general term does not approach 0 as n → ∞ in any of these three cases, so each of the series is divergent. 8. sn = n/(n + 1) = 1/(1 + 1/n) → 1 as n → ∞. The infinite series converges to 1.
6.6 2. Offer (a) is best. The second offer has present value 4600
1 − (1.06)−5 ≈ 20 540. 1 − (1.06)−1
12 000 · 1.115 [1 − (1.115)−8 ] ≈ 67 644.42. 0.115 (1.115)12 − 1 = 66, 384.08. Thus the contract The present value of the contract in (c) is 22, 000 + 7000 0.115(1.115)12 in (c) is the best in any case. When the interest rate becomes 12.5 %, contracts (b) and (c) have present values equal to 65907.61 and 64374.33, respectively.
4. Schedule (b) has present value
6. This is a geometric series with first term a = D/(1 + r) and quotient k = (1 + g)/(1 + r). It converges D/(1 + r) D a = = . iff k < 1, i.e. iff 1 + g < 1 + r, or g < r. The sum is 1−k 1 − (1 + g)/(1 + r) r −g
6.7 2. |(x + 1)3 − 1| = |x 3 + 3x 2 + 3x + 1 − 1| ≤ |x|3 + 3|x|2 + 3|x| ≤ |x| + 3 + 3|x| + 3|x| = 7|x|. Let > 0. Choose δ = /7. Then according to [6.24], limx→0 (x + 1)3 = 1. Because f (0) = 1, f is continuous at x = 0. 4x 2 − 100 4. (a) 4x 2 − 100 = 4(x 2 − 25) = 4(x + 5)(x − 5), so for x = 5, = 4(x + 5). Hence, for x = 5, x−5 4x 2 − 100 − 40 = |4(x + 5) − 40| = 4|x − 5|. Let > 0. Choose δ = /4. According to [6.24], the x−5 conclusion follows. (b) |x 2 − π 2 | = |(x + π )(x − π )|, so for x = −π , |(x 2 − π 2 )/(x + π ) + 2π | = |x − π + 2π| = |x + π |. Let > 0. Choose δ = . According to [6.24], the conclusion follows. © Knut Sydsæter and Peter Hammond 2010
16
CHAPTER 7
IMPLICATIONS OF CONTINUITY AND DIFFERENTIABILITY
Chapter 7 Implications of Continuity and Differentiability 7.1 2. For x = 0, we have f (x) = x 3 + ax 2 + bx + c = x 3 (1 + a/x + b/x 2 + c/x 3 ), which shows that f (x) is (large) positive if x is sufficiently large, and that f (x) is (large) negative if x is sufficiently large and negative. Hence Theorem 7.2 implies that f (x) = 0 has a solution. A similar argument applies to the general case, when the order n of the polynomial is odd. If n is even, there may be no real roots. For instance, x 2 + 1 = 0 has no real roots. 4. Recall that, arbitrarily close to any given real number, there are rational as well as irrational numbers. The function f is continuous at a = 0, because |f (x) − f (0)| = |f (x) − 0| = |f (x)| ≤ |x| for any x, so f (x) → f (0) as x → 0. If a = 0 is rational, then |f (x) − f (a)| = |f (x)|, which is equal to |x| when x is irrational. But if a = 0 is irrational, then |f (x) − f (a)| = |f (a)| whenever x is rational. In either case, f (x) does not approach 0 as x approaches a. It follows that f is discontinuous for all x = 0. We prove that g is discontinuous by using the εδ definition of continuity. Let a be an arbitrary number and choose ε = 21 . For each positive number δ, there is a rational number x1 and an irrational number x2 in the interval (a−δ, a+δ). Then g(x1 ) = 1 and g(x2 ) = 0. If a is rational, then |g(x2 )−g(a)| = |0−1| = 1, and if a is irrational, then |g(x1 ) − g(a)| = |1 − 0| = 1. In both cases there is a number x arbitrarily close to a for which |g(x) − g(a)| = 1 > ε. Hence g is discontinuous at a.
7.2 2. (a) No, because f (x) can get arbitrarily close to 1, by letting x be sufficiently close to 1. But there is no value of x for which f (x) = 1. Similarly, there is no minimum, because f (x) can get arbitrarily close to −1 by letting x be sufficiently close to −1, yet there is no value of x for which f (x) = −1. y (b) No, not at x = −1 and x = 1. 1
yx1 y
y 1
y x 2 3
3
1
y x 3
2
1
1 1
1
x
Figure 7.3.2(a)
1
2
3
4
5
Figure 7.3.2(b)
1
2
x
x Figure 7.3.2(c)
7.3 2. In (a), f is not differentiable at x = 0; in (b), f is not differentiable at x = 3; in (c), f is not even defined at x = 1. See Fig. 7.3.2(a)–(c). 4. There is at least one point where you must be heading in the direction of the straight line joining A to B (even if that straight line hits the shore).
7.4
√ 3 2 2. (a) 25 = 3(1 − 2/27)1/3 ≈ 3(1 − 13 27 − √ 5 1 1/5 (b) 33 = 2(1 + 1/32) ≈ 2(1 + 5·32 −
© Knut Sydsæter and Peter Hammond 2010
1 4 9 272 ) ≈ 2.924 2 1 25 322 ) ≈ 2.0125
CHAPTER 7
IMPLICATIONS OF CONTINUITY AND DIFFERENTIABILITY
17
4. g(0) = g(x) = 0 follows immediately from [1] and [2]. Because f is thrice continuously differentiable, S(x) and g(x) are differentiable in (0, x). The formula for g (t) follows easily, and the condition g (c) = 0 yields S(x) = f (c). The conclusion follows.
7.5 x 4 − 4x 3 + 6x 2 − 8x + 8 “0” 4x 3 − 12x 2 + 12x − 8 “0” = = lim = = x→2 x→2 x 3 − 3x 2 + 4 0 3x 2 − 6x 0 12x 2 − 24x + 12 12 lim = =2 x→2 6x − 6 6 1/2 2(1 + x) − 2 − x (1 + x)−1/2 − 1 “0” “0” = = lim = = (b) lim 2 −1/2 x→0 2(1 + x + x 2 )1/2 − 2 − x x→0 0 (1 + 2x)(1 + x + x ) −1 0
2. (a) lim
lim
x→0
− 21 (1 + x)−3/2 2(1 + x + x 2 )−1/2 + (1 + 2x)2 (− 21 )(1 + x + x 2 )−3/2
=−
1 3
1 − (1 + v β )−γ γ (1 + v β )−γ −1 βv β−1 “0” = = lim+ . If β = 1, then v→0 v→0 v 0 1 G = γ . If β > 1, then G = 0, and if β < 1, then G = ∞.
4. G = lim+
f (x) f (1/t) “0” f (1/t)(−1/t 2 ) f (1/t) f (x) = lim+ = = lim+ = lim+ = lim 2 x→∞ g(x) x→∞ g (x) t→0 g(1/t) t→0 g (1/t)(−1/t ) t→0 g (1/t) 0
6. lim 8.
f (x) 1/g(x) “0” −1/(g(x))2 g (x) = lim = · = lim x→a g(x) x→a 1/f (x) x→a −1/(f (x))2 f (x) 0 (f (x))2 g (x) g (x) 1 2 = lim · lim = L = L2 lim x→a (g(x))2 f (x) x→a f (x) x→a f (x)/g (x)
L = lim
The conclusion follows. (Here, we have ignored problems with “division by 0”, when either f (x) or g (x) tends to 0 as x tends to a.)
7.6 2. p = (157.8/D)10/3 4. f (x) = x 2 = (−x)2 is not one-to-one on (−∞, ∞), and therefore has no inverse. On [0, ∞), f is √ strictly increasing and has the inverse g(x) = x. 6. f (x) = 7x 6 + 25x 4 + 2 > 0 for all x, so f has an inverse g. g (−2) = 1/f (0) = 1/2. 8. (a) See Fig. 7.6.8(a).
(b) Triangles OBA and OBC in Fig. 7.6.8(b) are congruent.
10. See Fig. 7.6.10. (This example shows that the commonly seen statement: “if the inverse function exists, the original and the inverse function must both be monotonic” is wrong. This claim is correct for continuous functions, however.) 12. Differentiating f (g(x)) = x yields (∗) f (g(x) g (x) = 1, so g (x) = −1/f (g(x) . Differentiating (∗) yields f g(x) g (x)g (x) + f g(x) g (x) = 0. Solving for g (x) gives 3 2 g (x) = −f g(x) g (x) /f (g(x) = −f g(x) / f g(x) . The conclusions follow. © Knut Sydsæter and Peter Hammond 2010
18
CHAPTER 8
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
y
y y=x D
y y=x
C = (b, a)
(3, 5) B (1, 3)
(5, 3) A = (a, b)
(3, 1) O
x Figure 7.6.8(a)
E
x
x
Figure 7.6.8(b)
Figure 7.6.10
−1/2 > 0 for all x > 1. Hence, f is strictly increasing in [1, ∞), and has an 14. f (x) = 21 (x+1)−1/2 + 21 (x−1) √ √ inverse g. Because f (1) = 2 and f (x) → ∞ as x → ∞, the range of f is [ 2, ∞). To find a formula √ √ √ √ for g, let (1) y = x + 1 + x − 1. Interchanging x and y, we have (2) x = y + 1 + y − 1. The √ √ √ next step is to solve (2) for y, with x ∈ [ 2, ∞) and y ∈ [1, ∞). We obtain x = y + 1 + y − 1 ⇐⇒ √ √ √ √ x − y + 1 = y − 1 ⇒ (x − y + 1 )2 = y − 1 ⇐⇒ x 2 − 2x y + 1 + y + 1 = y − 1 ⇐⇒ √ 2x y + 1 = x 2 + 2 ⇒ 4x 2 (y + 1) = x 4 + 4x 2 + 4 ⇐⇒ y = 41 x 2 + √ 1/x 2 . √ √ 2 2 Thus, x = y + 1 + y − 1 with y ≥ 1 implies y = x /4 + 1/x , x ≥ 2. We prove the converse by substitution: √ √ 2 2 2 2 4 2 +4 x 4 −4x 2 +4 y + 1 + y − 1 = x4 + x12 + 1 + x4 + x12 − 1 = x +4x + = x 2x+2 + x 2x−2 = x. 2 4x 4x 2 √ We conclude that g(x) = 41 x 2 + 1/x 2 , x ∈ [ 2, ∞) is the required inverse function.
Chapter 8 Exponential and Logarithmic Functions 8.1 x
2. (a) ee ex = ee
x +x
(b) 21 (et/2 − e−t/2 ) (c) −
et − e−t (et + e−t )2
(d) z2 ez (ez − 1)−2/3 3
3
4. p (x) = kce−cx , p (x) = −kc2 e−cx . The graph is shown in Fig. 8.1.4. y 3 2 y = 21 (ex + e−x )
y = 21 (ex − e−x )
1
p a+k
-3
-2
-1
1 -1
p(x) = a + k(1 − e−cx )
-2 a
-3 x
Figure 8.1.4
© Knut Sydsæter and Peter Hammond 2010
Figure 8.1.6
2
3
x
CHAPTER 8
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
19
6. The graphs are in Fig. 8.1.6. Verifying the identities is straightforward. We prove only equality (a): cosh x cosh y + sinh x sinh y = 21 (ex + e−x ) 21 (ey + e−y ) + 21 (ex − e−x ) 21 (ey − e−y ) = 1 x+y + ex−y + e−x+y + e−x−y + ex+y − ex−y − e−x+y + e−x−y ) = 21 (ex+y + e−x−y ) = cosh(x + y) 4 (e 8. f (z + x) = a z+x = a z a x = f (z)f (x). Differentiating w.r.t. z while x is fixed gives f (z + x) = f (z)f (x). Putting z = 0 yields f (x) = f (0)f (x), which is [8.2].
8.2 2. (a) ln 3x = x ln 3 = ln 8, so x = ln 8/ ln 3. (b) x = e3 (c) x 2 − 4x + 5 = 1, or√x 2 − 4x + 4 = 0, √ so (x − 2)2 = 0. Thus x = 2. (d) x(x − 2) = 1, so x = 1 − 2 or x = 1 + 2. (e) x = 0 or √ ln(x + 3) = 0, so x = 0 or x = −2. (f) x − 5 = 1, so x = 36. √ 4. (a) t = (ln x − b)/a (b) t = (ln 2)/a (c) t = ± ln(8/ 2π ) = ± 25 ln 2 − 21 ln π 6. (a) True (π e ≈ 22.5, eπ ≈ 23.1) (b) True because (a) gives (π e )1/eπ < (eπ )1/eπ . 8. (a) Wrong. (Put A = B = C = 1.) (b) Correct by rule [8.7] (b). (c) Correct. (Use [8.7] (b) twice.) (d) Wrong. (If A = e and p = 2, then the equality becomes 0 = ln 2.) (e) Correct by [8.7](c). (f) Wrong. (Put A = 2, B = C = 1.) 10. (a) 1/(x + 1)
(b) 1/x
(c) ln x + 1
(d)
ln x − 1 (ln x)2
12. (a) (i) y = x − 1 (ii) y = 2x − 1 − ln 2 (iii) y = x/e (b) (i) y = x (ii) y = 2ex − e (iii) y = −e−2 x − 4e−2 14. (a) Let f (x) = ex − (1 + x + x 2 /2). Then f (0) = 0 and f (x) = ex − (1 + x), which is positive for all x > 0, as shown in the problem. Hence f (x) > 0 for all x > 0, and the inequality follows. (b) Let f (x) = ln(1 + x) − 21 x. Then f (0) = 0 and f (x) = 1/(x + 1) − 21 = (1 − x)/2(x + 1) > 0 in (0, 1), so ln(1 + x) > 21 x in (0, 1). To prove the other inequality, let g(x) = x − ln(1 + x). Then g(0) = 0 and g (x) = 1 − 1/(x + 1) = x/(1 + x) > 0 in (0, 1), so x > ln(1 + x) in (0, 1). (c) Let f (t) = ln[(1 + t)/(1 − t)] − 2t. Then f (0) = 0 and f (t) = 2/[(1 + t)(1 − t)] − 2 = 2t 2 /(1 − t 2 ) > 0 in (0, 1), so the required inequality follows. 16. (a) x − x = 0
(b) 4 ln x − x
(c) x 2 /y 2
18. (a) x
(b) 1/ ln x
(c) x ln a √ √ ln x + 2 20. (a) ln y = x ln x implies y /y = 21 x −1/2 ln x + x −1/2 so y = x x √ 2 x √ (b) 21 ( x )x (ln x + 1) (c) Differentiate ln ln y = x ln x + ln ln x to get
1 x y /(y ln y) = ln x + 1 + 1/(x ln x) and so y = x x +x (ln x)2 + ln x + x
22. First take logarithms to obtain the expression: ln a + (v/α)[α ln N + α ln K − ln(N α + bK α ) = B − (v/α) ln(N α + bK α ) for ln F (α), where B does not depend on α. Differentiation yields: N α ln N + bK α ln K −2 α α F (α) = vα F (α) ln(N + bK ) − α N α + bK α
8.3
2. (a) x = 4
(b) x = e
(c) x = 1/27
© Knut Sydsæter and Peter Hammond 2010
(d) x = −1050 and x = 1050
20
CHAPTER 8
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
4. (a) ex+1−4/x = e1 , so x + 1 − 4/x = 1, hence x = ±2. (b) Putting u = ln(x + e), the equation becomes u3 − 4u2 − u + 4 = 0, so (u − 4)(u2 − 1) = 0, implying that u = ±1 or 4. Because x = eu − e, one has x = 0, x = −e + 1/e, or x = e4 − e. 6. (a) 1 (b) Does not exist. (c) 4/5 (d) 1 (x 1/x = (eln x )1/x = eln x/x , and [8.19].) (e) 0. (x ln x = ln x/(1/x), and then l’Hôpital’s rule.) (f) 1. (x x = (eln x )x = ex ln x , and then use (e).) 4 3 8. (a) xex ≈ x + x 2 + 21 x 3 (b) √e2x √ ≈ 1 + √2x + 2x 2 + 3x √ √ 1 9 2 1 3 1 3 1 1 + 2 x + 8 x + 48 x (d) ex + 1 ≈ 2 + 4 2x + 32 2x 2 + 384 (7 2)x 3
1
(c) x 2 + e 2 x
≈
10. For x = 0, one has f (x) = x −3 2e−1/x , f (x) = x −6 (−6x 2 + 4)e−1/x , and therefore f (x) = 2 2 x −9 (24x 4 − 36x 2 + 8)e−1/x . Hence, f (k) = x −3k pk (x)e−1/x is correct for k = 1, 2, 3. The general formula is proved by induction. (Actually, all we need in the following is the fact that pk (x) is a polynomial. 2 Its degree is of no importance.) Now, f (0) = limx→0 [f (x) − f (0)]/x = limx→0 e−1/x /x = 0 √ 2 because, in general, (∗) e−1/x /x p → 0 as x → 0 for any natural number p. (Substitute x = 1/ t and use [8.18].) The general case is proved by induction: Suppose that f (k) (0) = 0 for an arbitrary natural number k. Then, by the definition of the derivative, f (k+1) (0) = limx→0 [f (k) (x) − f (k) (0)]/x = 2 2 limx→0 x −3k pk (x)e−1/x /x = limx→0 x −3k−1 pk (x)e−1/x = 0, using (∗) again. (Note that pk (x), as a polynomial, approaches a constant as x → 0.) 2
2
8.4 2. (a) S(t) = S0 e−at
(b) t = (ln 2)/a is the time it takes for sales to halve.
4. k = 0.1 ln(705/641) ≈ 0.0095. P (15) ≈ 739, P (40) ≈ 938. 6. (a) In 1950 there were about 276 thousand. In the next 10 years the number increased by 155 thousand. (b) y → 479.36 as t → ∞. The graph is sketched in Fig. 8.4.6. lny
Tractors (in 1000)
8
500 6 400 4
300 200
2 100 0 0 (1950)
5
10 (1960)
15
Figure 8.4.6
20 (1970)
t
2
4
6
8
lnx
Figure 8.4.10
8. h = −f K/f 2 = −rf (1 − f/K)K/f 2 = −r(K/f − 1) = −rh. Hence h = Ae−rt for some constant A. But then −1 + K/f = Ae−rt , and solving for f yields [8.24]. © Knut Sydsæter and Peter Hammond 2010
CHAPTER 9
SINGLE-VARIABLE OPTIMIZATION
21
10. (a) See Fig. 8.4.10 and the following table: ln x
3.00
3.69
4.09
4.38
4.61
4.79
4.94
5.08
5.19
5.30
ln y
7.44
6.31
5.44
4.79
4.32
4.09
3.81
3.56
3.22
3.00
(b) ln y = ln A+a ln x. We see from the graph in Fig. 8.4.10 that a ≈ −2. Assuming that the graph passes through (say) (ln x, ln y) = (5.19, 3.22), we obtain 3.22 = ln A + (−2)5.19, implying that ln A = 13.6, so A = e13.6 ≈ 806 129. The required formula is y = 806 129x −2 . 12. (a) y = 2.81x 0.83
(b) a =
ln y2 − ln y1 , A = y1 x1−a ln x2 − ln x1
14. Take ln of each side and solve for t0 .
8.5 2. (a) A = 4, k = 0.25. The price after 5 years is f (5) = 4e1.25 ≈ 14. (b) Price controls are needed after ≈ 6 years. The doubling time before price controls is 2.8 years, after price controls it is 7.3 years.
Chapter 9 Single-Variable Optimization 9.2
√ 3x)(2 + 3x) 2. = . Note that h(x) → 0 as x ± ∞. The function has a maximum at (3x 2 + 4)2 √ √ √ √ √ √ x = 2 3/3 and a minimum at x = −2 3/3, with h(2 3/3) = 2 3/3 and h(−2 3/3) = −2 3/3. h (x)
8(2 −
√
4. (a) f (x) = [4x(x 4 + 1) − 2x 2 4x 3 ]/(x 4 + 1)2 , then simplify and factor. (b) f has maximum 1 at x = 1, because f (x) increases in [0, 1] and decreases in [1, ∞). We see that f (−x) = f (x) for all x, so f (x) is symmetric about the y-axis. On (−∞, ∞), f has maximum 1 at x = −1 and at x = 1. 6. (a) f (x) = −48x(3x 2 + 4)−2 , so f (x) > 0 for x < 0 and f (x) < 0 for x > 0. Hence, f has a maximum at x = 0. (b) g (x) = −2(x − 2), so g (x) > 0 for x < 2 and g (x) < 0 for x > 2. Hence, g has a maximum at x = 2. (c) h (x) = 20(x + 2)3 , so h (x) < 0 for x < −2 and h (x) > 0 for x > −2. Hence, h has a minimum at x = −2. (d) F (x) = 4x/(2 + x 2 )−2 , so F (x) < 0 for x < 0 and √ F (x) > 0 for x > 0. Hence, F has a minimum at x = 0. (e) G (x) = 1/2 1 − x > 0 for x < 1, so G has a maximum at x = 1. (f) H (x) = −4x 3 /(1 + x 4 )2 , so H (x) > 0 for x ∈ [−1, 0), and H (x) < 0 for x ∈ (0, 1]. Hence, H has a maximum at x = 0, and minima at x = ±1. 8. Here d (x) = 2(x − a1 ) + 2(x − a2 ) + · · · + 2(x − an ) = 2[nx − (a1 + a2 + · · · + an )]. So d (x) = 0 for x = x, ¯ where x¯ = n1 (a1 + a2 + · · · + an ), the arithmetic mean of a1 , a2 , . . . , an . If x > x, ¯ then d (x) > 0, and if x < x, ¯ then d (x) < 0. We conclude that x¯ minimizes d(x).
9.3 2. In all cases the maximum and minimum exist by the extreme value theorem. Follow the recipe in [9.5]. (a) f (x) is strictly decreasing. Maximum −1 at x = 0, minimum −7 at x = 3. © Knut Sydsæter and Peter Hammond 2010
22
CHAPTER 9
SINGLE-VARIABLE OPTIMIZATION
(b) Maximum 10 at x = −1 and x = 2. Minimum 6 at x = 1. (f (x) = 3x 2 − 3 = 0 at x = ±1.) (c) Maximum 2.5 at x = 1/2 and x = 2. Minimum 2 at x = f (x) = 1−1/x 2 = 0 √1. (f√(x) = x+1/x and 2 at x = ±1.) (d) Maximum 4 at x = −1. Minimum −6 3 at 3. (f (x) = 5x (x 2 − 3).) (e) Maximum 4.5 · 109 at x = 3000. Minimum 0 at x = 0. (f (x) = 3(x 2 − 3000x + 2 · 106 = 3(x − 500)(x − 2000).) 4. (a) Total costs when there are 61, 70, and 80 passengers are: $ 48,190; $ 49,000; $ 48,000. (b) C(x) = (60 + x)(800 − 10x) = 48, 000 + 200x − 10x 2 , x ∈ [0, 20] (c) Maximum cost is with 70 travelers (x = 10). √ √ √ 6. h (t) = 1/2 t − 21 = (1 − t)/2 t. We see that h (t) ≥ 0 in [0, 1] and h (t) ≤ 0 in [1, ∞), so t = 1 maximizes h(t). The plant is highest after 1 month, when it is 1/2 meter tall. 8. Because A (Q) = [C (Q)Q − C(Q)]/Q2 , the result follows from Example 4.20 in Section 4.6. 1/b . 10. A (Q) = a(b − 1)Qb−2 − c/Q2 , so there is a minimum at Q = c/a(b − 1)
9.4 2. (a) No local extreme points. (b) Local maximum 10 at x = −1. Local minimum 6 at x = 1. (c) Local maximum −2 √1. √ √ at x = −1.√Local minimum 2 at x = (d) Local maximum 6 3 at x = − 3. Local minimum −6 3 at x = 3. (e) No local maximum point. Local minimum 0.5 at x = 3. (f) Local maximum 2 at x = −2. Local minimum −2 at x = 0. 4. (a) (a) We find f (t) = 0.05(t + 5)(35 − t)e−t . Obviously, f (t) > 0 for t < 35 and f (t) < 0 for t > 35, so t = 35 maximizes f (with f (35) ≈ 278). (b) f (t) → 0 as t → ∞. The graph is shown in Fig. 9.4.4. y 10
y 5 200
√2 3
100
2
1
1 √2 2
3
x
5 30
60
90
Figure 9.4.4
t Figure 9.4.8
6. (a) a > 0, b = 0 (f (0) = 0 ⇒ b = 0. f (0) ≥ 0 ⇒ a ≥ 0. If a = b = 0, then f (x) = x 3 + c has no loc. min. at x = 0.) (b) a = −6, b = 9. (f (1) = 0 and f (3) = 0 give 3 + 2a + b = 0 and 27 + 6a + b = 0.) 3 √ 2 − x2 (x) = 2x − 12x − 12 ; x = 8. (a) f (x) = ; f 2 is a local maximum point with 2 2 2 3 (x + 3x + (x + 3x + 2) √ √ 2) √ f ( 2) ≈ 0.17; x = − 2 is a local minimum point with f (− 2) ≈ 5.83. (b) There are no global extreme points. The graph is shown in Fig. 9.4.8. √ (c) ex = 2 and so x = 21 ln 2 gives a global maximum, but g has no global minimum. © Knut Sydsæter and Peter Hammond 2010
CHAPTER 10
INTEGRATION
23
10. f (x) = x 3 + ax + b → ∞ as x → ∞, and f (x) → −∞ as x → −∞. Thus f (x) has at least one real root. We have f (x) = 3x 2 + a. Thus, for a ≥ 0, f (x) > 0 for all x = 0, so f is strictly increasing, and there is only one real root. Note that for a ≥ 0, 4a 3 + 27b2 ≥ 0. Assume next that a < 0. Then f (x) = 0 √ √ √ √ for x = ± −a/3 = ± p, where p = −a/3 > 0. Then f has a local maximum at (− p, b + 2p p) √ √ and a local minimum at ( p, b − 2p p). If one of the local extreme values is 0, the equation has a double root, and this is the case iff 4p 3 = b2 , that is, iff 4a 3 + 27b2 = 0. The equation has three real roots iff the local maximum value is positive and the local minimum value is negative. This occurs iff √ |b| < 2p p or iff b2 < 4p3 or iff 4a 3 + 27b2 < 0.
9.5 2. (a) f (x) = 3(x − 1)(x + 2), f (x) = 6x + 3 (b) Stationary points: x = −2 and x = 1. f increases in (−∞, −2) and in (1, ∞). (c) x = −1/2 is the only inflection point. 4. x = −2 and x = 4 are minimum points, whereas x = 2 is a maximum point. Moreover, x = 0, x = 1, x = 3, and x = 5 are inflection points. 6. a = −2/5, b = 3/5 (f (−1) = 1 gives −a + b = 1. Moreover, f (x) = 3ax 2 + 2bx and f (x) = 6ax + 2b, so f (1/2) = 0 yields 3a + 2b = 0.) 8. π(Q) is stationary at Q = (P /ab)1/(b−1) . Moreover, π (Q) = −ab(b − 1)Qb−2 < 0 for all Q > 0, so this is a maximum point.
9.6 2. a ≥ 0, b arbitrary.
¯ where y¯ = 4. (a) Since u(c) is concave, T1 Tt=1 u(ct ) ≤ u T1 Tt=1 ct ≤ u T1 Tt=1 yt = u(y), 1 T ¯ t = 1, 2, . . . , T , this holds with equality.) t=1 yt . Thus y¯ is optimal. (For ct = y, T T −t −t (b) Put λt = (1 + r) / t=1 (1 + r) . Then Tt=1 λt = 1 and T max Tt=1 (1 + r)−t u(ct ) = Tt=1 (1 + r)−t max Tt=1 λt u(ct ). Also, Tt=1 λt u(ct ) ≤ u t=1 λt ct ≤ T T u t=1 λt yt . So ct = c¯ = t=1 λt yt (t = 1, 2, . . . , T ) solves the problem.
Chapter 10 Integration 10.1 2. See Figs. 10.1.2(a) to 10.1.2(d). (a) y
y
2
1
2 0
3x 2 dx =
2 0
x3 = 8
y
(b) 1/7
(c) e − 1/e
(d) 9/10
y 1
2 1
1
2
x
Figure 10.1.2(a)
1
x
Figure 10.1.2(b)
© Knut Sydsæter and Peter Hammond 2010
1
1
x
Figure 10.1.2(c)
1
Figure 10.1.2(d)
10
x
24
CHAPTER 10
4. A =
10.2
1 2
1
−1 (e
x
INTEGRATION
+ e−x ) dx =
1 1 x 2 −1 (e
− e−x ) = e − e−1
(t 3 + 2t − 3) dt = t 3 dt + 2t dt − 3 dt = 41 t 4 + t 2 − 3t + C d (b) (x −1)2 dx = (x 2 −2x +1) dx = 13 x 3 −x 2 +x +C. Alternative: Since (x −1)3 = 3(x −1)2 , dx we have (x − 1)2 dx = 13 (x − 1)3 + C1 . This agrees with the first answer, with C1 = C + 1/3. (c) (x − 1)(x + 2) dx = (x 2 + x − 2) dx = 13 x 3 + 21 x 2 − 2x + C
2. (a)
3 3 2 (d) Either first evaluate (x + 2) = x + 6x + 12x + 8, to get (x + 2)3 dx = 41 x 4 + 2x 3 + 6x 2 + 8x + C, or: (x + 2)3 = 41 (x + 2)4 + C1 . (e) (e3x − e2x + ex ) dx = 31 e3x − 21 e2x + ex + C 3 4 x − 3x + 4 dx = x2 − 3 + dx = 13 x 3 − 3x + 4 ln |x| + C (f) x x 4. (a) Differentiate the right-hand side. √ 1 (2x + 1)5 + C (ii) 23 (x + 2)3/2 + C (iii) −2 4 − x + C (b) (i) 10 6. (a) F (x) = ( 21 − 2x) dx = 21 x − x 2 + C. F (0) = 21 implies C = 21 . 5 (b) F (x) = (x − x 3 ) dx = 21 x 2 − 41 x 4 + C. F (1) = 12 implies C = 16 . 1 4 x + Ax + B. If we require that 8. The general form for f is f (x) = 13 x 3 + A, so that for f is f (x) = 12 1 4 f (0) = 1 and f (0) = −1, then B = 1 and A = −1, so f (x) = 12 x − x + 1.
10.3 2. (a) −12/5 4. (a) 6/5
(b) 41/2
(b) 26/3
3 1 (c) + t dt = t −1 2
(c) α(eβ − 1)/β
(d) − ln 2
3 2
5 ln(t − 1) + 21 t 2 = ln 2 + 2 6. (a) 32/15 ≈ 2.133. See Fig. 10.3.6. y
y
y=
2 2
√ x
4/3 1
1
1 2 Figure 10.3.6
x
1
x∗
3
4
x
Figure 10.3.10
b + c √ + d ln |b| (c) 2 2 − 3/2 (b) A b − 1 + (b − c) ln 1+c x 10. (a) Given F (x) = a f (t) dt, according to the mean-value theorem, there exists a number x ∗ ∈ (a, b) b such that [F (b) − F (a)]/(b − a) = F (x ∗ ). Here F (b) = a f (t) dt, F (a) = 0, and F (x ∗ ) = f (x ∗ ), so the conclusion follows. (b) f (x ∗ ) = 4/3 at x ∗ = 16/9. See Fig. 10.3.10, in which the two shaded areas are equal. 8. (a) 59/30
© Knut Sydsæter and Peter Hammond 2010
CHAPTER 11
FURTHER TOPICS IN INTEGRATION
25
10.4 2. (a) Let n be the total The number of individuals with income in the interval [b, 2b] 2b number of individuals. 2b 2b nB −2 −1 is then N = n . Their total income is M = n Br dr = n −Br = Br −2 r dr = 2b b b b 2b 2b Br −1 dr = n B ln r = nB ln 2. Hence the mean income is m = M/N = 2b ln 2. n b b 2b 2b 2b nD(p, r)f (r) dr = nAp γ r δ Br −2 dr = nABp γ r δ−2 dr = (b) Total demand is x(p) = b
nABp
b
15
4. PDV =
b
r δ−1 2δ−1 − 1 = nABp γ bδ−1 . δ−1 δ−1
2b γ
0
b
15 −1 −0.06t 500
500e−0.06t dt = 500 0 = e 1 − e−0.9 ≈ 4945.25. 0.06 0.06
FDV = e0.06·15 PDV = e0.9 PDV ≈ 2.4596 · 4945.25 ≈ 12163.3.
Chapter 11 Further Topics in Integration 11.1
1
1 1 1 x ln(x + 2) dx = −1 21 x 2 ln(x + 2) − −1 21 x 2 x+2 dx 1 1 1 4 3 = 2 ln 3 − 2 −1 x − 2 + x+2 dx = 2 − 2 ln 3 d x 2 = 2x ln 2, and therefore 2x / ln 2 is the indefinite integral of 2x . If follows that (b) Recall that dx 2 2 x 2 2 2 2x 2x 8 8 3 x2x dx = x = − dx = − − 2 ln 2 ln 2 (ln 2)2 0 0 ln 2 0 ln 2 0 (ln 2) (c) First use integration by parts on the indefinite integral: with f (x) = x 2 and g(x) = ex , (∗) x 2 ex dx = x 2 ex − 2xex dx. To evaluate the lastintegral we must use integration by parts once more: with f (x) = 2x and g(x) = ex , 2xex dx = 2xex − 2ex dx = 2xex −(2ex +C). Inserted into (∗) this gives x 2 ex dx = 1 1 x 2 ex −2xex +2ex +C, and hence, 0 x 2 ex dx = 0 (x 2 ex −2xex +2ex ) = (e−2e+2e)−(0−0+2) = e−2. Alternatively, more compactly using formula [11.2]: 1 1 1 1 1 1 2 x 2 x x x x e dx = x e − 2 xe dx = e − 2 xe − ex dx = e − 2[e − 0 ex ] = e − 2
2. (a)
−1
0
0
0
0
0
4. Using integration by parts, we have
t1
t1 F (t)μ(t) ˙ dt = F (t)μ(t) − t0
t0
t1
F˙ (t)μ(t) dt
t0
= F (t1 )μ(t1 ) − F (t0 )μ(t0 ) −
t1
F˙ (t)μ(t) dt = −
t0
t1
F˙ (t)μ(t) dt
t0
6. Using [11.2], we have
T
U (C(t))e 0
−rt
T −rt dt = − U (C(t))(1/r)e + 0
which becomes the required expression when U (C(0)) = 0. © Knut Sydsæter and Peter Hammond 2010
T 0
(d/dt)U (C(t))(1/r)e−rt dt
26
CHAPTER 11
FURTHER TOPICS IN INTEGRATION
11.2 2. (a) 19 (x 2 + 1)9 + C. (Substitute u = x 2 + 1, du = 2x dx.) (b) With u = x 3 + 2, du = 3x 2 dx and 1 u 2 x 3 +2 1 u 1 x 3 +2 dx = + C. (c) First attempt: u = x + 2, which gives x e 3 e du = 3 e + C = 3 e ln(x + 2) ln u du = dx and dx = du. Not promising. A better idea: Substitute u = ln(x + 2). 2x + 4 2u dx ln(x + 2) 2 2 1 1 1 Then du = and dx = 2 u du = 4 (u) + C = 4 (ln(x + 2)) + C. x+2 2x + 4 √ √ (d) Attempt: u = 1 + x. Then, du = dx, and x 1 + x dx = (u − 1) u du = (u3/2 − u1/2 ) du = √ 2 5/2 − 23 u3/2 + C = 25 (1 + x)5/2 − 23(1 + x)3/2 + C. Second attempt: u = 1+ x. Then u2 = 1 + x 5u √ and 2udu = dx. Then the integral is x 1 + x dx = (u2 − 1)u2u du = (2u4 − 2u3 ) du e.t.c. Check that you get the same answer. Actually, even integration by parts works in this case. Put f (x) = x √ and g (x) = 1 + x, and choose g(x) = 23 (1 + x)3/2 . (The answer looks different, but is not.) x3 x2 · x 2 2 dx = dx = (e) With u = 1 + x , x = u − 1, and du = 2xdx, so (1 + x 2 )3 (1 + x 2 )3 u−1 −1 1 1 du = 21 (u−2 − u−3 ) du = − 21 u−1 + 41 u−2 + C = + C. + 2 2) 2 2 u3 2(1 + x 4(1 + x) √ (f) With u = 4 − x 3 , u2 = 4−x 3 , and 2udu = −3x 2 dx, so x 5 4 − x 3 dx = x 3 4 − x 3 x 2 dx = 2 5 2 (4−u2 ) u (− 23 )u du = (− 83 u2 + 23 u4 ) du = − 89 u3 + 15 u +C = − 89 (4−x 3 )3/2 + 15 (4−x 3 )5/2 +C
x −2x 2t − 2 ln u = ln 13 (x 2 − 2x). Thus, dt = 2 3 t − 2t 3 the equation implies that 13 (x 2 − 2x) = 23 x − 1. Hence, x 2 − 4x + 3 = 0, with solutions x = 1 and x = 3. Here x = 1 is impossible, because the integral is not defined when x = 1. So the solution is x = 3.
4. Substituting u = t 2 − 2t and assuming x > 2 gives
x
2
6. Substitute z = x(t). Then dz = x (t)dt, and the result follows.
b+λ 8. (a) Introduce z = x − λ as a new variable in the right-hand side integral. Then dz = dx, and a+λ f (x − b λ) dx = a f (z) dz. Then replace the dummy variable z by x in the last integral. (b) Introduce z = x/λ as a new variable in the right-hand side integral. u8 1/6 du. Here u8 : (−u2 +1) = −u6 −u4 −u2 −1+1/(−u2 +1). 10. Substitute u = x . Then I = 6 1 − u2 It follows that I = − 67 x 7/6 − 65 x 5/6 − 2x 1/2 − 6x 1/6 − 3 ln |1 − x 1/6 | + 3 ln |1 + x 1/6 | + C
11.3
+∞
b 1 1 1 dx = (b − a) = 1 x= b−a a b−a a b−a −∞ +∞ b b 1 1 1 1 (b) xf (x) dx = x dx = x2 = (b2 − a 2 ) = (a + b) b − a 2(b − a) 2(b − a) 2 −∞ a a b 1 b3 − a 3 1 1 2 3 2 x = (c) = (a + ab + b ) 3(b − a) a 3 b−a 3 b b 4. Divergence because 0 x/(1 + x 2 ) dx = 0 21 ln(1 + x 2 ) = 21 ln(1 + b2 ) → ∞ as b → ∞. But b b 1 2 2 −b x/(1 + x ) dx = −b 2 ln(1 + x ) = 0 for all b, so the limit is 0.
2. (a)
f (x) dx =
b
© Knut Sydsæter and Peter Hammond 2010
CHAPTER 12
6.
1 1+x 2
≤
1 x2
for x ≥ 1, and
b
dx =
1 1 x2
LINEAR ALGEBRA: VECTORS AND MATRICES
b 1 (−1/x)
27
= 1 − 1/b → 1 as b → ∞, so by
Theorem 11.1 the given integral converges. 8. (a) z = rτ1 (1 − e−rτ ) (b) z = rτ2 1 − rτ1 (1 − e−rτ ) √ √ √ 10. Integrating by parts, ln x/ x dx = 2 x ln x − 4 x + C. Hence, 1 1 √ √ √ √ √ ln x/ x dx = (2 x ln x − 4 x) = −4 − (2 h ln h − 4 h) → −4 as h → 0+ , so the given h h √ integral converges to −4. ( h ln h → 0 by l’Hôpital’s rule.) ∞ ∞ p p p 12. See Fig. 11.3.12. If p > 1, then ∞ n=1 (1/n ) = 1 + n=2 (1/n ) is finite because n=2 (1/n ) is the sum of the shaded rectangles, and this sum is certainly less than the area under the curve y = 1/x p over p [1, ∞), which is equal to 1/(p − 1). If p ≤ 1, the sum ∞ n=1 (1/n ) is the sum of the larger rectangles in the figure, and this sum is larger than the area under the curve y = 1/x p over [1, ∞), which is unbounded p when p ≤ 1. Hence, ∞ n=1 (1/n ) diverges in this case. y
y = 1/x p
1
1
2
3
4
x
Figure 11.3.12
11.4 2. The Gini coefficients are approximately: 0.34 (US 1980), 0.37 (US 1990), 0.35 (Netherlands 1959), 0.27 (Netherlands 1985), 0.69 (World 1989). (These numbers are obtained by approximating the Lorenz curve with a broken line through the points corresponding to the observations, then computing the area between this broken line and the diagonal of the square. Note that this underestimates the Gini coefficients.)
Chapter 12 Linear Algebra: Vectors and Matrices 12.1 2. (a) No sector delivers to itself. (b) The total amount of good i needed to produce one unit of each good. (c) This collection gives the number of units of each good which are needed to produce one unit of good j . (d) No economic interpretation. (The goods are usually measured in different units, so it is meaningless to add them together.) 4. The Leontief model for this three sector model is as follows: 0.9x1 − 0.2x2 − 0.1x3 = 85 −0.3x1 + 0.8x2 − 0.2x3 = 95 −0.2x1 − 0.2x2 + 0.9x3 = 20 which does have the claimed solution. © Knut Sydsæter and Peter Hammond 2010
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12.2 2. a + b + c = (−1, 6, −4), a − 2b + 2c = (−3, 10, 2), 3a + 2b − 3c = (9, −6, 9), −a − b − c = −(a + b + c) = (1, −6, 4) 4. 3(x, y, z) + 5(−1, 2, 3) = (3x − 5, 3y + 10, 3z + 15) = (4, 1, 3) provided that 3x − 5 = 4, 3y + 10 = 1, and 3z + 15 = 3. So x = 3, y = −3, z = −4. 6. (a)
3x − 2y = −1 −4x + 3y = 2
, x = 1, y = 2
(b)
2x + 4y = 1 −3x − 6y = 0
, which have no solution.
8. We need to find numbers t and s such that t (2, −1)+s(1, 4) = (4, −11). This vector equation is equivalent to (2t + s, −t + 4s) = (4, −11), which in turn is equivalent to the equation system (i) 2t + s = 4 (ii) −t + 4s = −11. This system has the solution t = 3, s = −2, so (4, −11) = 3(2, −1) − 2(1, 4).
12.3 2. (a) λ = 0 gives x = (3, 1), λ = 1/4 gives x = (2, 5/4), λ = 1/2 gives x = (1, 3/2), λ = 3/4 gives x = (0, 7/4), λ = 1 gives x = (−1, 2). See Fig. 12.3.2. (b) When λ runs through [0, 1], then x will run through all points on the line segment between a and b. 4. See Fig. 12.3.4. (The point R should be one unit further down.) z S (3,2, 4)
y λ 1
b
λ 3/4
λ 1/2
λ 1/4
1
1
y
P
a 1
Q
λ 0 x
2
3
R
x
Figure 12.3.2
Figure 12.3.4
12.4 2. (a) a · b = b · a = −6, (a + b) √· c = a · c + b · c = 9, a · (3b) = 3a · b = −18. (b) a = 3, b = 3, c = 29 , (c) |ab| = 6 ≤ ab = 9 4. The scalar product of the two vectors is x 2 +(x −1)x +3·3x = x 2 +x 2 −x +9x = 2x 2 +8x = 2x(x +4), which is 0 for x = 0 and x = −4. 6. x = (5, 7, 12), u = (20, 18, 25), x · u = 526 8. ||a + b||2 = (a + b) · (a + b) = a · a + 2a · b + b · b ≤ ||a||2 + 2||a||||b|| + ||b||2 = (||a|| + ||b||)2 . The conclusion follows because ||a + b|| and ||a|| + ||b|| are both nonnegative. ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 8 2 6λ + 2 x1 10. (a) Total output is ⎝ x2 ⎠ = λ ⎝ 4 ⎠ + (1 − λ) ⎝ 6 ⎠ = ⎝ −2λ + 6 ⎠. 4 10 −6λ + 10 x3 © Knut Sydsæter and Peter Hammond 2010
CHAPTER 12
LINEAR ALGEBRA: VECTORS AND MATRICES
29
To produce the output in (i) we can put λ = 1/2. To produce the output in (ii) would require a value of λ such that 6λ+2 = 7, −2λ+6 = 5, and −6λ+10 = 5, and the second of these equations gives a different value for λ from the other two. (b) In (a) we saw that (i) can be produced even without throwing away outputs. For (ii) to be possible if we were allowed to throw away output, there must exist a λ ∈ [0, 1] such that 6λ + 2 ≥ 7, −2λ + 6 ≥ 5, and −6λ + 10 ≥ 5. These inequalities reduce to λ ≥ 5/6, λ ≤ 1/2, λ ≤ 5/6, which are incompatible. (c) Revenue = R(λ) = p1 x1 + p2 x2 + p3 x3 = p1 (6λ + 2) + p2 (−2λ + 6) + p3 (−6λ + 10) = (6p1 − 2p2 − 6p3 )λ + 2p1 + 6p2 + 10p3 . If the constant slope 6p1 − 2p2 − 6p3 is > 0, then R(λ) is maximized at λ = 1; if 6p1 − 2p2 − 6p3 is < 0, then R(λ) is maximized at λ = 0. Only in the special case where 6p1 − 2p2 − 6p3 = 0 can the two plants both remain in use.
12.5 2. (a) Direct verification. (To show that a lies on L, put t = 0.) (b) The direction of L is given by (−1, 2, 1). (c) The equation of plane is (−1)(x1 − 2) + 2(x2 − (−1)) + 1 · (x3 − 3) = 0, or −x1 + 2x2 + x3 = −1. (d) We must have 3(−t + 2) + 5(2t − 1) − (t + 3) = 6, and so t = 4/3. Thus P = (2/3, 5/3, 13/3). (b) x1 = −2 − t, x2 = 1 + 2t, x3 = −1 + 3t.
4. (a) Direct verification.
12.6 2. (a) A =
12.7
2 3
3 4
4 5
(b) A =
1 −1 −1 1
1 −1
4. A + B =
⎛
⎞ ⎛ ⎞ ⎛ 4 1 −1 −2 3 −5 5 2. A + B = ⎝ 9 2 7 ⎠, A − B = ⎝ 1 −2 −3 ⎠, AB = ⎝ 19 3 −1 4 −1 −1 −2 1 ⎛ ⎞ ⎛ ⎞ 0 4 −9 23 8 25 BA = ⎝ 19 3 −3 ⎠, (AB)C = A(BC) = ⎝ 92 −28 76 ⎠ 5 1 −3 4 − 8 −4 ⎛ ⎞ 0.2875 ⎝ 4. T(Ts) = 0.2250 ⎠ 0.4875
12.8
1 7
0 0 , 3A = 5 6
3 −5 −3
3 9
⎞ 3 16 ⎠, 0
⎛
⎞⎛ ⎞ ⎛ ⎞ a d e x ax + dy + ez 2. We start by performing the multiplication ⎝ d b f ⎠ ⎝ y ⎠ = ⎝ dx + by + f z ⎠. Next, e f c z ex + fy + cz ⎛ ⎞ ax + dy + ez (x, y, z) ⎝ dx + by + f z ⎠ = (ax 2 + by 2 + cz2 + 2dxy + 2exz + 2fyz) ex + fy + cz which is a 1 × 1 matrix.
4. Equality in both [1] and [2] ⇔ AB = BA.
© Knut Sydsæter and Peter Hammond 2010
⎞ 1 6. (a) x0 = ±(1/ 3) ⎝ −1 ⎠ 1 √
⎛
(b) An x0 = x0 for all n.
30
CHAPTER 13
DETERMINANTS AND MATRIX INVERSION
a 2 + bc ab + bd 8. (a) Direct verification yields (1) = (a + d)A − (ad − bc)I2 = ac + cd bc + d 2 3 (b) Multiplying (1) by A and using A = 0 yields (2) (a + d)A2 = (ad − bc)A, and further, (3) 0 = (a + d)A3 = (ad − bc)A2 . If ad − bc = 0, (3) yields A2 = 0. If ad − bc = 0, 2 (2) yields (a + d)A2 = 0, and if a +d = 0, again A = 0. Finally, if ad − bc = a + d = 0, 1 1 then (1) implies A2 = 0. (c) Let A = . Then A2 = 0, and also (of course) A3 = 0. −1 −1
A2
12.9 2. A =
AB =
−1 0 2 3 1 −6 , B = , (A + B) = , (αA) = 5 2 2 4 7 −4 4 10 4 10 −2 4 , (AB) = = B A , A B = 10 8 10 8 10 14
3 2
2 , −10
4. Symmetry if a 2 − 1 = a + 1 and a 2 + 4 = 4a, which reduces to a = 2. 6. (a) (A1 A2 A3 ) = (A1 (A2 A3 )) = (A2 A3 ) A1 = (A3 A2 )A1 = A3 A2 A1 . (b) Easy induction proof. 8. (a) tr(A + B) = ni=1 (aii + bii ) = ni=1 aii + ni=1 bii = tr(A) + tr(B) (b) tr(cA) = ni=1 caii = c ni=1 aii = c tr(A) (c) tr(AB) = ni=1 jn=1 aij bj i , whereas tr(BA) = n n n n i=1 j =1 bij aj i = j =1 i=1 bj i aij , because the indices i and j can be interchanged. The two are equal because the order of multiplication and summation in each double sum is irrelevant. (If this is hard to understand, write out the sums in full for the cases when n = 2 and/or n = 3.) (d) tr(A ) = ni=1 aii = tr(A)
Chapter 13 Determinants and Matrix Inversion 13.1
3 2. See Fig. 13.1.2. The shaded area is 18 = 2
0 . 6
(2, 6)
(3, 0) Figure 13.1.2
a11 b11 + a12 b21 a11 b12 + a12 b22 , implying that a21 b11 + a22 b21 a21 b12 + a22 b22 |AB| = (a11 b11 + a12 b21 )(a21 b12 + a22 b22 ) − (a11 b12 + a12 b22 )(a21 b11 + a22 b21 ). On the other hand, |A||B| = (a11 a22 − a12 a21 )(b11 b22 − b12 b21 ). A tedious computation shows that the two expressions are equal.
4. The matrix product is AB =
© Knut Sydsæter and Peter Hammond 2010
CHAPTER 13
6. We write the system as
Y − C = I0 + G0 −bY + C = a
I0 + G0 −1 a 1 Y = 1 −1 −b 1
=
DETERMINANTS AND MATRIX INVERSION
31
. Then Cramer’s rule yields
a + I 0 + G0 , 1−b
1 I 0 + G0 −b a C= 1 −1 −b 1
=
a + b(I0 + G0 ) 1−b
The expression for Y is most easily found by substituting the second equation into the first, and then solving for Y . Then use C = a + bY to find C. 8. |A(t)| = 2t (1 − t), which is 0 for t = 0 and t = 1.
13.2 ⎛
−1 2. AB = ⎝ 7 5
⎞ −1 −1 13 13 ⎠, |A| = −2, |B| = 3, |AB| = |A| · |B| = −6 9 10
4. By Sarrus’ rule the determinant is (1 + a)(1 + b)(1 + c) + 1 + 1 − (1 + b) − (1 + a) − (1 + c), which reduces to the given expression. 6. Direct verification using [13.7] or Sarrus’ rule.
13.3 2. +a12 a23 a35 a41 a54 (Four lines between pairs of elements rise as one goes to the right.)
13.4 ⎛
2 ⎝ 2. A = 1 3
1 0 1
⎞ 1 2 ⎠, |A| = |A | = −2 5
4. (a) The first and the second columns are proportional. (b) Add the second column to the third. Then the first and the third columns are proportional. (c) The term x − y is a common factor for each entry in the first row. If x = y, all elements in the first row are 0. If x − y = 0, we divide this row by x − y, and the resulting determinant has rows 1 and 2 identical, so it is 0. 6. |AB| = |A||B| = −12, 3|A| = 9, | − 2B| = (−2)3 (−4) = 32, |A| + |B| = −1, whereas |A + B| is not determined. 8. (a) Because A2 = In it follows from part 8 of Theorem 13.1 that |A|2 = 1, and so |A| = ±1. (b) Direct verification. (c) (In − A)(In + A) = In2 + In A − AIn − A2 = In − A2 = 0 ⇐⇒ A2 = In © Knut Sydsæter and Peter Hammond 2010
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DETERMINANTS AND MATRIX INVERSION
10. Start by adding the sum of the last n − 1 rows to the first row. Then na + b is a common factor in the first row, which we take out. Next, add the first row multiplied by −a to all the other n − 1 rows. The upper triangular matrix resulting from these operations obviously has the determinant bn−1 : a + b a Dn = .. . a
a a+b .. .
··· a ← ··· a 1 . .. . .. ··· a
a 1 1 a a + b = (na + b) .. .. . . a a
··· ← na + b na + b · · · ··· a a + b ··· . = .. .. .. . . . . . a a ··· 1 −a · · · −a ··· 1 1 ··· 0 ··· a ← . = (na + b) . .. .. .. . . . . 0 ··· a + b ←
na + b a .. . a+b 1 b .. . 0
1 0 .. . ··· b
··· ··· .. .
According to [13.14], the last determinant is bn−1 . Thus Dn = (na + b)bn−1 .
13.5 2. In each of these cases we keep expanding by the last (remaining) column. The answers are: (b) abcd (c) 6 · 4 · 3 · 5 · 1 = 360
(a) −abc
13.6 2. Multiply the two matrices to get I3 . −1 −4 3 2 −3 = , we get: (a) x 4. Using −3 2 3 −4 ⎛ ⎞ ⎛ 0 1 1 1 1 6. (a) |A| = 1, A2 = ⎝ 1 1 2 ⎠, A3 = ⎝ 2 2 1 1 1 1 2 (b) By (a), (A − I)2 A = I.
(c) P = (A − I)−1
= 3, y = 1
(b) x = 1, y = −2
(c) x = y = 0
⎞ 2 3 ⎠, then direct verification of the equality. 2⎛ ⎞ 0 0 1 = ⎝ 1 0 1 ⎠ does the job. So does −P. 0 1 0
8. (a) A2 = (PDP−1 )(PDP−1 ) = PD(P−1 P)DP−1 = PDIDP−1 = P(D)2 P−1 . (b) Suppose the formula is valid for m = k. Then Ak+1 = AAk = PDP−1 (PDk P−1 ) = PD(P−1 P)Dk P−1 = PDIDk P−1 = PDDk P−1 = PDk+1 P−1 . 10. (a) If C2 + C = I, then C(C + I) = I, and so C−1 = C + I = I + C. (b) Because C2 = I − C, so C3 = C2 C = (I − C)C = C − C2 = C − (I − C) = −I + 2C. Moreover, C4 = C3 C = (−I + 2C)C = −C + 2C2 = −C + 2(I − C) = 2I − 3C. ⎛ ⎞ s + 17 4t − 16 0 1 ⎝ 12. (a) AT = 12 2s + 10 5t − 8 0 ⎠. We see that AT = I3 and so T = A−1 for s = −5, t = 4. 3s + 15 4t − 16 12 (b) BX = 2X + C ⇐⇒ BX − 2X = C ⇐⇒ (B − 2I)X = C. But it is⎛easy to see that B − 2I =⎞A, −1/2 0 0 1/6 so BX = 2X + C ⇐⇒ AX = C ⇐⇒ X = A−1 C = TC. Hence, X = ⎝ 1/2 3 −3 1/6 ⎠ . 1/2 −1 2 1/6 © Knut Sydsæter and Peter Hammond 2010
CHAPTER 14
13.7
⎛ C 1 ⎝ 11 2. C12 |A| C13
C21 C22 C23
⎞ ⎛ C31 −3 1 ⎠ ⎝ C32 = 18 72 6 C33
5 −6 14
FURTHER TOPICS IN LINEAR ALGEBRA
⎛
⎞ 9 18 ⎠. −18
3. (I − A)−1
18 5 ⎝ = 62 2 4
33
⎞ 16 10 19 8⎠ 7 16
4. Let B denote the n × p matrix whose ith column has the elements bi1 , bi2 , . . . , bin . The p systems of n equations in n unknowns can be expressed as AX = B, where A is n × n and X is n × p. Following the method in Example 13.20, exactly the same row operations that transforms the n × 2n matrix (A : I) into (I : A−1 ) will also transform the n × (n + p) matrix (A : B) into (I : B∗ ), where B∗ is the matrix with elements bij∗ . Because these row operations are together equivalent to premultiplcation by A−1 , it must ∗ ∗ ∗ be true that B∗ = A−1 B. When i = k, the solution to the system is x1 = bk1 , x2 = bk2 , . . . , xn = bkn .
13.8 2. The determinant of the system is equal to −10, so the solution is unique. The determinants in (2) are b1 D1 = b2 b 3
1 −1 3
0 2 −1
,
3 D2 = 1 2
b1 b2 b3
0 2 −1
,
3 D3 = 1 2
1 b1 −1 b2 3 b3
Expanding each of these determinants by the column (b1 , b2 , b3 ), we find that D1 = −5b1 + b2 + 2b3 , 1 3 D2 = 5b1 − 3b2 − 6b3 , D3 = 5b1 − 7b2 − 4b3 . Hence, x1 = 21 b1 − 10 b2 − 15 b3 , x2 = − 21 b1 + 10 b2 + 53 b3 , 1 7 2 x3 = − 2 b1 + 10 b2 + 5 b3 .
Chapter 14 Further Topics in Linear Algebra 14.1 2. Only the vectors in (b) are linearly independent. 4. If x(1, 1, 1) + y(2, 1, 0) + z(3, 1, 4) + w(1, 2, −2) = (0, 0, 0), then x, y, z, and w satisfy the equations x + 2y + 3z + w = 0, x + y + z + 2w = 0, x + 4z − 2w = 0. One solution is x = −2, y = −1, z = 1, w = 1, so the vectors are linearly dependent. 6. (a) Suppose αa + βb + γ c = 0. Taking the scalar product of each side with a yields a · (αa + βb + γ c) = a · 0 = 0. But a · b = a · c = 0, so α(a · a) = 0. Because a = 0, α = 0. In a similar way we can prove that β = γ = 0, so a, b, and c are linearly independent. (b) Generalize the proof in (a).
14.2 2. (a) By cofactor expansion along the first row, the determinant of the matrix A is |A| = x · (−1) − 0 · 1 + (x 2 − 2) · 1 = x 2 − x − 2 = (x + 1)(x − 2). If x = −1 and x = 2, then |A| = 0, so the rank of A equals 3. If x = −1 or x = 2, then |A| = 0 and r(A) ≤ 2. On the other hand, the minor we get if we 0 1 = 1 = 0 for all x, so r(A) can never be strike out the first row and the third column in A is −1 x less than 2. Thus, r(A) = 2 if x = −1 or x = 2, r(A) = 3 otherwise. (b) A little calculation shows that the determinant is t 3 + 4t 2 − 4t − 16, and if we note that this expression has t + 4 as a factor, it follows that the determinant is t 3 + 4t 2 − 4t − 16 = t 2 (t + 4) − 4(t + 4) =
© Knut Sydsæter and Peter Hammond 2010
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CHAPTER 14
FURTHER TOPICS IN LINEAR ALGEBRA
(t 2 − 4)(t + 4) = (t + 2)(t − 2)(t + 4). Thus, if t is not any of the numbers −2, 2, and −4, the rank of the matrix is 3. If we strike out the second row and the first column of the matrix, we get the minor 5 6 1 t + 4 = 5t + 14, which is different from 0 for all the three special values of t that we found above, and thus the rank of the matrix is 2 if t = −4, −2, or 2, and 3 otherwise. (c) The first and third rows are identical, as are the second and fourth. But the first two rows are always linearly independent. So the rank is 2 for all values of x, y, z, and w.
14.3 2. Unique solution for c = 3/2: x = (3 − kc)/(3 − 2c), y = (k − 2)/(3 − 2c). For c = 3/2, k = 2 the solutions are x = (−3/2)a + 1, y = a, a arbitrary. For c = 3/2, k = 2 there are no solutions. 4. A((1 − λ)x1 + λx2 ) = (1 − λ)Ax1 + λAx2 = (1 − λ)b + λb = b. This shows that if x1 and x2 are different solutions, then so are all points on the straight line through x1 and x2 . 6. (a) |At | = (t − 2)(t + 3), so At has an inverse for t = 2 and t = −3. all t except 1 3 is always nonsingular, r(A2 ) = 2, r(A−3 ) = 2. (b) r(At ) = 3 for t = 2 and for t = −3. Because 2 5 (c) x1 = −46 + 19t, x2 = 19 − 7t, x3 = t, with t arbitrary. (d) Any z = 0 satisfying zA2 = 0 has this property. In particular, z = (10a, −7a, a) for a = 0.
14.4 2. |A − λI| = 0 iff |(A − λI) | = 0 iff |A − λI| = 0 using [12.44] in Section 12.9 and Thm. 13.1 [2] in Section 13.4. The conclusion follows. ⎞ ⎞ ⎛ 2 ⎛ 3 2a 2a 2 0 4a 4a 3 0 4. (a) X AX = (ax 2 + ay 2 + bz2 + 2axy), A2 = ⎝ 2a 2 2a 2 0 ⎠, A3 = ⎝ 4a 3 4a 3 0 ⎠ 0 0 b2 0 0 b3 (b) λ = 0, λ = 2a, λ = b (c) p(λ) = −λ(b − λ)(2a − λ) = −λ3 + (2a + b)λ2 − 2abλ. Using A2 and A3 from part (a), it is easy to show that p(A) = −A3 + (2a + b)A2 − 2abA = 0. 6. Since A−1 exists, |A| = 0. So λ is an eigenvalue for AB ⇐⇒ |AB − λI| = 0 ⇐⇒ |A(B − λA−1 )| = 0 ⇐⇒ |B − λA−1 | = 0 ⇐⇒ |(B − λA−1 )A| = 0 ⇐⇒ |BA − λI| = 0 ⇐⇒ λ is an eigenvalue for BA. (Using [14.11] in the next section, here is a simpler argument: AB and A−1 (AB)A = BA have the same eigenvalues.)
14.5 2. (a) No, because the eigenvectors are not linearly independent. linearly independent.
14.6
(b) Yes, because the eigenvectors are
1 − λ 2 2. (a) |A − λI| = = λ2 + λ − 6 = 0, with the real eigenvalues λ1 = −3 and λ2 = 2. One 2 −2 − λ 1 2 set of linearly independent eigenvectors is and . These two eigenvectors are orthogonal. −2 1
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Thus, Theorem 14.7 (not 14.6!) is confirmed. Further, let 1/5 1 2 −1 P= for which P = 2/5 −2 1
−2/5 1/5
35
It is easy to verify that P−1 AP is the diagonal matrix diag (−3, 2), which confirms Theorem 14.8 (not 14.6!). −λ 0 1 (b) Here |A − λI| = 0 −λ 1 = −λ(λ − 1)(λ + 2) = 0, with the real eigenvalues λ1 = −2, 1 1 −1 − λ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 1 1 ⎝ ⎠ ⎝ ⎠ ⎝ λ2 = 0 and λ3 = 1. One set of linearly independent eigenvectors is 1 , −1 , and 1 ⎠. −2 0 1 Obviously, these eigenvectors are mutually orthogonal. Thus Theorem 14.7 is confirmed. Further, let ⎛
1 P=⎝ 1 −2
1 −1 0
⎞ 1 1⎠ 1
⎛
for which
P−1
1/6 = ⎝ 1/2 1/3
1/6 −1/2 1/3
⎞ −1/3 0 ⎠ 1/3
It is easy to verify that P−1 AP is the diagonal matrix diag (−2, 0, 2), which confirms Theorem 14.8.
Chapter 15 Functions of Several Variables 15.1 2. f (1, 1) = 2, f (−2, 3) = 51, f (1/x, 1/y) = 3/x 2 − 2/xy + 1/y 3 , [f (x + h, y) − f (x, y)]/ h = 6x − 2y + 3h, [f (x, y + k) − f (x, y)]/k = −2x + 3y 2 + 3yk + k 2 √ √ 4. (a) F (1, 1) = 10, F (4, 27) = 60, F (9, 1/27) = 10, F (3, 2 ) = 10 3 · 21/6 , F (100, 1000) = 1000, F (2K, 2L) = 25/6 F (K, L) (b) a = 5/6 6. (a) y gets 21.053 ≈ 2.07 times larger. (b) ln y = ln 2.9 + 0.015 ln x1 + 0.25 ln x2 + 0.35 ln x3 + 0.408 ln x4 + 0.03 ln x5 8. (a) x + y > 0 (b) Must require 2 − (x 2 + y 2 ) ≥ 0, i.e. x 2 + y 2 ≤ 2. (c) Put a = x 2 + y 2 . We must require (4 − a)(a − 1) ≥ 0, i.e. 1 ≤ a ≤ 4. (Use a sign diagram.) 10. Total output is ni=1 (T /ti ) = T ni=1 (1/ti ). If all n of the machines were equally efficient, and their production per unit were t¯H , then each machine would produce T /t¯H units. Total output is the same as before if T ni=1 (1/ti ) = nT /t¯H . So t¯H = n/ ni=1 (1/ti ), the harmonic mean of t1 , . . . , tn .
15.2
√ √ 2 2. x 2 + y 2 = y 2 = 6, then f (x, y) = 6 − √ 6 is a level curve of f at height c = 6 − 4, because when x + 6 + 2 = 6 − 4. In the general case, a level curve is defined by f (x, y) = x 2 + y 2 − (x 2 + y 2 ) + 2 = c for some constant c. Let x 2 + y 2 = u2 , where u > 0. Then u − u2 + 2 = c. Solving for u yields √ u = 21 ± 21 9 − 4c (provided that c ≤ 9/4) and thus x 2 + y 2 is a constant, which is the equation for a √ circle centered at (0, 0). If 9 − 4c > 0 and 1 − 9 − 4c > 0, i.e. 2 < c < 9/4, there are two circles. If c ≤ 2 or c = 9/4, there is only one positive value of u, so the level curve is one circle.
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4. Generally, the graph of g(x, y) = f (x) in 3-space consists of a surface traced out by moving the graph of z = f (x) parallel to the y-axis in both directions. The graph of g(x, y) = x is the plane through the y-axis at a 45◦ angle with the xy-plane. The graph of g(x, y) = −x 3 is shown in Fig. 15.2.4. (Only a portion of the unbounded graph is indicated, of course.) z
y x
Figure 15.2.4
15.3 2. (a) f1 (x, y) = 7x 6 , f2 (x, y) = −7y 6 , and f12 (x, y) = 0 4 5 (b) f1 (x, y) = 5x ln y, f2 (x, y) = x /y, and f12 (x, y) = 5x 4 /y (c) f1 (x, y) = 5(x 2 − 2y 2 )4 2x = 10x(x 2 − 2y 2 )4 , f2 (x, y) = 5(x 2 − 2y 2 )4 (−4y) = −20y(x 2 − 2y 2 )4 , and f12 (x, y) = 40x(x 2 − 2y 2 )3 (−4y) = −160xy(x 2 − 2y 2 )3
4. (a) FS = 2.26·0.44S −0.56 E 0.48 = 0.9944S −0.56 E 0.48 , FE = 2.26·0.48S 0.44 E −0.52 = 1.0848S 0.44 E −0.52 (b) SFS + EFE = S · 2.26 · 0.44S −0.56 E 0.48 + E · 2.26 · 0.48S 0.44 E −0.52 = 0.44 F + 0.48 F = 0.92 F , so k = 0.92. = 2, z = 4e2y , z = z = 0 (b) z = y/x, z = ln x, z = −y/x 2 , 6. (a) zx = 2x, zy = 2e2y , zxx yy xy yx x y xx = −y 2 exy , z = 2x−x 2 exy ,z = zyy = 0, zxy = zyx = 1/x (c) zx = y 2 −yexy , zy = 2xy−xexy , zxx yy xy = 2y − xyexy − exy zyx
8. Here ∂z/∂x = x/(x 2 + y 2 ), ∂z/∂y = y/(x 2 + y 2 ), ∂ 2 z/∂x 2 = (y 2 − x 2 )/(x 2 + y 2 )2 , and ∂ 2 z/∂y 2 = (x 2 − y 2 )/(x 2 + y 2 )2 . Thus, ∂ 2 z/∂x 2 + ∂ 2 z/∂y 2 = 0. 10. ux = au/x and uy = bu/y, so uxy = auy /x = abu/xy. Hence, uxy /ux uy = 1/u (u = 0). Then, uxy uxy 1 ∂ 1 −ux −1 1 ∂ = · 2 = 2 = ux ∂x ux uy ux u u uy ∂y ux uy
15.4 1 2 2. F (1, 0) = F (0, 0) + 0 F1 (x, 0) dx ≥ 2, F (2, 0) = F (1, 0) + 1 F1 (x, 0) dx ≥ F (1, 0) + 2 1 1 F (0, 1) = F (0, 0) + 0 F2 (0, y) dy ≤ 1, F (1, 1) = F (0, 1) + 0 F1 (x, 1) dx ≥ F (0, 1) + 2 1 F (1, 1) = F (1, 0) + 0 F2 (1, y) dy ≤ F (1, 0) + 1 4. With u = y/x, zx = f (u) − uf (u) and zy = f (u). The equation for the tangent plane at (x, y, z) is Z − xf (u) = [f (u) − uf (u)](X − x) + f (u)(Y − y). This goes through the origin (0, 0, 0) iff © Knut Sydsæter and Peter Hammond 2010
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−xf (u) = [f (u) − uf (u)](−x) + f (u)(−y) iff 0 = xuf (u) + f (u)(−y), which is satisfied because xu = y.
15.5 2. F1 (x, y, z) = 2xexz + x 2 zexz + y 4 exy , so F1 (1, 1, 1) = 4e; F2 (x, y, z) = 3y 2 exy + xy 3 exy , so F2 (1, 1, 1) = 4e; F3 (x, y, z) = x 3 exz , so F3 (1, 1, 1) = e. 4. First-order partials: w1 = 3yz + 2xy − z3 , w2 = 3xz + x 2 , w3 = 3xy − 3xz2 . Second-order partials: w11 = 2y, w12 = w21 = 3z + 2x, w13 = w31 = 3y − 3z2 , w22 = 0, w23 = w32 = 3x, w33 = −6xz. 6. For (x, y) = (0, 0), f1 = y(x 4 + 4x 2 y 2 − y 4 )(x 2 + y 2 )−2 . Thus, for y = 0, f1 (0, y) = −y. This is also correct for y = 0, because f1 (0, 0) = limh→0 [f (h, 0) − f (0, 0)]/ h = 0. Similarly, for (x, y) = (0, 0), f2 = x(x 4 − 4x 2 y 2 − y 4 )(x 2 + y 2 )−2 , and f2 (x, 0) = x for all x. It follows that f12 (0, 0) = −1 and 6 4 2 2 4 6 f21 (0, 0) = 1. Also, for (x, y) = (0, 0), f12 = f21 = (x + 9x y − 9x y − y )(x 2 + y 2 )−3 . Thus, f12 (0, y) = −1, f12 (x, 0) = 1, and f12 (x, y) has no limit as (x, y) → (0, 0). Similarly for f21 .
15.6 2. (a) YK = aAK a−1 and YK = aBLa−1 , so KYK + LYL = aAK a + aBLa = a(AK a + BLa ) = aY (b) KYK + LYL = KaAK a−1 Lb + LAK a bLb−1 = aAK a Lb + bAK a La = (a + b)AK a Lb = (a + b)Y 2aKL5 − bK 4 L2 2bK 5 L − aK 2 L4 and Y = , so KYK + LYL = (c) YK = L (aL3 + bK 3 )2 (aL3 + bK 3 )2 K 2 L2 (aL3 + bK 3 ) 2aK 2 L5 − bK 5 L2 + 2bK 5 L2 − aK 2 L5 K 2 L2 = = = Y . (According to 3 3 3 2 3 2 (aL + bK ) (aL + bK ) aL3 + bK 3 Section 16.5 these functions are homogeneous of degrees a, a + b, and 1, respectively, so the results are immediate consequences of Euler’s Theorem.) 4. ∂D/∂p and ∂E/∂q are normally negative, because the demand for a commodity goes down when the price of that commodity increases. ∂D/∂q and ∂E/∂p are (usually) positive, because the demand for a commodity increases when the price of a substitute increases.
−(m/ρ)−1
−(m/ρ)−1 , YL = mbL−ρ−1 Aeλt aK −ρ + bL−ρ . 6. YK = maK −ρ−1 Aeλt aK −ρ + bL−ρ
−ρ −(m/ρ)−1 −ρ −ρ λt −ρ Thus, KYK + LYL = m(aK + bL )Ae aK + bL = mY . (This function is homogeneous of degree m, so the result is an immediate consequences of Euler’s Theorem.)
15.7 2. If the monopolist is not allowed to discriminate, he must charge the same price P in both markets. Thus, from [1], P = a1 − b1 Q1 = a2 − b2 Q2 . Substituting Q 1 = (a1 − P )/b1 and Q2 = (a2 − P )/b2 in the expression for profit gives the quadratic function π = − (a1 b2 +a2 b1 )α −[a1 b2 +a2 b1 +(b1 +b2 )α]P + (b1 + b2 )P 2 /b1 b2 . This is maximized when P = 21 α + [a1 b2 + a2 b1 )/2(b1 + b2 )], giving maximum profit of [a1 b2 + a2 b1 − α(b1 + b2 )]2 /4b1 b2 (b1 + b2 ). If this expression is subtracted from the (higher) maximum profit in the example, we eventually find that the loss of profit is (a1 − a2 )2 /4(b1 + b2 ). Note that the loss is 0 if a1 = a2 . 2 4. Lˆ = σyy − σxy /σxx
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15.8 2. (a) ξ = −5/4, η = −1/2, d = −7/8 (b) ξ = 3/5, η = −1/5, d = 26/5 a b x ax + by a b x 4. = , so (x, y) = x(ax + by) + y(bx + cy), which reduces b c y bx + cy b c y to ax 2 + 2bxy + cy 2 .
15.9 2. (a) (x, y)
1 1 x 1 1 y
⎛
3 (b) (x1 , x2 , x3 ) ⎝ −1 3/2
⎞⎛ ⎞ 3/2 x1 −2 ⎠ ⎝ x2 ⎠ 3 x3
−1 1 −2
4. Let λ1 , . . . , λn be the eigenvalues of A. If A is positive semidefinite and symmetric, then according to Theorem 15.2(b), all the eigenvalues are nonnegative. Because |A| is not 0, 0 is not an eigenvalue, so all the eigenvalues must be positive. Then according to Theorem 15.2(a), A is positive definite.
Chapter 16 Tools for Comparative Statics 16.1 dz dy ∂z ∂z dy ∂z ∂z = F1 (t, y) + F2 (t, y) = + . If z = t 2 + yey and y = t 2 , then = 2t, = ey + yey , dt dt ∂t ∂y dt ∂t ∂y dz dy 2 2 = 2t. Hence we get = 2t (1 + et + t 2 et ). and dt dt 4. The usual rules for differentiating (a) a sum; (b) a difference; (c) a product; (d) a quotient; (e) a composite function of one variable.
2.
6. (a) F is defined for x > 0, y > 0. F (x, y) = 0 ⇐⇒ y = 1/x 2 . See Fig. 16.1.6. 2 + e−3t dz = → 2 as t → ∞. (b) dt 1 + e−3t y
2
F (x, y) > 0
1 y = 1/x 2 1
2
x
Figure 16.1.6
8. From [16.1], dz/dt = F1 (x, y)dx/dt + F2 (x, y)dy/dt. Differentiating w.r.t. t yields d 2 z/dt 2 = (x, y)dx/dt + (d/dt)[F1 (x, y)dx/dt] + (d/dt)[F2 (x, y)dy/dt]. Here (d/dt)[F1 (x, y)dx/dt] = [F11 2 2 F12 (x, y)dy/dt]dx/dt+F1 (x, y)d x/dt . Also, (d/dt)[F2 (x, y)dy/dt] = [F21 (x, y)dx/dt+F22 (x, y)dy/dt]dy/dt+ 2 2 F2 (x, y)d y/dt . Assuming F12 = F21 , the conclusion follows. © Knut Sydsæter and Peter Hammond 2010
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16.2 2. ∂z/∂t1 = F (x)∂x/∂t1 , ∂z/∂t2 = F (x)∂x/∂t2 4. (a) Let u = ln v, where v = x 3 + y 3 + z3 − 3xyz. Then ∂u/∂x = (1/v)(∂v/∂x) = (3x 2 − 3yz)/v. Similarly, ∂u/∂y = (3y 2 − 3xz)/v, and ∂u/∂z = (3z2 − 3xy)/v. Hence, x
∂u ∂u ∂u x y z 3v +y +z = (3x 2 − 3yz) + (3y 2 − 3xz) + (3z2 − 3xy) = =3 ∂x ∂y ∂z v v v v
which proves (i). Equation (ii) is then proved by elementary algebra. (b) Here ∂z/∂x = 2xyf (x 2 y) and ∂z/∂y = x 2 f (x 2 y), so x∂z/∂x = 2x 2 yf (x 2 y) = 2y∂z/∂y. 1 1 2 2 6. F (α) = 0 (∂/∂α)(xeαx ) dx = 0 x 3 eαx dx. Introduce u = αx 2 as a new variable. Then the integral 1 2 α α is (1/2α 2 ) 0 ueu du = (1/2α 2 )0 (ueu − eu ) = (αeα − eα + 1)/2α 2 . Also, F (α) = (1/2α)0 eαx = (eα − 1)/2α, so F (α) = (αeα − eα + 1)/2α 2 is correct. t n(τ )(−δ)(1 − T˙ )e−δ(t−T ) dτ = 8. With T = T (t), N˙ (t) = n(t)e−δ(t−T ) − n(t − T )e−δ(t−T ) (1 − T˙ ) + t−T
[n(t) − (1 − T˙ )n(t − T )]e−δ(t−T ) − δ(1 − T˙ )N (t). τ 2t 10. We have z(t) = t F (τ, t) dτ , where F (τ, t) = x(τ )e− t r(s) ds . Leibniz’s formula gives z˙ (t) = 2t τ 2t 2t 2F (2t, t) − F (t, t) + t (∂F (τ, t)/∂t) dτ = 2x(2t)e− t r(s) ds − x(t) + t x(τ )e− t r(s) ds r(t) dτ = 2t 2x(2t)p(t) − x(t) + t F (τ, t)r(t) dτ = 2p(t)x(2t) − x(t) + r(t)z(t), and therefore z˙ (t) − r(t)z(t) = 2p(t)x(2t) − x(t). Q a 12. (a) g (Q) = c + h 0 f (D) dD − p Q f (D) dD and g (Q) = (h + p)f (Q) ≥ 0 for all Q, so g Q∗ a a is convex. (b) g (Q∗ ) = 0 yields c + h 0 f (D) dD − p Q∗ f (D) dD = 0. Here Q∗ f (D) dD = Q∗ a ∗ ∗ 0 f (D) dD − 0 f (D) dD = 1 − F (Q ). It follows that F (Q ) = (p − c)/(h + p).
16.3 2. (a) With F (x, y) = xy, F1 = y, F2 = x, F11 = F22 = 0, and F12 = 1. Hence, [16.9] and [16.12] yield y 1 2yx 2 y = − , y = − 3 (−2 · 1 · yx) = 3 = 3 x x x x
(b) y = −(1 + 3y)/(3x − 1), y = 6(1 + 3y)/(3x − 1)2 (c) y = 5x 4 /6y 5 , y = −(10/3)x 3 y −5 − (125/36)x 8 y −11 4. (a) With F (x, y) = 2x 2 + xy + y 2 , y = −F1 /F2 = −(4x + y)/(x + 2y) = −4 at (2, 0). Moreover, y = −(28x 2 + 14y 2 + 14xy)/(x + 2y)3 = −14 at (2, 0). The tangent has the equation y = −4x + 8. (b) y = 0 requires y = −4x. Inserting this into the original equation gives the two points. 6. For x = −1 and y = 2 both LHS and RHS are −1, so the curve passes through (−1, 2). Implicit differentiation yields 2yy + 5 = ex(y−2) + xex(y−2) (y − 2 + xy ). At (−1, 2) we find y = −4/3, so the equation for the tangent is y = 13 (−4x + 2).
16.4 2. Let z = ug with u = a1 x1d + a2 x2d + a3 x3d . Then, for i = 1, 2, 3, Eli z = Elu ug Eli u = g(xi /u)ai dxid−1 = dgai xid /u, so El1 z + El2 z + El3 z = dga1 x1d /u + dga2 x2d /u + dga3 x3d /u = dg. (This result follows © Knut Sydsæter and Peter Hammond 2010
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easily from the fact that the function is homogeneous of degree dg (see Problem 2(b) in Section 16.6) and from the elasticity form [16.28] of the Euler equation.) 4. (a) Elx z = 20, Ely z = 30, Elt x = t/(t + 1), Elt y = 2t/(t + 1), so, according to [16.15], Elt z = Elx z Elt x + Ely z Elt y = 20t/(t + 1) + 30 · 2t/(t + 1) = 80t/(t + 1). 2y 2 2x 2 1 2[ln t + t 4 e−2t (2 − t)] (b) Elt z = 2 + · · (−t + 2) = x + y 2 ln t x2 + y2 (ln t)2 + t 4 e−2t 6. Taking logarithms yields ln z = ax + by and a ln x + b ln y + c ln z = 0. Differentiating w.r.t. x yields zx /z = a +byx and a/x +byx /y +czx /z = 0. Solving for yx and zx , and then multiplying the expressions −a(1 + cx) a(x − y) obtained by x/y and x/z respectively, yields finally Elx y = , Elx z = . b(1 + cy) 1 + cy 8. The easiest way to solve this problem is to note that KL = KL(aL2/3 + bK 2/3 )−3/2 = (aK −2/3 + bL−2/3 )−3/2 2/3 (aL + bK 2/3 )3/2 Comparing this with Example 16.20, we see that the production function is actually a CES-function with A = A0 e0.021t and ρ = 2/3. According to [2] in that example, σKL = 1/(2/3 + 1) = 3/5. = a(a − 1)F /K 2 , FKL = abF /KL, and 10. With F (K, L) = AK a Lb , FK = aF /K, FL = bF /L, FKK 2 3 /KL. FLL = b(b−1)F /L . Also, −F F (KF +LF ) = −(aF /K)(bF /L)(a+b)F = −ab(a+b)F K L K L
2 + (FK )2 FLL = −ab(a + b)F 3 /KL, so σKL = 1. Moreover, KL (FL ) FKK − 2FK F2 FKL
16.5 2. x(tp, tr) = A(tp)−1.5 (tr)2.08 = At −1.5 p −1.5 t 2.08 r 2.08 = t −1.5 t 2.08 Ap −1.5 r 2.08 = t 0.58 x(p, r), so the function is homogeneous of degree 0.58 . 4. By using [16.18] we find that f is homogeneous of degree 0, and using the results in Example 15.13(b) in Section 15.3 we confirm that xf1 (x, y) + yf2 (x, y) = 0. 6. f (tx, ty) = a ln[g(tx, ty)/tx] = a ln[tg(x, y)/tx] = a ln[g(x, y)/x] = f (x, y), so f is homogeneous of degree 0. 8. Suppose F is homogeneous of degree k = 0. Then F −1 is homogeneous of degree 1/k iff F −1 (tx) = t 1/k F −1 (x) for all t > 0. But this equation is equivalent to F (F −1 (tx)) = F (t 1/k F −1 (x)). Here the LHS is tx, and because F is homogeneous of degree 1/k, the RHS is (t 1/k )k F (F −1 (x)) = tx. The √ function F (x) = x defined for x ≥ 0 with range [0, ∞) has an inverse F −1 = x 2 , defined for x ≥ 0. F is homogeneous of degree 1/2, while F −1 is homogeneous of degree 2.
16.6 2. (a)
1 (tx1 tx2 tx3 )2 1 1 + + F (tx1 , tx2 , tx3 ) = (tx1 )4 + (tx2 )4 + (tx3 )4 tx1 tx2 tx3 1 1 t 6 (x1 x2 x3 )2 1 1 = tF (x1 , x2 , x3 ) + + = 4 4 x2 x3 t (x1 + x24 + x34 ) t x1
The function is homogeneous of degree 1.
g g
(b) G(tx1 , tx2 , tx3 ) = a(tx1 )d + b(tx2 )d + c(tx3 )d = t d (ax1d + bx2d + cx3d ) = (t d )g G(x1 , x2 , x3 ) = t dg G(x1 , x2 , x3 ), so G is homogeneous of degree dg. © Knut Sydsæter and Peter Hammond 2010
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41
4. All are homogeneous of degree 1, as is easily checked by using [16.23]. 6. When F (x, y) = a ln x + b ln y, xF1 + yF2 = x(a/x) + y(b/y) = a + b, which is not equal to kF (x, y) for any choice of k, so F is not homogeneous of any degree. 8. (a) f ((tx1 )m , . . . , (txn )m ) = f (t m x1m , . . . , t m xnm ) = (t m )r f (x1m , . . . , xnm ) = t mr h(x1 , . . . , xn ), so h is homo. of degree mr. (b) Homo. of degree sp. (c) Homo. of degree r for r = s, not homo. for r = s. (d) Homo of degree r + s. (e) Homo. of degree r − s.
16.7 2. Differentiating partially w.r.t. x yields (∗) 3x 2 + 3z2 zx − 3zx = 0, so zx = x 2 /(1 − z2 ). Similarly, or by , differentiate (∗) partially w.r.t. y to obtain 6zz z + 3z2 z − symmetry, zy = y 2 /(1 − z2 ). To find zxy y x xy 2 2 2 3 3zxy = 0, so zxy = 2zx y /(1 − z ) . 4. Implicit differentiation gives fP (r, P )Pw = gw (w, P ) + gP (w, P )Pw . Hence −gw (w, P ) Pw = < 0, because gw > 0, gP > 0, and fP < 0. gP (w, P ) − fP (r, P ) 6. Differentiating the equation w.r.t. x gives 1 − azx = f (y − bz)(−bzx ). Differentiating w.r.t. y gives −azy = f (y − bz)(1 − bzy ). Simple algebra yields azx + bzy = 1.
16.8 2. Write the function on the form g ∗ (μ, ε) = (1 + μ)a (1 + ε)αa − 1, where a = 1/(1 + β). Then ∂g ∗ (μ, ε)/∂μ = a(1 + μ)a−1 (1 + ε)αa ,
∂g ∗ (μ, ε)/∂ε = (1 + μ)a αa(1 + ε)αa−1
Hence, g ∗ (0, 0) = 0, ∂g ∗ (0, 0)/∂μ = a, ∂g ∗ (0, 0)/∂ε = αa, and g ∗ (μ, ε) ≈ aμ + αaε. 4. f (0.98, −1.01) ≈ −4.97. The exact value is −4.970614, so the error is 0.000614. 6. (a) and (b): dz = (y 2 + 3x 2 )dx + 2xy dy 8. (a) dU = 2a1 u1 du1 + · · · + 2an un dun −ρ −ρ −ρ−1 −ρ−1 (b) dU = A(δ1 u1 + · · · + δn un )−1−1/ρ (δ1 u1 du1 + · · · + δn un dun ) 10. Let T (x, y, z) = [x 2 + y 2 + z2 ]1/2 = u1/2 , where u = x 2 + y 2 + z2 . Then dT = 21 u−1/2 du = 1 −1/2 (2xdx + 2ydy + 2zdz). For x = 2, y = 3, and z = 6, we have u = 49, T = 7 and dT = 2u 1 (xdx +ydy +zdz) = 17 (2dx +3dy +6dz). Accordingly, T (2+0.01, 3−0.01, √ 6+0.02) ≈ T (2, 3, 6)+ 7 1 1 7 [2 · 0.01 + 3(−0.01) + 6 · 0.02] = 7 + 7 · 0.11 ≈ 7.016. (The exact value is 49.2206 ≈ 7.015739.) 12. (a) dX = AβN β−1 eρt dN + AN β ρeρt dt (b) dX1 = BEX E−1 N 1−E dX + B(1 − E)X E N −E dN 14. g(0) = f (x10 , . . . , xn0 ). Using the result in part (a) of Problem 11 in Section 16.2 gives g (0) = f1 (x10 , . . . , xn0 ) dx1 + · · · + fn (x10 , . . . , xn0 ) dxn . The approximation g(1) ≈ g(0) + g (0) now yields, using vector notation, f (x0 + dx) ≈ f (x0 ) + f1 (x0 ) dx1 + · · · + fn (x0 ) dxn .
16.9 2. By differentiation, 3 dx1 + 2x2 dx2 − dy1 − 9y22 dy2 = 0, 3x12 dx1 − 2dx2 + 6y12 dy1 − dy2 = 0. Letting dx2 = 0 and solving for dy1 and dy2 leads to dy1 = Adx1 and dy2 = Bdx1 , where A = ∂y1 /∂x1 = J −1 (3 − 27x12 y22 ) and B = ∂y2 /∂x1 = J −1 (3x12 + 18y12 ) with J = 1 + 54y12 y22 . © Knut Sydsæter and Peter Hammond 2010
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4. Differentiation w.r.t. x yields y + ux v + uvx = 0, u + xux + yvx = 0. Solving this system for ux and vx yields u2 − y 2 xy − uv u2 − y 2 2xy − 1 ux = = , vx = = yv − xu 2yv yv − xu 2yv where we substituted xu = −yv and uv = 1 − xy. Now, ∂ 2u ∂ 2uux 2yv − (u2 − y 2 )2yvx u = = ∂x 2 ∂x x 4y 2 v 2 2u(u2 − y 2 ) − (u2 − y 2 )(2xy − 1)/v (u2 − y 2 )(4uv − 1) = = 2 2 4y v 4y 2 v 3 (The answer to this problem can be expressed in many different ways.) ∂w ∂z ∂w ∂w 6. From w = G(x, y, z) we have (∗) ( )y = Gx + Gz ( )y . Here, ( )y,z = Gx and ( )x,y = Gz . ∂x ∂x ∂x ∂z Substituting these expressions into (∗) yields the result.
16.10 2. Denote the left hand sides of the three equations by determinant of these functions w.r.t. u, v, and w is ∂f1 /∂u ∂f1 /∂v ∂f1 /∂w 1 ∂f2 /∂u ∂f2 /∂v ∂f2 /∂w = 1 ∂f /∂u ∂f /∂v ∂f /∂w −1 3
3
3
f1 , f2 , and f3 , respectively. Then the Jacobian −1 3v 2 −1
−3w2 −1 = 2(3w2 − 1) 3w 2
At P the value of this determinant is 4 = 0, and the functions f1 , f2 , and f3 are C 1 functions, so according to the implicit function theorem the system defines u, v, and w as differentiable functions of x, y, and z in a neighborhood of P . Differentiating the system w.r.t. x (holding y and z constant) yields ux − vx − 3w 2 wx = ux + 3v 2 vx − wx −ux − vx + 3w 2 wx
=
0 2
= −2x
Substituting (x, y, z, u, v, w) = (1, 1, 0, −1, 0, 1) and solving the system yields ux = 5/2, vx = 1, and wx = 1/2.
Chapter 17 Multivariable Optimization 17.1 2. First-order conditions: f1 (x, y) = −4x − 2y + 36 = 0, f2 (x, y) = −2x − 4y + 42 = 0, with the solution x = 5, y = 8. 4. (a) P (10, 8) = P (12, 10) = 98 (b) First-order conditions: Px = −2x + 22 = 0, Py = −2y + 18 = 0. It follows that x = 11 and y = 9, with P (11, 9) = 100. 6. Solving the constraint for z yields z = 4x + 2y − 5, so we have to minimize P (x, y) = x 2 + y 2 + (4x + 2y − 5)2 w.r.t. x and y. We find P (x, y) = 17x 2 + 5y 2 + 16xy − 40x − 20y + 25. The first-order conditions are P1 = 34x + 16y − 40 = 0, P2 = 16x + 10y − 20 = 0, with the solution x = 20/21, y = 10/21. The minimum value is 525/441. © Knut Sydsæter and Peter Hammond 2010
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43
8. FK = −2(K −3)−(L−6) and FL = −4(L−6)−(K −3), so here [17.3] yields −2(K −3)−(L−6) = 0.65, −4(L − 6) − (K − 3) = 1.2. The solution is K = 2.8, L = 5.75. 10. x = am/(a + b + c)p, y = bm/(a + b + c)q, z = cm/(a + b + c)r. (The constraint yields z = (m − px − qy)/r, so we maximize P = Ax a y b [(m − px − qy)/r]c w.r.t. x and y. Keeping in mind that z depends on x and y, the first-order conditions are Px = Ax a−1 y b zc−1 [az − c(px/r)] = 0 and Py = Ax a y b−1 zc−1 [bz − c(qy/r)] = 0. These give px = (a/c)rz and qy = (b/c)rz. Inserted into the constraint and solved for z, these yield the given value for z, and the other two follow easily.)
17.2 2. Use Theorem 9.2 with F (u) = ln u. 4. By the chain rule, g1 (x, y) = F (f (x, y))f1 (x, y) and g2 (x, y) = F (f (x, y))f2 (x, y). Because F > 0, the conclusion follows. 6. (a) S = {(x, y) : g(x, y) = c}, where g(x, y) = x 2 + xy + y 2 , c = 3, with g continuous. (b) The given equation is symmetric in x and y, so −2 ≤ x ≤ 2. (c) The set is closed because g(x, y) = x 2 + xy − y 2 is continuous. The pair given in the hint does satisfy the equation, so the conclusion follows.
17.3 2. (a) f1 (x, y) = 3x 2 − 9y and f2 (x, y) = 3y 2 − 9x. The only stationary point in the interior of S is (3, 3). ((0, 0) is a stationary point on the boundary of S.) Along √ the four parts √ of the boundary we find the following candidates for extreme points: (0, 0), (4, 0), (4, 12), (4, 4), ( 12, 4), and (0, 4). Comparing the values of f at all these points, we find that f has mimimum 91 at (0, 4) and at (4, 0), minimum 0 at (3, 3). (A maximum and a minimum exist by the extreme-value theorem.) √ √ (b) Maximum 9/4 at (−1/2, 3/2) and at (−1/2, − 3/2). Minimum −1/4 at (1/2, 0). (c) Maximum 3 at (0, 0) and at (1, 0). Minimum 2 at (0, −1) and at (0, 1). (d) Maximum 2e4 at (0, 0). Minimum 0 for all (x, 1/2) where x ∈ [0, 2], and for all (2, y) where y ∈ [0, 1/2]. 1 2 4. (a) The domain D is shown in Fig. 17.3.4. The first order partials are f1 (x, y) = e− 4 (x+y ) 1 − 41 (x + y) , 1 2 f2 (x, y) = e− 4 (x+y ) 1 − 21 y(x + y) . (b) The point (7/2, 1/2) gives the maximum value 4e−15/16 . 6. (a) f1 = 9x(x 2 + y 2 )1/2 − 4x(x 2 + y 2 )−1/2 , f2 = 9y(x 2 + y 2 )1/2 − 4y(x 2 + y 2 )−1/2 + 1. At a stationary point, f1 = 0 implies that x = 0 or x 2 + y 2 = 4/9. But x 2 + y 2 = 4/9 implies f2 = 1, so x = 0 at the stationary√point. Then f2√= 9y 2 − 3 if y > 0, whereas f2 = −9y 2 + 5 if y < 0. The stationary points are (0, 13 3) and (0, − 13 5). (b) S is closed and bounded and f is continuous, so extreme points exist according to the extreme-value theorem. There are no stationary points in the interior of S, so the extreme points must be on the boundary. Along x = 0, y ∈ [−1, 1], f (0, y) = 3y 3 − 3y if y ≥ 0, and f (0, y) = −3y 3 + 5y if y ≤ 0. Along √ this part of √ the boundary f has its largest value 0 at (0, 1) and at (0, 0), and its smallest value −10 5/9 at (0, − 13 5). Along x 2 + y 2 = 1, y ∈ [−1, 1], f (x, y) = y − 1, which has its largest value 0 at (0, 1) and its smallest value −2 at (0, −1). √ We conclude √ that the maximum value is 0 at (0, 1) and at (0, 0), and that its smallest value is −10 5/9 at (0, − 13 5). © Knut Sydsæter and Peter Hammond 2010
44
CHAPTER 17
MULTIVARIABLE OPTIMIZATION y = 2x 2
y
z ln(1 + x 2 y)
z= y
2
y
y = x2
x
1
y = kx
D y =1−x x Figure 17.3.4
−1 Figure 17.4.2(f)
1
x
Figure 17.4.8
17.4 2. (a) (0, 0) saddle point, (1, 1) local minimum. (b) (−4/3, 1/3) local minimum. (c) (2, 3) local maximum, (a, 0) and (0, b) no decision (a and b arbitrary numbers). √ (d) (0, 0) saddle point, (a, 0) and (−a, 0) local maxima where a = 3/2. (e) (0, 0) saddle point, (1/2, 1/4) and (−1/2, −1/4) local mininma. (f) (0, b) no decision (b an arbitrary number). (The graph of this function is shown in Fig. 17.4.2(f). f is defined for x = 0 and for y > −1/x 2 . f1 (x, y) = 2xy/(1 + x 2 y) and f2 (x, y) = x 2 /(1 + x 2 y). Here f1 = f2 = 0 at all points (0, b) with b ∈ ⺢. Because AC − B 2 = 0 when (x, y) = (0, b), the second-derivative test fails. But by studying the function directly, we see that (0, b) is a local maximum point if b < 0; a saddle point if b = 0; and a local minimum point if b > 0. See Fig. 17.4.2(f).) 4. (a) The level curves are x = 1 and y = 0. g(x, y) is positive if y > 0 and x = 1. (b) g1 = y(x 2 − 1)ex+3y , g2 = (x − 1)2 (1 + 3y)ex+3y . (−1, −1/3) is a local minimum. (1, y0 ) is a local minimum for y0 > 0, a local maximum for y0 < 0, and a saddle point for y0 = 0. (Use the results in (a).) (c) g(x, 1) = (x −1)2 ex+3 → ∞ as x → ∞, so g has no maximum; g(x, −1) = −(x −1)2 ex−3 → −∞ as x → ∞, so g has no minimum either. 6. We find easily in all three cases that (0, 0) is a stationary point where z = 0 and AC − B 2 = 0. In case (a), z ≤ 0 for all (x, y), so the origin is a maximum point. In case (b), z ≥ 0 for all (x, y), so the origin is a minimum point. In (c), z takes positive and negative values at points arbitrarily close to the origin, so it is a saddle point. 8. (a) Fig. 17.4.8. The domain where f (x, y) is negative is shaded. (b) (0, 0) is seen to be a stationary point at which z = 0. As Fig. 17.4.8 shows, this function takes positive and negative values for points arbitrary close to (0, 0), so the origin is a saddle point. (c) g(t) = f (th, tk) = (tk−t 2 h2 )(tk−2t 2 h2 ) = 2h4 t 4 − 3h2 kt 3 + k 2 t 2 , so g (t) = 8h4 t 3 − 9h2 kt 2 + 2k 2 t and g (t) = 24h4 t 2 − 18h2 kt + 2k 2 . Then g (0) = 0 and g (0) = 2k 2 , so t = 0 is a minimum point for k = 0. For k = 0, g(t) = 2t 4 h4 , which has a minimum at t = 0.
17.5 2. (a) The set is convex; see Fig. 17.5.2(a). (b) The set is convex; see Fig. 17.5.2(b). (c) The set is not convex; see Fig. 17.5.2(c). (d) The set is convex; see Fig. 17.5.2(d). (e) The set is not convex; see Fig. 17.5.2(e). (f) The set is not convex; see Fig. 17.5.2(f). ((4, 0) and (0, 4) belong to the set, but not 21 (4, 0) + 21 (0, 4) = (2, 2).) © Knut Sydsæter and Peter Hammond 2010
CHAPTER 17 y
y
y 4 3 2 1
3 2 1 -3 -2 -1 -1
1
2
3
x
x
-2 -3 Figure 17.5.2(a) Convex.
Figure 17.5.2(b) Convex.
y
-4 -3 -2 -1 -1 -2 -3 -4
1 2 3 4
x
Figure 17.5.2(c) Not convex.
y
y
3
3
4
2 2
3
1
1
-3 -2 -1
1
2
3
x
-1 x -1
45
MULTIVARIABLE OPTIMIZATION
1
2
3
-1 Figure 17.5.2(d) Convex.
2 1
-2 -3 Figure 17.5.2(e) Not convex.
1
2
3
4
x
Figure 17.5.2(f) Not convex.
4. To prove that aS + bT is convex, take x ∈ aS + bT , y ∈ aS + bT , and a number λ ∈ [0, 1]. Because x ∈ aS + bT , there are points x1 ∈ S and y1 ∈ T such that x = ax1 + by1 . Similarly, there are points x2 ∈ S and y2 ∈ T such that y = ax2 + by2 . Then (1 − λ)x + λy = (1 − λ)(ax1 + by1 ) + λ(ax2 + by2 ) = a[(1 − λ)x1 + λx2 ] + b[(1 − λ)y1 + λy2 ]. This belongs to aS + bT because S and T are convex, so (1 − λ)x1 + λx2 ∈ S and (1 − λ)y1 + λy2 ∈ T . Therefore aS + bT is convex.
17.6 2. If f is linear, then f is both concave and convex according to Example 17.12. On the other hand, if f is both concave and convex on the convex set S, then [17.15] is satisfied with equality for any λ ∈ (0, 1) and for all x, x0 in S. Thus the graph of f contains any line segment between two points on its graph. So the graph of f is a hyperplane, and f is linear. 4. We prove the first statement in [17.16]. The second statement is proved in the same way. Suppose first that f is concave. Take any two points (x1 , y1 ) and (x2 , y2 ) in Mf . Then y1 ≤ f (x1 ) and y2 ≤ f (x2 ). Let λ ∈ [0, 1]. Then (1−λ)(x1 , y1 )+λ(x2 , y2 ) = ((1−λ)x1 +λx2 , (1−λ)y1 +λy2 ). Here (1−λ)x1 +λx2 ∈ S, because S is convex. Moreover, by the concavity of f , f ((1 − λ)x1 + λx2 ) ≥ (1 − λ)f (x1 ) + λf (x2 ) ≥ (1 − λ)y1 + λy2 . This inequality implies that (1 − λ)(x1 , y1 ) + λ(x2 , y2 ) belongs to Mf , so Mf is convex. Suppose on the other hand that Mf is convex. Take any x1 , x2 in S and let λ ∈ [0, 1]. Of course, f (x1 ) ≤ f (x1 ) and f (x2 ) ≤ f (x2 ), so (x1 , f (x1 )) and (x2 , f (x2 )) both belong to Mf . Because Mf is convex, (1 − λ)(x1 , f (x1 )) + λ(x2 , f (x2 )) = ((1 − λ)x1 + λx2 , (1 − λ)f (x1 ) + λf (x2 )) ∈ Mf . This implies that (1 − λ)f (x1 ) + λf (x2 ) ≤ f ((1 − λ)x1 + λx2 ), and so f is concave. © Knut Sydsæter and Peter Hammond 2010
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17.7 2. Part (a) remains valid iff either a or b is strictly positive. This follows immediately from the arguments in the proof of part (a) of Theorem 17.6. (Of course, if a = b = 0, af + bg is certainly not strictly concave.) It follows from the (c) thatif F is strictly concave, then the inequality proof of part 0 0 F f ((1 − λ)x + λx) ≥ F (1 − λ)f (x ) + λf (x) becomes strict for x0 = x, so U is strictly concave. There is no need to assume that f is strictly concave. For part (e), suppose f and g are strictly concave over the convex set S. Let λ ∈ (0, 1) and let x = y. Then, because f and g are strictly concave, f ((1 − λ)x + λy) > (1 − λ)f (x) + λf (y), and g((1 − λ)x + λy) > (1 − λ)g(x) + λg(y). It follows that h((1 − λ)x + λy) = min{f (1 − λ)x + λ(y) , g (1 − λ)x + λy } > min{(1 − λ)f (x) + λf (y), (1 − λ)g(x) + λg(y)} But (1 − λ)f (x) + λf (y) ≥ (1 − λ) min{f (x), g(x)} + λ min{f (y), g(y)} = (1 − λ)h(x) + λh(y) Similarly, (1−λ)g(x)+λg(y) ≥ (1−λ)h(x)+λh(y). Therefore, h((1−λ)x+λy) > (1−λ)h(x)+λh(y).
17.8 2 2. (a) f11 = 2a, f12 = 2b, f22 = 2c, and f11 f22 − (f12 ) = 2a2c − (2b)2 = 4(ac − b2 ). The result follows from Theorem 17.10. (b) According to Theorem 17.9, f is concave iff a ≤ 0, b ≤ 0, and ac − b2 ≥ 0; f is convex iff a ≥ 0, b ≥ 0, and ac − b2 ≥ 0. (x, y) = −ex − ex+y , f12 (x, y) = −ex+y , and f22 (x, y) = −ex+y , so f11 (x, y) < 0 and 4. (a) f11 2 2x+y f11 f22 − (f12 ) = e > 0. Hence, f is strictly concave. 2 (b) g is strictly convex. (g11 = ex+y + ex−y > 0 and g11 g22 − (g12 ) = 4e2x > 0.) 2 = 2 − 6x ≤ 0, f22 = −2 ≤ 0, and f11 f22 − (f12 ) = −2(2 − 6x) − 1 ≥ 0. 6. f is concave where f11 Here 2 − 6x ≤ 0 iff x ≥ 1/3 and −2(2 − 6x) − 1 ≥ 0 iff x ≥ 5/12. We conclude that the largest convex domain on which f is concave is S = {(x, y) : x ≥ 5/12}. 8. f1 = −4x + 4, f2 = −2y + 4, f11 = −4, f12 = 0, f22 = −2. Clearly (1, 2) is a stationary point for f and the conditions in Theorem 17.11(a) are satisfied, so (1, 2) is the maximum point.
10. (a) The firms choose p and q to maximize px+qy−(5+x)−(3+2y) = 26p+24q−5p 2 −6q 2 +8pq−69. This is a strictly concave function. The maximum is given by p = 9, q = 8, x = 16, y = 4. A’s profit is 123, whereas B’s is 21. (b) Firm A’s profit is now πA (p) = px − 5 − x = p(29 − 5p + 4q) − 5 − 29 + 5p − 4q = 34p − 5p 2 + 4pq − 4q − 34, with q fixed. This quadratic polynomial is maximized at p = pA (q) = 15 (2q +17). Likewise, firm B’s profit is now πB (q) = qy −3−2y = 28q −6q 2 +4pq −8p−35, with p fixed. This quadratic polynomial is maximized at q = qB (p) = 13 (p + 7). (c) Equilibrium occurs where p = pA (q) and q = qB (p). So 5p = 2q + 17 and 3q = p + 7. These give p = 5, q = 4, x = 20, y = 12. A gets 75, B gets 21. (d) The process starts at (9, 8) and converges to (5, 4). See Fig. 17.8.10. © Knut Sydsæter and Peter Hammond 2010
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47
q p = pA (q)
10 8
(9, 8)
6
q = qB (p)
4
(5, 4)
2 2
4
6
8
10 12 14
p
Figure 17.8.10
17.9 2. (a) Stationary points must satisfy the three equations: f1 = 2x + 2y + 2z = 0, f2 = 4y + 2x = 0, and f3 = 6z + 2x = 0. The only solution is (0, 0, 0). The Hessian matrix is ⎛ ⎞ ⎛ ⎞ f13 2 2 2 f11 f12 ⎟ ⎜ ⎟ = ⎜ f23 ⎠ ⎝2 4 0⎠ ⎝ f21 f22 f31
f32
2 0 6 f33 2 2 2 2 2 = 4, 2 4 0 = 8. So (0, 0, 0) is a local minimum point. The leading principal minors are 2, 2 4 2 0 6 (b) f has two stationary points, (3, 3, 3) and ( 13 y02 , y0 , y0 ), where y0 ≈ 1.63 is the other real root (apart from y = 3) of the equation y 4 − 54y + 81 = 0. (There are two complex roots.) The Hessian matrix is ⎛ ⎞ 6x −9 −9 ⎜ ⎟ H = ⎝ −9 6y 0⎠
−9
0
6z 6x −9 = 36xy − 81, D3 = |H| = The leading principal minors areD1 = 6x, D2 = −9 6y 54(4xyz − 9y − 9z). At (3, 3, 3) the leading principal minors are all positive, so this is a local minimum point. At ( 13 y02 , y0 , y0 ), D1 > 0, D2 < 0, and D3 < 0. According to [17.27], this is a saddle point. (c) The stationary points are the solutions of the equation system f1 = 8x2 − 8x1 = 0, f2 = 20 + 8x1 − 12x22 = 0, f3 = 48 − 24x3 = 0, f4 = 6 − 2x4 The first equation gives x2 = x1 , and then the second equation gives 12x12 − 8x1 − 20 = 0 with the two solutions x1 = 5/3 and x1 = −1. The last two equations determine x3 and x4 . There are two stationary points, (5/3, 5/3, 2, 3) and (−1, −1, 2, 3). The ⎛ Hessian matrix is ⎞ −8 8 0 0 ⎜ 8 −24x2 0 0⎟ ⎟ f (x1 , x2 , x3 , x4 ) = ⎜ ⎝ 0 0 −24 0⎠ 0 0 0 −2 and the leading principal minors of the Hessian are D1 = −8, D2 = 192x2 −64 = 64(3x2 −1), D3 = −24D2 , and D4 = 48D2 . At (−1, −1, 2, 3) we get D2 < 0, so this point is a saddle point. The other stationary point, (5/3, 5/3, 2, 3), we get D1 < 0, D2 > 0, D3 < 0, and D4 > 0, so this point is a local maximum point. © Knut Sydsæter and Peter Hammond 2010
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17.10 2. If f is increasing, or decreasing, then the upper level sets {x : f (x) ≥ a} are all intervals, which are convex sets, so f is quasi-concave. 4. Let x, x0 be any pair of vectors in S, and λ ∈ [0, 1]. Let z = (1 − λ)x + λx0 . Now, if f (z) ≥ min{f (x), f (x0 )}, then f (x) ≥ f (x0 ) implies f (z) ≥ f (x0 ), as in [17.31]. Hence, Theorem 17.15 implies that f is quasi-concave. Conversely, suppose f is quasi-concave. If f (x) ≥ f (x0 ), then [17.31] implies f (z) ≥ f (x0 ). On the other hand, if f (x) ≤ f (x0 ), then interchanging x and x0 in [17.31] and replacing λ by 1 − λ implies that f (z) ≥ f (x). In either case, f (z) ≥ min{f (x), f (x0 )}. 6. (a) This is mainly an exercise in manipulating determinants. If you feel that the calculations below look = ai aj z/xi xj for frightening, try to write them out in full for the case k = 3 (or k = 2). Note that zij 2 i = j , and zii = ai (ai − 1)z/xi . Rule 3 in Theorem 13.1 tells us that a common factor in any column (or row) in a determinant can be “moved outside”. Therefore, a1 (a1 − 1) a 1 a2 a 1 ak z z ··· z 2 x1 x2 x1 xk x1 . . . z1k z11 z12 a2 a1 a2 (a2 − 1) a 2 ak z z . . . z z z · · · z 21 22 2k 2 = x2 x1 x 2 xk x Dk = . 2 . . . .. .. .. .. .. .. .. . . . . . . zk1 zk2 . . . zkk ak a1 a k a2 ak (ak − 1) z z ··· z xk x1 xk x2 xk2 a1 − 1 a1 a1 ··· x1 x1 x1 a − 1 a1 ··· a1 1 a2 − 1 a2 a2 a2 − 1 · · · a2 (2) a1 a2 . . . ak k a2 ··· (1) a1 a2 . . . ak k x2 x2 x2 = = z z . .. .. .. x1 x2 . . . xk . (x1 x2 . . . xk )2 .. . . . . . . .. .. .. .. ak ak · · · ak − 1 ak ak − 1 ak ··· xk xk xk where equality (1) holds because aj z/xj is a common factor in column j for each j and equality (2) holds because 1/xi is a common factor in row i for each i. a1 a2 . . . ak k (b) Let sk = ki=1 ai = a1 + · · · + ak , and Pk = z . We use the expression for Dk that (x1 x2 . . . xk )2 we found in part (a), and add rows 2, 3, . . . , k to the first row. Then each entry in the first row becomes equal to sk − 1. Afterwards we take the common factor sk − 1 in row 1 and move it outside. s − 1 s − 1 ··· s − 1 1 1 ··· 1 k k k a2 − 1 · · · a2 a2 a2 a2 a2 − 1 · · · Dk = Pk .. .. .. = (sk − 1)Pk .. .. .. .. .. . . . . . . . . ak ak ak · · · ak − 1 ak · · · ak − 1 Now subtract column 1 from all the other columns. Rule 7 in Theorem 13.1 says that this does not change 1 0 · · · 0 a2 −1 · · · 0 the value of the determinant, so Dk = (sk − 1)Pk .. .. .. = (−1)k−1 (sk − 1)Pk . .. . . . . ak 0 · · · −1 © Knut Sydsæter and Peter Hammond 2010
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(c) By assumption, ai > 0 for all i, so if ni=1 ai < 1, then sk = ki=1 ai < 1 for all k. Therefore Dk has the same sign as (−1)k . Combining Theorem 15.3 and Theorem 17.14, it follows that f is strictly concave.
Chapter 18 Constrained Optimization 18.1 2. (a) When f (x, y) = 3xy, g(x, y) = x 2 + y 2 , and c = 8, equation [18.3] reduces to 3y/3x = 2x/2y, and so x 2 = y 2 . Inserted into the constraint this yields x 2 = 4, and so x = ±2. The solution candidates are therefore: (2, 2), (2, −2), (−2, 2), (−2, −2). Here f (2, 2) = f (−2, −2) = 12 and f (−2, 2) = f (2, −2) = −12. So (2, 2) and (−2, −2) are the only possible solutions of the maximization problem, and (−2, 2) and (2, −2) are the only possible solutions of the minimization problem. (The extreme-value theorem ensures that we have found the solutions, because f is continuous and the constraint curve is a closed bounded set (a circle).) (b) With L = x + y − λ(x 2 + 3xy + 3y 2 − 3), the first-order conditions are 1 − 2λx − 3λy = 0 and 1 − 3λx − 6λy = 0. From these equations we get 2λx + 3λy = 3λx + 6λy, or λ(3y + x) = 0. Here λ = 0 is impossible, so x = −3y. Inserted into the constraint we have (3, −1) and (−3, 1) as the only possible solutions of the maximization and minimization problems, respectively. The extreme value theorem ensures that solutions exist. (The objective function is continuous and the constraint curve is a closed bounded set (an ellipse).)
18.2 2. (a) x = a/5, y = 2a/5 solve the problem. Solving x + 2y = a for y yields y = 21 a − 21 x, and then x 2 + y 2 = x 2 + ( 21 a − 21 x)2 = 45 x 2 − 21 ax + 41 a 2 . This quadratic function certainly has a minimum at x = a/5. (b) L(x, y) = x 2 + y 2 − λ(x + 2y − a). The necessary conditions are L1 = 2x − λ = 0, L2 = 2y − 2λ = 0, implying that 2x = y. From the constraint, x = a/5 and then y = 2a/5, λ = 2a/5. (c) Find the point on the line x+2y = a which has the minimal distance from the origin. The corresponding maximization problem has no solution. (See Fig. 18.2.2.) (d) f ∗ (a) = (a/5)2 + (2a/5)2 = a 2 /5, so df ∗ (a)/da = 2a/5, which is also the value of the Lagrangean multiplier. Equation [18.7] is confirmed. y a/2
(a/5, 2a/5)
a
x
Figure 18.2.2
4. L(2, 2) = 2 > L(1, 1) = 1. (In fact, L(x, y) has a saddle point at (1, 1).) © Knut Sydsæter and Peter Hammond 2010
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6. (a) L(x, y) = 100−e−x −e−y −λ(px +qy −m), so the first-order conditions are: (1) Lx = e−x −λp = 0 and (2) Ly = e−y − λq = 0. These imply that −x = ln(λp) = ln λ + ln p, −y = ln λ + ln q. Inserting the expressions for x and y into the constraint, and solving for ln λ, yields m + q ln q − q ln p ln λ = −(m + p ln p + q ln q)/(p + q). Therefore x(p, q, m) = and p+q m + p ln p − p ln q y(p, q, m) = . p+q tm + tq ln(tq) − tq ln(tp) tm + tq(ln t + ln q) − tq(ln t + ln p) (b) x(tp, tq, tm) = = tp + tq t (p + q) = x(p, q, m), so x is homogeneous of degree 0. The constraint px + qy = m remains unchanged when x, y, m are all multiplied by t, so the maximization of U is over the same constraint set in both cases. Hence, the optimal values of x and y have to be unchanged. y y y 8. (a) g(x, y) = 0 xt (y − t) dt = 0 (xyt − xt 2 ) dt = 0 ( 21 xyt 2 − 13 xt 3 ) dt = 16 xy 3 . Similar computations 1 yield f (x, y) = − 16 αxy 5 + 12 xy 4 + 16 xy 3 . The solution of [∗] is x = 384α 3 M, y = 1/4α. (Note that, 1 3 because 6 xy = M, the problem reduces to that of maximizing M + 21 My − αMy 2 for y ≥ 0.) (b) As α tends to 0, the cost of extraction tends to infinity. Because the sales price increases as t increases, it is reasonable to expect that the terminal time y tends to infinity. (c) The value function is f ∗ (M) = 1 4 M + M/16α. The Lagrange multiplier must satisfy the equation f1 − λg1 = 0, or − 16 αy 5 + 12 y + 16 y 3 − 1 3 λ 6 y = 0. Inserting the optimal values of x and y and solving for λ yields λ = 1 + 1/16α. Clearly, ∂f ∗ (M)/∂M = λ.
18.3 2. (a) (i) The constraint implies that x = y 2 /8 ≥ 0, so the problem reduces to that of minimizing h(x) = (x − 1)2 + 8x for x ≥ 0. Because h (x) = 2x + 6 ≥ 6 for all x ≥ 0, the minimum value is 1 at x = 0. (ii) The problem reduces to minimizing k(y) = ( 18 y 2 − 1)2 + y 2 . Here k (y) = 2( 18 y 2 − 1) 41 y + 2y = 1 2 y( 16 y + 23 ). Thus k (y) < 0 for y < 0, and k (y) > 0 for y > 0. We conclude that k(y) is minimized at y = 0, which gives the same solution as in (i). (b) For the Lagrangean L = (x−1)2 +y 2 −λ(y 2 −8x), the first-order conditions are: (1) 2(x−1)+8λ = 0; (2) 2y − 2λy = 0. From (2), either λ = 1 or y = 0. But inserting λ = 1 into (1) yields x = −3, and then no value of y satisfies the constraint. So y = 0, implying the solution x = y = 0. (c) The problem is to find which point on the parabola y 2 = 8x is closest to (1, 0).
18.4 2. In this case, f (x, y) = x 2 + y 2 , g(x, y) = x + 2y, so 0 D(x, y) = g1 g 2
g1 f11 − λg11 f21
−
λg21
0 = f12 − λg12 1 f − λg 2 g2
22
22
1 2 0
2 0 = −10 2
By [18.14], there is a local minimum at the stationary point (a/5, 2a/5).
18.5 2. Here L = x + 4y + 3z − λ(x 2 + 2y 2 + 13 z2 − b). So necessary conditions are: (1) L1 = 1 − 2λx = 0; (2) L2 = 4 − 4λy = 0; (3) L3 = 3 − 23 λz = 0. It follows that λ = 0, © Knut Sydsæter and Peter Hammond 2010
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2 and so x √ = 1/2λ, y = 1/λ, z = 9/2λ. Inserting these values into the constraint yields √ λ = 9/b, so λ = ±3/ b. The value of the criterion function is x + 4y + 3z = 18/λ, so λ = −3/ b minimizes the √ criterion function. The minimum point is (a, 2a, 9a), where a = − b/6. 1 11 4. With x = 16 15 + h, y = 3 + k, and z = − 15 + , constraints [1] and [2] reduce to h + 2k + l = 0 and 2h − k − 3 = 0, so k = −h and = h. But then easy algebra shows that f (x, y, z) = (16/15)2 + (1/3)2 + (−11/15)2 + 3h2 . Because 3h2 ≥ 0 for all h, (16/15)2 + (1/3)2 + (−11/15)2 must then be the maximum value of f (x, y, z) subject to the constraints. The maximum value is obviously attained when h = k = = 0, or when (x, y, z) = (16/15, 1/3, −11/15).
6. (a) Here L = x + y + z − λ(x 2 + y 2 + z2 − 1) − μ(x − y − z − 1). Necessary conditions are: (1) L1 = 1 − 2λx − μ = 0; (2) L2 = 1 − 2λy + μ = 0; (3) L3 = 1 − 2λz + μ = 0. Adding (1) and (2) yields λ(x + y) = 1. Adding (1) and (3) yields λ(x + z) = 1. It follows that λ = 0 and also that z = y. Substituting z = y in the two constraints yields: (4) x 2 + 2y 2 = 1; (5) x − 2y = 1. From (5), x = 2y + 1, so (4) gives 6y 2 + 4y = 0, with solutions y = 0 and y = −2/3. The only possible solutions to the problem are therefore (1, 0, 0) and (−1/3, −2/3, −2/3). In fact, (1, 0, 0) solves the maximization problem, while (−1/3, −2/3, −2/3) solves the minimization problem. (b) The constraints are represented geometrically by a closed bounded curve which is the intersection between a sphere (x 2 + y 2 + z2 = 1) and a plane (x − y − z = 1). The continuous function f (x, y, z) = x + y + z must therefore attain maximum and minimum values subject to the constraints, and in (a) we have found the extreme points. 8. Here L = a12 x12 + · · · + an2 xn2 − λ(x1 + · · · + xn − 1). Necessary conditions are that Lj = 2aj2 xj − λ = 0, j = 1, . . . , n, and so xj = λ/2aj2 . Inserted into the constraint, this implies that 1 = 21 λ(1/a12 +· · ·+1/an2 ). Thus, for j = 1, . . . , n, we have xj =
1 1 = 2 n aj2 (1/a12 + · · · + 1/an2 ) aj i=1 (1/ai2 )
(Implicitly, this solution assumes that each ai = 0. If at least one ai is 0, the minimum value is 0, which is attained by letting a corresponding xi be 1, with the other xj all equal to 0.) 10. (a) The Lagrangean is here L = Ax1a1 · · · xnan − λ(p1 x1 + · · · + pn xn − m), and so Li = ai U/xi − λpi = 0, i = 1, . . . , n. In particular, p1 x1 = a1 U/λ, and so λ = a1 U/p1 x1 . It follows that pi xi = (ai /a1 )p1 x1 . Inserted into the budget constraint, we have p1 x1 + (a2 /a1 )p1 x1 + a1 m · · · + (an /a1 )p1 x1 = m, which implies that x1 = . The demand functions are given p1 (a1 + · · · + an ) ai m by xi = for i = 1, . . . , n. pi (a1 + · · · + an ) (b) The Lagrangean is L = x1a + · · · + xna − λ(p1 x1 + · · · + pn xn − m), and so Li = axia−1 − λpi = 0. −a/(1−a) 1/(1−a) In particular, λ = ax1a−1 /p1 . Algebraic manipulations yield pi xi = pi p1 x1 , i = 1, . . . , n. Using the budget constraint equation gives an expression for x1 , and subsequently we find the other −1/(1−a) mp demand quantities xi . The consumer’s demand functions are xi = n i , for i = 1, . . . , n. " −a/(1−a) pj j =1
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12. For n = 2, the Lagrangean is L = a11 x12 + 2a12 x1 x2 + a22 x22 − λ(x12 + x22 − 1). The first-order conditions for x1 = x10 , x2 = x20 , with λ = λ0 to solve the problem are: 2a11 x10 + 2a12 x20 − 2λ0 x10 = 0
(∗)
2a12 x10 + 2a22 x20 − 2λ0 x20 = 0
Cancelling the common factor 2 and writing (∗) as a matrix equation yields
a11 a21
a12 a22
x10 x20
= λ0
x10 x20
or, in obvious matrix notation, (∗∗) Ax0 = λ0 x0 . Thus, if x0 solves the problem, then x0 is an eigenvector of A, with λ0 as an eigenvalue. Multiplying (∗∗) from the left by x0 = (x10 , x20 ) then implies that x0 Ax0 = x0 (λ0 x0 ) = λ0 x0 x0 = λ0 , because x0 x0 = 1. We conclude that the maximum value of x Ax subject to x x = 1 must be the largest eigenvalue of A, whereas the minimum value must be the smallest eigenvalue. Both a maximum and a minimum value do exist because of the extreme-value theorem. The argument in the general case of n variables is entirely similar. In fact, equating the partials of the Lagrangean to 0 leads to a system of equations like (∗), which is then easily seen to reduce to (∗∗)
18.6 2. (a) The Lagrangean is L = 4z − x 2 − y 2 − z2 − λ(z − xy). The first-order conditions are −2x + λy = 0, −2y + λx = 0, 4 − 2z − λ = 0, and z = xy. The only triples (x, y, z) that satisfy these conditions are (0, 0, 0) with λ = 4, (1, 1, 1) with λ = 2, and (−1, −1, 1) with λ = 2. (b) f ∗ ≈ λ·0.1 = 2·0.1 = 0.2
18.7 2. If L(x, c) = f (x) −
m
j =1 λj (gj (x)
− cj ), then ∂L/∂cj = λj , so [18.24] follows from [18.31].
18.8 2. (a) Writing the constraints as g1 (x, y) = x + e−x − y ≤ 0 and g2 (x, y) = −x ≤ 0, the Lagrangean is L = 21 x − y − λ1 (x + e−x − y) − λ2 (−x). The Kuhn–Tucker conditions are then: ∂L 1 = − λ1 (1 − e−x ) + λ2 = 0 ∂x 2 ∂L = −1 + λ1 = 0 ∂y λ1 ≥ 0 (= 0 if x + e−x < y)
(1) (2) (3)
λ2 ≥ 0
(4)
(= 0 if x > 0)
From (2) and (3), we conclude that λ1 = 1 and so x + e−x = y. Either x = 0 or x > 0. In the latter case, (4) implies λ2 = 0. Then (1) implies or = 21 . Hence x = ln 2, and so y = x + e−x = ln 2 + 21 . e−x
If x = 0, then (1) implies λ2 = − 21 , which contradicts λ2 ≥ 0. © Knut Sydsæter and Peter Hammond 2010
1 2
− (1 − e−x ) = 0,
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We conclude that (x, y) = (ln 2, ln 2 + 21 ) is the only point satisfying the Kuhn–Tucker conditions. (By sketching the constraint set and studying the level curves 21 x − y = c, it is easy to see that the point we found solves the problem.) (b) With the Lagrangean L = x 2 + 2y − λ1 (x 2 + y 2 ) − λ2 (−y), the Kuhn–Tucker conditions are: ∂L = 2x − 2λ1 x = 0 ∂x ∂L = 2 − 2λ1 y + λ2 = 0 ∂y λ1 ≥ 0 (= 0 if x 2 + y 2 < 5)
(1) (2) (3)
λ2 ≥ 0
(4)
(= 0 if y > 0)
Note that the second constraint cannot be active, because if y = 0, then (2) implies λ2 = −2, which contradicts (4). So y > 0, and from (4), λ2 = 0. Then from (2), λ1 y = 1. Hence λ1 > 0, and so from (3), x 2 + y 2 = 5. Also, (1) evidently implies that x = 0 or λ1 = 1. √ √ If x = 0, then x 2√+ y 2 = 5 implies y = ± 5.√With y√ > 0, only y = 5 is possible. Then (2) implies that λ1 = 1/ 5. Thus (x, y, λ1 , λ2 ) = (0, 5, 1/ 5, 0) satisfies (1) to (4) and is a solution candidate. If λ1 = 1, then λ1 y = 1 implies y = 1 and from x 2 +y 2 = 5 it follows that x = ±2. Thus (2, 1, 1, 0) and (−2, 1, 1, 0) are both solution candidates. √ √ Now, f (0, 5) = 2 5, while f (2, 1) = f (−2, 1) = 6. The constraint set is closed and bounded (as a half-disk), so the extreme-value theorem applies. Thus, (2, 1) and (−2, 1) both solve the problem, with λ1 = 1 and λ2 = 0. 4. (a) The Lagrangean is L = x + ay − λ(x 2 + y 2 − 1) + μ(x + y). There must exist numbers λ and μ such that: (1) 1 − 2λx + μ = 0; (2) a − 2λy + μ = 0; (3) λ ≥ 0 (= 0 if x 2 + y 2 < 1); (4) μ ≥ 0 (= 0 if x + y > 0). √ √ √ (b) For a ≥ −1, the point√(1/ 1 +√a 2 , a/ 1 + a 2 ) solves the problem, with λ√= 21 1 + a 2 , μ = 0. For a < −1, the point (1/ 2, −1/ 2) solves the problem, with λ = (1 − a)/2 2, μ = − 21 (1 + a). 6. The Lagrangean is L = ln(x 2 + 2y) − 21 x 2 − y − λ1 (2 − xy) + λ2 (x − 1) + λ3 (y − 1). The Kuhn–Tucker conditions are that there must exist numbers λ1 ≥ 0, λ2 ≥ 0, and λ3 ≥ 0 such that
(3)
2x − x + λ 1 y + λ2 = 0 x 2 + 2y 2 L2 (x, y) = 2 − 1 + λ 1 x + λ3 = 0 x + 2y λ1 (2 − xy) = 0
(4)
λ2 (1 − x) = 0
(5)
λ3 (1 − y) = 0
(1) (2)
L1 (x, y) =
(We have deliberately formulated the conditions in a slightly different way than was done for Problems 2 and 4. See [18.37] in the main text.) © Knut Sydsæter and Peter Hammond 2010
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Suppose first that λ1 = 0. Then (1) and (2) imply 2x − x + λ2 = 0, x 2 + 2y
2 − 1 + λ3 = 0 x 2 + 2y
Because x and y are both ≥ 1, we get λ3 = 1 − 2/(x 2 + 2y) ≥ 1 − 23 > 0 and λ2 = x − 2x/(x 2 + 2y) = xλ3 > 0. It follows from (4) and (5) that x = y = 1, which contradicts xy ≥ 2. Thus λ1 > 0 and so xy = 2. Suppose next that λ2 > 0. Then x = 1 (because of (4)) and y = 2/x = 2, so λ3 = 0. From (2) it follows that λ1 = 3/5. Then (1) implies that λ2 = −3/5, contradicting λ2 ≥ 0. Hence, λ2 = 0. Suppose now that λ3 > 0. Then (5) implies y = 1, and so x = 2/y = 2. From (1) it follows that λ1 = 4/3. Then, from (2), λ3 = −1/3 + 1 − 8/3 < 0, which is impossible. Thus λ3 = 0. So far we have proved that λ2 = λ3 = 0, λ1 > 0, and xy = 2. Equations (1) and (2) now yield x2
2x = x − λ1 y, + 2y
x2
2 = 1 − λ1 x + 2y
It follows that x − λ1 y = x(1 − λ1 x). But λ1 > 0 and so y = x 2 . Because xy = 2, we must have x 3 = 2. Hence, x = 21/3 and y = 2x −1 = 22/3 is the only point which satisfies the Kuhn–Tucker conditions. (The corresponding nonnegative multipliers are λ1 = 2−1/3 − 13 , λ2 = λ3 = 0.) Why is this a maximum? In the feasible set, x 2 +2y ≥ 3, and f1 (x, y) and f2 (x, y) are both negative. For a given value of x, f (x, y) decreases as y increases, and for a given value of y, f (x, y) decreases as x increases. It follows that the constrained maximum lies on the part of the curve xy = 2 between the two points (1, 2) and (2, 1). (See Fig. 18.8.4.) This is a closed and bounded set, so the extreme-value theorem implies that f (x, y) has a maximum on this part of the curve. This point must be a maximum point for f over the whole feasible set. y
4
3 2 1 (1, 1) 1
2
3
4
x
Figure 18.8.4
18.9 2. The Lagrangean is L = 9x + 8y − 6(x + y)2 − λ1 (x − 5) − λ2 (y − 3) − λ3 (−x + 2y − 2) and the Kuhn–Tucker conditions are: (1)
L1 = 9 − 12(x + y) − λ1 + λ3 ≤ 0 (= 0 if x > 0)
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(2)
L2 = 8 − 12(x + y) − λ2 − 2λ3 ≤ 0 (= 0 if y > 0)
(3)
λ1 ≥ 0 (= 0 if x < 5)
(4)
λ2 ≥ 0 (= 0 if y < 3)
(5)
λ3 ≥ 0 (= 0 if −x + 2y < 2)
55
We go through the systematic procedure explained at the end of Section 18.8. (Shortcuts are possible. For instance, one could start by proving that x = 5 is impossible, then that y cannot be positive, etc.) x < 5, y < 3, −x + 2y < 2. Then λ1 = λ2 = λ3 = 0. If x = 0, then from (1), y ≥ 3/4 > 0, so from (2), 8 − 12y = 0. Thus y = 2/3, which contradicts y ≥ 3/4. So x > 0. Then (1) implies 9 = 12(x + y). If y > 0, then (2) implies 8 = 12(x + y), a contradiction. Thus y = 0. But then 9 = 12x, so x = 3/4. So a solution candidate is (x, y) = (3/4, 0), with λ1 = λ2 = λ3 = 0. x = 5, y < 3, −x + 2y < 2. Then λ2 = λ3 = 0 and (1) implies λ1 = 9 − 12(5 + y) < 0, a contradiction. x < 5, y = 3, −x + 2y < 2. Then λ1 = λ3 = 0 and (2) implies λ2 = 8 − 12(x + 3) < 0, a contradiction. x < 5, y < 3, −x + 2y = 2. Then λ1 = λ2 = 0 and 2y = 2 + x > 0, so y > 0 and (2) imply 8 − 12(x + y) = 2λ3 . It follows that λ3 = 4 − 6[x + 21 (2 + x)] = −2 − 9x < 0, a contradiction. x < 5, y = 3, −x + 2y = 2. Then x = 4 and λ1 = 0. Now (1) gives λ3 = 75, and then (2) gives λ2 = −226, contradicting λ2 ≥ 0. x = 5, y < 3, −x + 2y = 2. Then the first and second equalities yield y = 3.5, which contradicts y < 3. x = 5, y = 3, −x + 2y < 2. Then λ3 = 0. Now (1) gives λ1 = −87, contradicting λ1 ≥ 0. x = 5, y = 3, −x + 2y = 2. Obviously impossible. We conclude that the only admissible pair satisfying (1) to (5) is (x, y) = (3/4, 0). The extreme-value theorem implies that we have found the solution. (Theorem 18.4 of the next section also shows optimality.) 4. Using (a) in Theorem 18.3 with x = xˆ implies that f (x0 ) −
m j =1
λj gj (x0 ) ≥ f (ˆx) −
m j =1
λj gj (ˆx). But,
because xˆ also solves [18.41], f (ˆx) = f (x0 ). Thus, because λj ≥ 0 and gj (ˆx) ≤ cj , j = 1, . . . , m, m m m m and also because of [18.45], we have (∗) λj cj ≥ λj gj (ˆx) ≥ λj gj (x0 ) = λj cj . Here the j =1
j =1
j =1
j =1
two middle terms, being squeezed between two equal numbers, must themselves be equal. Therefore m m m f (ˆx) − λj gj (ˆx) = f (x0 ) − λj gj (x0 ) ≥ f (x) − λj gj (x) for all x ≥ 0. Also, if λk > 0 and j =1
gk (ˆx) < ck for any k, then
m j =1
j =1
j =1
λj (gj (ˆx) − cj ) < 0, which contradicts (∗). Thus xˆ satisfies (a) and (b) in
Theorem 18.3. 6. (a) As suggested by the hint, L(x∗ , λ∗ ) ≤ L(x∗ , λ) for all λ ≥ 0 implies that (1)
m " j =1
λj∗ (gj (x∗ ) − cj ) ≥
m "
λj (gj (x∗ ) − cj )
j =1
for all λ ≥ 0. If gk (x∗ ) > ck for any k, then jm=1 λj (gj (x∗ ) − cj ) can be made arbitrarily large by choosing λk large and λj = 0 for all j = k. Hence, gj (x∗ ) ≤ cj , j = 1, . . . , m. Now, however, because © Knut Sydsæter and Peter Hammond 2010
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one can have λj = 0, j = 1, . . . , m, (1) implies that mj =1 λj∗ (gj (x∗ ) − cj ) ≥ 0. Yet λj∗ ≥ 0 and m ∗ ∗ ∗ gj (x ) ≤ cj , j = 1, . . . , m, so j =1 λj (gj (x ) − cj ) = 0. Finally, whenever gj (x) ≤ cj , j = 1, . . . , m, the inequality L(x, λ∗ ) ≤ L(x∗ , λ∗ ) implies that f (x) − f (x∗ ) ≤
m "
λj∗ [gj (x) − gj (x∗ )] =
j =1
m "
λj∗ [gj (x) − cj ] ≤ 0
j =1
Therefore f (x) ≤ f (x∗ ), so x∗ solves [18.41]. (b) First, part (b) of the question should be corrected to read as follows: “Suppose that there exist x∗ ≥ 0 and λ∗ ≥ 0 satisfying both gj (x∗ ) ≤ cj and gj (x∗ ) = cj whenever > 0 for j = 1, . . . , m, as well as L(x, λ∗ ) ≤ L(x∗ , λ∗ ) whenever x ≥ 0. Show that L(x, λ) has a saddle point at (x∗ , λ∗ ) in this case.” λj∗
Then an answer is as follows: The first inequality in [∗] is satisfied by assumption. As for the second, under the indicated hypotheses, one has L(x∗ , λ∗ ) − L(x∗ , λ) =
m "
(λj − λj∗ )[gj (x∗ ) − cj ] ≤ −
j =1
m "
λj∗ [gj (x∗ ) − cj ] = 0
j =1
whenever λ ≥ 0. Thus, [∗] has been verified. y
P
x
2x y c 2 2x y c 1 Figure 18.10.2
18.10 2. (a) As shown in Fig. 18.10.2, the feasible set is the intersection of the two circular disks with the non2 = 4 = negative orthant. The solution is at the point P , where the circles intersect. Thus, (x + 1)2 + y√ 1 x 2 + (y + 1)2 . It follows that x = y and so 2x 2 + 2x + 1 = 4. The (relevant) positive root is 2 ( 7 − 1), √ √ so the solution is 21 ( 7 − 1), 21 ( 7 − 1) . (b) The Kuhn–Tucker conditions are: (1)
2 − λ1 2(x + 1) − λ2 2x ≤ 0 (= 0 if x > 0)
(2)
1 − λ1 2y − λ2 2(y + 1) ≤ 0 (= 0 if y > 0)
(3)
λ1 ≥ 0 (= 0 if (x + 1)2 + y 2 < 4)
(4)
λ2 ≥ 0 (= 0 if x 2 + (y + 1)2 < 4) The objective function is linear and thus concave. The constraint functions are both sums of convex functions, so they are convex. The conditions (1) to (4) are therefore sufficient for an admissible pair to
© Knut Sydsæter and Peter Hammond 2010
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57
√ √ solve the problem. With x = y = 21 ( 7 − 1), equations (1) to (4) are satisfied with λ1 = (7 + 3 7)/28 √ and λ2 = (3 7 − 7)/28. (c) The change in the optimal value of 2x + y is approximately λ2 · 0.1 ≈ 0.003. 4. (a) If x ∗ = a, f (x ∗ ) ≤ 0; if x ∗ ∈ (a, b), f (x ∗ ) = 0; if x ∗ = b, f (x ∗ ) ≥ 0. (b) See Figs. 18.10.4. y
y
y
x* a
b
x
(a)
a
x*
b
x
(b)
a
x* b x
(c)
Figure 18.10.4
6. (a) See Fig. 18.10.6. The optimal solution is obviously at (x, y) = (1, 0). At this point the two constraints g(x, y) = y − (1 − x)3 ≤ 0 and h(x, y) = −y ≤ 0 are both binding. The gradient vector of g is (g1 , g2 ) = (3(1 − x)2 , 1) = (0, 1) at (1, 0), and the gradient of h is (h1 , h2 ) = (0, −1). Hence, the two gradients are linearly dependent at (1, 0) (in fact, (0, −1) = (−1)(0, 1)), and so the constraint qualification is violated at this point. y 1
y (1 x)3
1
x
3x y c Figure 18.10.6
(b) The Lagrangean is L = 3x + y − λ1 (y − (1 − x)3 ) + λ2 x + λ3 y, so the Kuhn–Tucker conditions are: (1)
L1 = 3 − 3λ1 (1 − x)2 + λ2 = 0
(2)
L2 = 1 − λ1 + λ3 = 0
(3)
λ1 ≥ 0 (= 0 if y < (1 − x)3 )
(4)
λ2 ≥ 0 (= 0 if x > 0)
(5)
λ3 ≥ 0 (= 0 if y > 0) Suppose x > 0. Then from (4), λ2 = 0, and (1) implies λ1 (1 − x)2 = 1. Thus λ1 = 0, so λ1 > 0 and then from (3), y = (1 − x)3 . From λ1 (1 − x)2 = 1 it follows also that x = 1, so y = (1 − x)3 = 0, and
© Knut Sydsæter and Peter Hammond 2010
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CHAPTER 19
LINEAR PROGRAMMING
hence y > 0. So from (5), λ3 = 0. Then (2) yields λ1 = 1. Therefore (1) implies that (1 − x)2 = 1. So x = 2, because x > 0. But for x = 2, one has y = (1 − x)3 = −1, a contradiction. Suppose x = 0. Then (1) implies 3 − 3λ1 + λ2 = 0, so 3λ1 = 3 + λ2 > 0. Thus, from (3), y = (1 − x)3 = 1, and so by (5), λ3 = 0. Then from (2), λ1 = 1. So all the conditions (1)–(5) are satisfied by x = 0, y = 1, with λ1 = 1 and λ2 = λ3 = 0. So the Kuhn–Tucker conditions have a unique solution, which is not optimal. Finally, the two constraints g(x, y) = y − (1 − x)3 ≤ 0 and k(x, y) = −x ≤ 0 are both binding at (0, 1). But (g1 , g2 ) = (3(1 − x)2 , 1) = (3, 1) and (k1 , k2 ) = (−1, 0) are linearly independent. So the constraint qualification is satisfied at (0, 1), where the Kuhn–Tucker conditions are met. The Kuhn–Tucker conditions can be violated at the optimum (1, 0), because the constraint qualification is not satisfied at that point.
Chapter 19 Linear Programming 19.1 2. (a) No maximum exists. Consider Fig. 19.1.2. By increasing c, the dashed level curve x1 + x2 = c for the criterion function moves to the north-east and so this function can obviously take arbitrary large values. (b) Yes, the maximum is at P = (1, 0). The level curves are the same as in (a), but the direction of increase is reversed. x2
y
x1 x2 1
4 3
x1 3x2 3
x1 x2 c
1000
700x 1000y c
2 1
P 1
2
3
4
5
6
Figure 19.1.2
x1
1000
2000
x
Figure 19.1.4
⎧ ⎨ 3x + 5y ≤ 3900 max 700x + 1000y subject to x + 3y ≤ 2100 ⎩ 2x + 2y ≤ 2200
4. The LP problem is: (∗)
x ≥ 0, y ≥ 0
The problem is represented geometrically in Fig. 19.1.4. The solution is obviously at P , where the two lines 3x + 5y = 3900 and 2x + 2y = 2200 intersect. Solving these equations yields x = 800 and y = 300. So the firm should produce 800 TV sets of type A and 300 sets of type B.
19.2
#
3u1 + u2 ≥ 3 u1 ≥ 0, u2 ≥ 0 2u1 + 4u2 ≥ 4 # x1 + 2x2 ≤ 10 x1 ≥ 0, x2 ≥ 0 (b) max 11x1 + 20x2 subject to 3x1 + 5x2 ≤ 27
2. (a) min 6u1 + 4u2 subject to
© Knut Sydsæter and Peter Hammond 2010
CHAPTER 19
19.3
#
2. max 300x1 + 500x2 subject to
10x1 + 25x2 ≤ 10 000 20x1 + 25x2 ≤ 8 000
LINEAR PROGRAMMING
59
x1 ≥ 0 , x2 ≥ 0
The solution can be found graphically. It is x1∗ = 0, x2∗ = 320, and the value of the criterion function is 160, 000, the same value found in Example 19.2 for the optimal value of the primal criterion function.
19.5 2. (a) Figure 19.5.2 shows a portion of the feasible set and two dashed level curves for the criterion function Z = y1 + 2y2 . We see that the minimum is attained at the point (y1∗ , y2∗ ) = (3, 2). (b) The dual is: x +x −x + x ≤1 1
max 15x1 + 5x2 − 5x3 − 20x4
2
3
4
6x1 + x2 + x3 − 2x4 ≤ 2
subject to
x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0 Because y1∗ and y2∗ are both positive, the first two constraints are satisfied with equality. Furthermore, the two last constraints in the primal problem are satisfied with strict inequality at the optimum. Hence x3∗ = x4∗ = 0. Thus we have x1∗ + x2∗ = 1, 6x1∗ + x2∗ = 2, so x1∗ = 1/5 and x2∗ = 4/5. The maximum is thus at (x1∗ , x2∗ , x3∗ , x4∗ ) = (1/5, 4/5, 0, 0). (c) If the constraint is changed to y1 + 6y2 ≥ 15.1, the solution to the primal (P) is still at the intersection of the lines (1) and (2) in Fig. 19.5.2, but with (1) shifted up slightly. The solution to the dual is completely unchanged. The optimal values in (P ) and (D) both increase by (15.1 − 15) · x1∗ = 0.02. y2 (4) 10 (3) 5
y1 + 2y2 = Z0
(2) (3, 2)
(1) 5
10
15
y1
Figure 19.5.2
4. The dual to problem [1] is
[2]
min q1 V1 + q2 V2 + r 1 x¯ 1 + · · · + r N x¯ N
⎧ 1 ξ1 q1 + ξ21 q2 + r 1 ⎪ ⎪ ⎪ ⎨ ξ 2 q1 + ξ 2 q2 +r 2 1 2 s.t. . .. ⎪ ⎪ ⎪ ⎩ N ξ1 q1 + ξ2N q2 +r N
≥1 ≥1 .. . ≥1
where the variables are all ≥ 0. The implications [i] to [vi] are all special cases of the complementary slackness conditions [19.13] and [19.14] in Theorem 19.4. © Knut Sydsæter and Peter Hammond 2010
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Chapter 20 Difference Equations 20.1 2. (a) Monotone convergence to x ∗ from below. (b) Damped oscillations around x ∗ . (c) Monotonically increasing towards ∞. (d) Explosive oscillations around x ∗ . (e) xt = x ∗ for all t. (f) Oscillations around x ∗ with constant amplitude. (g) Monotonically (linearly) increasing towards ∞. (h) Monotonically (linearly) decreasing towards −∞. (i) xt = x0 for all t. 4. (a) Because the parameters are positive, yk+1 is positive provided that yk > 0. Because y0 is positive, by induction, yt is positive for all t. (b) Substituting yt = 1/xt gives the equation xt+1 = (a/c)xt + b/c. When a = 2, b = 3, and c = 4, this equation reduces to xt+1 = (1/2)x
−1 t + 3/4. When x0 = 1/y0 = 2, the solution is xt = (1/2)t+1 + 3/2, and so yt = (1/2)t+1 + 3/2 . Clearly, yt → 2/3 as t → ∞. Wt+2 Pt+1 γ + βWt+1 = = , and [2] follows immediately.(b) According to [2], the Wt+1 Pt γ + βWt fraction Wt+1 /(γ + βWt ) is the same for all t. For t = 0 it is equal to W1 /P0 = c, and thus [3] follows. From [20.4], the solution is
6. (a) From [1],
Wt = W0 −
cγ cγ (cβ)t + 1 − cβ 1 − cβ
(c) The equation is stable iff |cβ| < 1. In this case, Wt →
(1 = cβ)
cγ as t → ∞. 1 − cβ
20.2 2. According to [4] in Example 20.5, the yearly repayment is z=
0.07 · 100, 000 ≈ 8058.64 1 − (1.07)−30
In the first year the interest payment is 0.07B = 7000, and so the principal repayment is ≈ 8058.64 − 7000 = 1058.64. In the last year, the interest payment is 0.07b29 ≈ 8058.64[1 − (1.07)−1 ] ≈ 527.20, and so the principal repayment is ≈ 8058.64 − 527.20 = 7531.44.
20.4 2. xt = A + B t is a solution: xt+2 − 2xt+1 + xt = A + B (t + 2) − 2[A + B (t + 1)] + A + B t = A + Bt + 2B − 2A − 2Bt − 2B + A + B t = 0. Substituting t = 0 and t = 1 in xt = A + B t yields A = x0 and A + B = x1 , with solution A = x0 and B = x1 − x0 . So xt = A + B t is the general solution of the given equation. 4. xt = (A + B t)2t + 1 is a solution: xt+2 − 4xt+1 + 4xt = [A + B (t + 2)]2t+2 + 1 − 4{[A + B (t + 1)]2t+1 + 1} + 4[(A + B t)2t + 1] = 4A2t + 4Bt2t + 8B2t + 1 − 8A2t − 8Bt2t − 8B2t − 4 + 4A2t + 4Bt2t + 4 = 1. Substituting t = 0 and t = 1 in xt = A 2t + B t2t + 1 yields A + 1 = x0 and 2A + 2B + 1 = x1 , with solution A = x0 − 1 and B = 21 x1 − x0 + 21 . So xt = A 2t + B t2t + 1 is the general solution of the given equation. © Knut Sydsæter and Peter Hammond 2010
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DIFFERENCE EQUATIONS
61
20.5 2. (a) The characteristic equation m2 + 2m + 1 = (m + 1)2 = 0 has the double root m = −1, so the general solution of the homogeneous equation is xt = (C1 + C2 t)(−1)t . We find a particular solution by inserting u∗t = A 2t . This yields A = 1, and so the general solution of the inhomogeneous equation is xt = (C1 + C2 t)(−1)t + 2t . (b) By using the method of undetermined coefficients to determine the constants A, B, and C in the 3 1 particular solution u∗t = A5t + B cos π2 t + C sin π2 t, we obtain A = 41 , B = 10 , and C = 10 . So the 1 t 3 π 1 π t general solution to the given equation is xt = C1 + C2 2 + 4 5 + 10 cos 2 t + 10 sin 2 t. √ 4. The characteristic equation is m2 −4(ab+1)m+4a 2 b2 = 0, with solutions m1,2 = 2(ab+1± 1 + 2ab ). The general solution is therefore Dn = C1 mn1 + C2 mn2 . 6. Inserting xt = ut (−a/2)t into [20.21], assuming that b = 41 a 2 , we obtain the equation xt+2 + axt+1 + 41 a 2 xt = ut+2 (−a/2)t+2 + aut+1 (−a/2)t+1 + 41 a 2 ut (−a/2)t = 41 a 2 (−a/2)t (ut+2 − 2ut+1 + ut ), which is 0 if ut+2 − 2ut+1 + ut = 0. The general solution of this equation is ut = A + B t, so xt = ut (−a/2)t = (A + B t)(−a/2)t , which is the result claimed for case 2 in the frame on page 751. 8. (a) The first two equations state that consumption and capital are proportional to the net national product in the previous period. The third equation states that net national product, Yt , is divided between consumption, Ct , and net investment, Kt − Kt−1 . (b) To derive a second-order difference equation, first replace t by t + 2 in the last displayed equation in the problem to obtain Yt+2 = Ct+2 + Kt+2 − Kt+1 . But Ct+2 = cYt+1 , Kt+2 = σ Yt+1 , and Kt+1 = σ Yt , so we obtain Yt+2 − (c + σ )Yt+1 + σ Yt = 0. Explosive oscillations occur when (c + σ )2 < 4σ and σ > 1. 10. From the quadratic formula, m2 + am + b has real zeros iff b ≤ 41 a 2 . Combining [20.27] and [20.28], we must prove that f (m) = am2 + bm + c = 0 has both roots in the interval (−1, 1) iff |a| < 1 + b and b < 1. For these roots to be in the interval (−1, 1), it is necessary that f (−1) > 0, f (1) > 0, f (−1) < 0, and f (1) > 0. Thus 1 − a + b > 0, 1 + a + b > 0, −2 + a < 0, and a + 2 > 0, which is equivalent to |a| < 1 + b and |a| < 2. But then, because the roots are real, b ≤ 41 a 2 < 1. Conversely, if |a| < 1 + b and b < 1, then f (−1) > 0, f (1) > 0, and |a| < 2. So f (−1) = −2 + a < 0 and f (1) = 2 + a > 0. (2) (1) 12. (a) Because u(1) t and ut are two linearly independent solutions of the homogeneous (!) equation, Aut + Bu(2) t is the general solution of the homogeneous equation. So, to prove that [∗∗] is the general solution of (1) t−1 t−1 ci u(2) (2) ci ui+1 i+1 [∗], it suffices to prove that Ut∗ = −u(1) t Di+1 + ut Di+1 is a particular solution of [∗]. Insertion i=0
i=0
in [∗], followed by tedious algebraic computation, eventually yields the result. t−1 t−i−1 " r2 − r1t−i−1 (b) pt = Ar1t + Br2t + ui+2 , t = 0, 1, . . . r2 − r 1 i=0 (First write the equation in the form pt+2 + λ1 pt+1 + λ2 pt = ut+2 , t = 0, 1, 2, . . .. Assuming that r1 , r2 are real(!) and different roots of the characteristic polynomial, two linearly independent solutions of the (2) i+1 i+2 i+2 i+1 t t homogeneous equation are u(1) = r1i+1 r2i+1 (r2 − r1 ). t = r1 and ut = r2 . So Di+1 = r1 r2 − r1 r2 © Knut Sydsæter and Peter Hammond 2010
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CHAPTER 21
DIFFERENTIAL EQUATIONS
Then −r1t
t−1 "
ui+2 r2i+1
i=0
r1i+1 r2i+1 (r2 − r1 )
+ r2t
t−1 "
ui+2 r1i+1
i=0
r1i+1 r2i+1 (r2 − r1 )
=
t−1 t−i−1 " r − r t−i−1 2
i=0
1
r2 − r 1
ui+2
is the particular solution in [∗∗] to the nonhomogeneous equation.)
Chapter 21 Differential Equations 21.1 2. (a) If x = Ct 2 , then x˙ = 2Ct, and so t x˙ = 2Ct 2 = 2x. so C = 2. Hence, x = 2t 2 is the desired solution.
(b) x = Ct 2 passes through (1, 2) if 2 = C · 12 ,
4. (a) Replace the problem by: If x = x(t) satisfies x 2 = 2 1 − t 2 , then x˙ =
−t . √ x 1 − t2 (The original problem makes no sense. The differential equation only makes sense if t ∈ (−1, 1), √ 2 2 2 2 and then x + 2 1 − t =0 is only satisfied for x = 0.) Differentiating each side of x = 2 1 − t w.r.t. t yields 2x x˙ = −2t/ 1 − t 2 , and so x˙ = −t/x 1 − t 2 . 2 (b) Differentiating each side yields 21 et 2t − e−x x(x ˙ + 1) + e−x x˙ = 0, and the result follows. 2 (c) Differentiating each side yields −x + 2(1 − t)x x˙ = 3t 2 . Simplifying and using (1 − t)x 2 = t 3 yields the given equation.
6. The point of this problem is that you do not have to solve the differential equation (which would involve the inverse tangent function, anyway). From x˙ = (1+x 2 )t, we see at once that x˙ < 0 for t < 0 and x˙ > 0 for t > 0. Thus t = 0 is a global minimum point for x(t), and because x(0) = 0, one has x(t) ≥ 0 for all t. Differentiating the given equation w.r.t. t yields x¨ = 2x xt ˙ + (1 + x 2 ) = 2xt (1 + x 2 )t + (1 + x 2 ) = 2 2 (1 + x )(2xt + 1). Clearly x¨ > 0 for all t, so x = x(t) is convex. (In fact, the equation does not have a solution on all of ⺢, so x(t) only has a minimum on an open interval around 0.)
21.2 2. See Fig. 21.2.2. (The solutions are the circles t 2 + x 2 = C, t = 0, with C an arbitrary nonnegative constant. This can be shown by direct differentiation, or by solving the separable equation x˙ = −t/x using the method set out in the next section.) x t 2 + x2 = 4 1 1
Figure 21.2.2
© Knut Sydsæter and Peter Hammond 2010
t
CHAPTER 21
DIFFERENTIAL EQUATIONS
63
21.3 2. (a) Direct integration yields x = 41 t 4 − 21 t 2 + C. (b) x = tet − et − 21 t 2 + C (c) ex (dx/dt) = t + 1, or ex dx = (t +1) dt, so that ex dx = (t +1) dt and ex = 21 t 2 +t +C. The solution is x = ln( 21 t 2 +t +C). 4. In both cases N˙ depends on both N and t. (In the first figure, for instance, N (t1 ) = N (t2 ), but N˙ (t1 ) = ˙ 2 ).) N(t 6. (a) x = Ct a
(b) x = Ct b eat
(c) x = Cbt b /(1 − aCt b )
8. If z = x/t, then x = zt, so x˙ = z˙ t + z, and the equation x˙ = g(x/t) becomes the separable equation 1x 1 x −2 x3 + t 3 = . The suggested substitution leads + t z˙ = g(z) − z. The special equation is x˙ = 3tx 2 3t 3 t dz 1 dt to t z˙ = 13 z−2 − 23 z. Separating the variables and integrating yields = . The integral −2 z − 2z 3 t z2 dz . It can be found by introducing the new variable u = 1 − 2z3 , implying that on the LHS is 1 − 2z3 du = −6z2 dz. The final answer is x = 3 21 t 3 + Ct .
21.4 2. (a) Because g(ξ ) = 1/ξ , f (s) = s, x0 = 1, and t0 =
√ x t 2, formula [21.3] yields ξ dξ = s ds, 1
√
2 x t √ or 1 21 ξ 2 = √2 21 s 2 , or 21 x 2 − 21 = 21 t 2 − 1. Thus, the required solution is x = t 2 − 1. (Actually, it is probably easier to find the general solution x 2 − t 2 = C first, and then determine the appropriate constant.) x t x t (b) Formula [21.3] gives 0 (ξ + 1)−2 dξ = 0 e−2s ds, so 0 − 1/(ξ + 1) = − 21 0 e−2s , implying that 1 − e−2t . 1 − 1/(x + 1) = 21 (1 − e−2t ). Hence, x = 1 + e−2t 4. Using the given identity, the separable differential equation [1] becomes
αy −1 1 + y 1 − αy
dy =
dx x
Integration yields ln y − (1/) ln |1 − αy | = ln x + C1 . Multiplying both sides by yields ln y − ln |1 − αy | = ln x + C1 , or ln
y = ln eC1 x , |1 − αy |
hence
y = Cx 1 − αy
with C = ±eC1
Putting β = 1/C and solving for y yields [2].
21.5 2. x = Ce−t/2 + 21 . The equilibrium state x ∗ = 1/2 is stable. See Fig. 21.5.2. 4. dx/dt = b − ax, so dx/(b − ax) = dt + A for some constant A. This implies that (−1/a) ln |b − ax| = t + A, so ln |b − ax| = −at − aA, and |b − ax| = e−at−aA = e−at e−aA . It follows that (∗) b − ax = ±e−at−aA = ±e−aA e−at = Ce−at for C = ±e−aA . The conclusion follows by solving (∗) for x. © Knut Sydsæter and Peter Hammond 2010
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CHAPTER 21
DIFFERENTIAL EQUATIONS x 1
x = 21 e−t/2 +
1 2
(C = 21 )
x = −e−t/2 +
1 2
(C = −1)
t
−1 Figure 21.5.2
˙ 6. From x = X/N , by logarithmic differentiation, x/x ˙ = X/X − N˙ /N . Moreover, [2] implies that ˙ X/X = a N˙ /N, so x/x ˙ = (a − 1)N˙ /N = (a − 1)[α − β(1/x)]. It follows that the differential equation
for x is x˙ = (a − 1)αx − (a − 1)β. The solution of this equation is x(t) = [x(0) − β/α]eα(a−1)t + β/α. Then [2] and x = X/N together imply that N (t) = [x(t)/A]1/(a−1) , X(t) = A[N (t)]a . For 0 < a < 1, x(t) → β/α, N(t) → (β/αA)1/(a−1) , and X(t) → A(β/αA)a/(a−1) as t → ∞.
21.6 2. Formula [21.12] yields
t
x=
s(1 + s 2 )e
t s
2ξ dξ
t
ds =
0
s(1 + s 2 )et
2 −s 2
ds
0 2
By substituting u = −t 2 , then integrating by parts, we eventually derive the solution x(t) = et −1− 21 t 2 . It follows that limt→∞ x(t) does not exist because x(t) → ∞. 4. Multiply the given equation by x −n to obtain x −n x˙ = Q(t)x 1−n + R(t). From z = x 1−n we find 1 z˙ = (1 − n)x −n x, ˙ so the new equation is 1−n z˙ = Q(t)z + R(t), which is a linear differential equation in the unknown function z. 6. Using the technique suggested by Problem 4 yields the solution 1/(1−b) K = Ce−αδ(1−b)t + αAna0 (1 − b)e(av+ε)t /(av + ε + αδ(1 − b))
21.7 2. (a) x˙ = (x − 1)(x + 1)2 . Here x = 1 is unstable, whereas x = −1 is neither stable nor unstable. (It is stable on the right, but unstable on the left.) See Fig. 21.7.2(a). (b) No equilibrium states. See Fig. 21.7.2(b). (c) The only equilibrium state x = 0 is unstable. See Fig. 21.7.2(c).
2
x˙
x˙
x˙
2
2
2
1
1
1
2
x
1 2
Figure 21.7.2(a)
© Knut Sydsæter and Peter Hammond 2010
1
1 1 2
Figure 21.7.2(b)
x
3 2
1 1 2
Figure 21.7.2(c)
x
CHAPTER 21
DIFFERENTIAL EQUATIONS
65
4. (a) x(t) = (1 + Aet )/(1 − Aet ), where A = (x0 − 1)/(x0 + 1). Another solution is x(t) = −1 for all t. For x0 < 1, x(t) → −1 as t → ∞. For x0 > 1, which occurs when 0 < A < 1, one has x(t) → ∞ as t → (− ln A)− , and x(t) → −∞ as t → (− ln A)+ . For x0 = 1, x(t) ≡ 1. See Fig. 21.7.4(a) for some integral curves. (b) x = −1 is stable; x = 1 is unstable. See Fig. 21.7.4(b). x˙
x 2 x≡1 1 t x ≡ −1 −2
Figure 21.7.4(a)
−1
1
2
x
Figure 21.7.4(b)
21.8 2. (a) It is easy to verify by direct differentiation that u1 = et and u2 = tet both satisfy x¨ − 2x˙ + x = 0. If tet = ket for all t, then t = k for all t, which is absurd. The general solution is therefore x(t) = Aet +Btet , where A and B are arbitrary constants. (b) A particular solution of the equation is obviously u∗ (t) = 3, so the general solution of the nonhomogeneous equation is x(t) = Aet + Btet + 3. 4. By direct differentiation, u1 = eat and u2 = eat/(1−α) are easily seen to be solutions. They are not proportional, so the general solution is x(t) = Aeat + Beat/(1−α) , where A and B are arbitrary constants.
21.9 2. (a) x = C1 et + C2 e−t − 21 sin t. Not stable. 2 3 (c) x = Ae5t + Bte5t + 75 t + 125 . Not stable. 4. x = Ae2t + Be7t/4 +
(b) x = C1 et + C2 e−t −
1 −t 2 te .
Not stable.
1 1 15 t + 2 + (2 sin t + 3 cos t) 14 14 65
6. (a) If x = tf (1/t), then x˙ = f (1/t) − (1/t)f (1/t) and x¨ = (1/t 3 )f (1/t). But then x¨ + t −n−2 x = (1/t 3 )f (1/t) + t −n−2 tf (1/t) = (1/t 3 )[f (1/t) + t −n+2 f (1/t)] = 0, because when x = f (t) satisfies x¨ + t n−2 x = 0, then f (u) + un−2 f (u) = 0 at u = 1/t in particular, and so f (1/t) + t −n+2 f (1/t) = 0. (b) Let n = 2. Then x¨ + t n−2 x = x¨ + x = 0, with general solution x = A sin t + B cos t. From (a), it follows that x¨ +t −n−2 x = x¨ +t −4 x = 0, or t 4 x¨ +x = 0, has the solution x = t[A sin(1/t)+B cos(1/t)]. √ 8. For λ = γ (a − α) > 0 the solution is p(t) = C1 ert + C2 e−rt − k/λ, where r = λ; for λ = 0 the solution is p(t) = C1 t + C2 + 21 kt 2 ; for λ < 0 the solution is p(t) = C1 cos st + C2 sin st − k/λ, where √ s = −λ . In no case is the solution ever stable. 10. If x = uert , then x˙ = ert (u˙ + ru) and x¨ = ert [u¨ + 2r u˙ + r 2 u]. Hence, x¨ + a x˙ + bx = ert [u¨ + (2r + a)u˙ + (r 2 + ar + b)] = ert u¨ because r = −a/2 and r satisfies the characteristic equation. We conclude that x = uert satisfies the given equation provided u¨ = 0. Then u = At + B, and the conclusion follows. © Knut Sydsæter and Peter Hammond 2010
66
APPENDIX B
SUMS, PRODUCTS, AND INDUCTION
Appendix B Sums, Products, and Induction B.1
√ √ √ √ √ √ √ √ √ 2. (a) 2 0 + 2 1 + 2 2 + 2 3 + 2 4 = 2(1 + 2 + 3 + 2) = 2(3 + 2 + 3) (b) (x + 0)2 + (x + 2)2 + (x + 4)2 + (x + 6)2 = 4(x 2 + 6x + 14) (c) a1i b2 + a2i b3 + a3i b4 + · · · + ani bn+1 (d) f (x0 )x0 + f (x1 )x1 + f (x2 )x2 + · · · + f (xm )xm 6 + 15 + 28 49 2·3+3·5+4·7 · 100 = · 100 = · 100 ≈ 144.12 1·3+2·5+3·7 3 + 10 + 21 34 6. (a) The total number of people moving from region i. (b) The total number of people moving to region j .
4.
B.2 2. (a + b)6 = a 6 + 6a 5 b + 15a 4 b2 + 20a 3 b3 + 15a 2 b4 + 6ab5 + b5 . (The coefficients are those in the sixth row of Pascal’s triangle in the text.) 8 8·7·6 8 8 4. (a) = = 56. Also, = = 56; 3 1 · 2 · 3 8 − 3 5 9·8·7·6 8 8 8·7·6·5 8+1 9 + = 56 + = 126 and = = = 126. 3 3+1 1·2·3·4 3+1 4 1·2·3·4 m m(m − 1) · · · (m − k + 1) m! m (b) = = = and k! (m − k)!k! m−k k m m m! m! m!(k + 1 + m − k) (m + 1)! + = + = = = (m − k)!k! (m − k − 1)!(k + 1)! (m − k)!(k + 1)! (m − k)!(k + 1)! k k + 1 m+1 k+1 6.
n−1 "
(a + id) = a + (a + d) + · · · + [a + (n − 2)d] + [a + (n − 1)d]
i=0
= [a + (n − 1)d] + [a + (n − 2)d] + · · · + (a + d) + a = 21 [2a + (n − 1)d]n = na + An alternative proof is this: n−1 n−1 n−1 " " " 1 n(n − 1)d (a + id) = a+d i = na + [1 + (n − 1)](n − 1) = na + . 2 2 i=0
i=0
n(n−1)d . 2
i=0
B.3 2.
i
sum of all the i numbers in the ith row, so in the first double sum we sum all the sums in j =1 aij is the the m rows. m i=j aij is the sum of all the m − j + 1 numbers in the j th column, so in the second double sum we sum all the sums in the m columns.
B.4 2. (a)
n % k=1 n %
% 4k 2 2k 2k 4 · 12 4 · 22 4n2 = = · · · · 2k − 1 2k + 1 4k 2 − 1 4 · 12 − 1 4 · 2 2 − 1 4n2 − 1 n
k=1
n−1 n n−1 " " % ai a1 a2 an (b) = ··· (c) a s bi = ai+1 ai+2 · · · an bi = b b1 b2 bn i=1 i i=1 s=i+1 i=1 (a2 a3 · · · an )b1 + (a3 a4 · · · an )b2 + · · · + an bn−1
© Knut Sydsæter and Peter Hammond 2010
APPENDIX C
TRIGONOMETRIC FUNCTIONS
67
B.5 2. We prove only [B.7]; the proof of [B.6] is very similar, but slightly easier. For n = 1 the LHS and the RHS of [B.7] are both equal to 1. Suppose [B.7] is true for n = k, so that & ' k " k(k + 1) 2 3 3 3 3 3 i = 1 + 2 + 3 + ··· + k = 2 i=1
Then k+1 " i=1
&
k(k + 1) i = 1 + 2 + 3 + · · · + k + (k + 1) = 2 3
3
3
3
3
3
'2 + (k + 1)3 = (k + 1)2 ( 41 k 2 + k + 1)
(k + 1)2 (k 2 + 4k + 4) (k + 1)(k + 2) 2 , which proves that [B.7] = But this last expression is equal to 4 2 is true for n = k + 1. By induction, we have proved [B.7]. 4. The claim is true for n = 1. Suppose k 3 + (k + 1)3 + (k + 2)3 is divisible by 9. Then (k + 1)3 + (k + 2)3 + (k + 3)3 = (k + 1)3 + (k + 2)3 + k 3 + 9k 2 + 27k + 27 = k 3 + (k + 1)3 + (k + 2)3 + 9(k 2 + 3k + 3) is divisible by 9, because the first three terms are divisible by 9 by the induction hypothesis, whereas the last term is also obviously divisible by 9. 6. s1 = 41, s2 = 43, s3 = 47, s4 = 53, s5 = 61 are all prime. For n = 41, we have n2 − n + 41 = (41)2 , which is not prime!
Appendix C Trigonometric Functions C.1 2. Look at Fig. C.1 on page 866 of the text. If x is changed to −x, the point P−x will have coordinates (u, −v), so sin(−x) = − sin x, and cos(−x) = cos x. Then by definition [C.2], tan(−x) = sin(−x)/ cos(−x) = − sin x/ cos x = − tan x. 4. cos(y − π/2) = sin y follows directly from [C.8]. Then, from the hints in the question, as well as [C.8] and the result of Problem 2, it follows that sin(x + y) = cos[x + (y − π/2)] = cos x cos[−(y − π/2)] + sin x sin[−(y − π/2)] = sin x cos y + cos x sin y Also, using the result of Problem 2 again, this last equation implies that sin(x − y) = sin x cos(−y) + cos x sin(−y) = sin x cos y − cos x sin y 6. (a) 1/2. (Either draw√a figure similar C.3 on page√868, or use the formula for sin(x − y) in √ to Figure √ Problem 4.) (b) − 3/2 (c) − 2/2 √ (d) − √ 2/2 (e) 3/3 (f) sin π/12 = sin(π/3 − π/4) = sin π/3 cos π/4 − cos π/3 sin π/4 = 41 ( 6 − 2). 8. Note that x + y = A and x − y = B imply x = 21 (A + B) and y = 21 (A − B). The claimed formula then follows easily from the hint. © Knut Sydsæter and Peter Hammond 2010
68
APPENDIX C
TRIGONOMETRIC FUNCTIONS
10. (a) See Fig. C.1.10(a). Period π, amplitude 1. (b) See Fig. C.1.10(b). Period 4π , amplitude 3. (c) See Fig. C.1.10(c). Period 2π/3, amplitude 2. y
y
y
3 2
1 −2π
π −π
x
4
1
4π
−1
Figure C.1.10(a)
−π
Figure C.1.10(b)
12. (a) y = 2 sin 41 x
(b) y = 2 + cos x
2
x
−1 −2 −3
π
2π x
Figure C.1.10(c)
(b) y = 2e−x/π cos x
14. The given equality is (cos x − cos y)2 + (sin x − sin y)2 = 1 + 1 − 2 cos(x − y). By expanding and using [C.7], we easily find formula [C.8].
C.2 sin x cos x cos x − sin x(− sin x) cos2 x + sin2 x gives y = = = 1 + tan2 x. Since cos x cos2 x cos2 x cos2 x + sin2 x = 1, we get y = 1/ cos2 x as an alternative answer.
2. y = tan x =
4. (a) a sin ax
(b) a sin bt + abt cos bt
(c) −a cos(at + b) sin[sin(at + b)] cos{cos[sin(at + b)]}
6. f (x) = 3(sin x − x − 1)2 (cos x − 1). In the open interval I = (0, 3π/2), f (x) = 0 only at cos x = 1, when x = π/2. (It is easy to see that sin x < x + 1 for all x ≥ 0 because the function defined by g(x) = x + 1 − sin x for all x ≥ 0 satisfies g(0) = 1 and g (x) = 1 − cos x ≥ 0 for all x ≥ 0.) Thus, the only possible extreme points are 0, π/2, and 3π/2. Comparing the function values at these points, we find that f (x) has its maximum −1 at x = 0, and its minimum −(2 + 3π/2)3 at x = 3π/2. (The extreme-value theorem ensures that extreme points do exist.) 8. Implicit differentiation yields 1 · cos y + x(− sin y)y − y sin x − y cos x = 0, so y =
cos y − y cos x . sin x + x sin y
At (π, π/2), y = 1/2, so the equation for the tangent is y = x/2. π/2 π/2 10. (a) − cos x + C (b) 0 cos x dx= 0 sin x = sin(π/2) − 0 = 1 (c) by parts yields Integrating 2 2 I = sin x dx = sin x(− cos x)− cos x(− cos x) dx = − sin x cos x + cos x dx = − sin x cos x + 1 2 (1−sin x) dx. Hence, π I = − sin π Solving for I gives I = − 2 (sin x cos x −x)+C1 . π πx cos x +x −I +C. (d) 0 x cos x dx = 0 x sin x − 0 sin x dx = 0 + 0 cos x = cos π − cos 0 = −2. 12. (a) π/4
(b) π/2
(c) π/6
(d) π/3. (You can read all these values off from Table C.1.)
14. Make use of [C.20] and the chain rule to derive y =
C.3 2. See Fig. C.3.2. © Knut Sydsæter and Peter Hammond 2010
1+
1 ex − e 2
· −x 2
ex + e−x . Then simplify. 2
69 √ 4. (a) 2√ 3(cos π/3 + i sin π/3) (b) cos π + i sin π (d) 2(cos 7π/4 + i sin 7π/4)
(c) 4(cos 4π/3 + i sin 4π/3)
Imaginary axis 4i 3i
w = 1 + 3i
2i z+w =3+i
i −1
1
3
−i −2i
Figure C.3.2
© Knut Sydsæter and Peter Hammond 2010
2
z = 2 − 2i
4
Real axis
70 TEST I (Elementary Algebra) A certain familiarity with elementary algebra is an essential prerequisite for reading the textbook (and for understanding most modern economics texts). This test is designed for students and instructors to discover whether the students have the proper background. (In a number of countries, many beginning economics students’ background in elementary algebra appears to have become much weaker during the last few years. In fact, lecturers using this test (or similar ones) have been shocked by the results, and have had to readjust their courses.) At the head of each problem, immediately after the number, the relevant sections of the introductory chapters in the book are given in parentheses, followed in square brackets by the number of points for a correct answer to each separate part of the problem. In a 20–30 minute test, any student who scores less than 50 (out of 100) has serious problems with elementary algebra. Such students definitely need to review the relevant section of Chapters 1 and 2, or consult other elementary material. The correct answers are given on a separate page following the test.
1
(1.2) [Points: (a) 2; (b) 2; (c) 3; (d) 3] Calculate/simplify: 7 3 · 72 −2 219 − 217 −2 −2 3 (a) (b) (5.5 − 3.5) (c) (d) 74 5 5 5 219 + 217
2
(1.2–1.4) [Points: (a) 2; (b) 2; (c) 2; (d) 4]
√ (c) 132 − 122 = ? √(b) 11 √% of 3500 is? 3+ 2 (d) Rationalize the denominator of √ √ (i.e. find a new fraction that is equal but has no square 3− 2 root in the denominator).
(a) If 2x 2 y = 5, then 4x 4 y 2 = ?
3
(1.3) [Points: (a) 2; (b) 2; (c) 2] Expand: (a) (x + 2y)2
4
(b) (2x − 3y)2
(c) (a + b)(a − b)
(1.3) [Points: (a) 2; (b) 3; (c) 3; (d) 4] Expand and simplify: (a) 5a − (3a + 2b) − 2(a − 3b) (c) (1 − x)2 (1 + x)2
(b) (x + 2)2 + (x − 2)2 − 2(x + 2)(x − 2) (d) (2 − a)3
5
(1.2) [Points: 4] If the GNP of a certain country in 2000 was 8 billion dollars, write down an expression for the GNP 6 years later if it increases by 5% each year.
6
(1.3) [Points: (a) 3; (b) 3; (c) 4] Factorize: (a) 5a 2 b + 15ab2
(b) 9 − z2
© Knut Sydsæter and Peter Hammond 2010
(c) p 3 q − 4p 2 q 2 + 4pq 3
71 7
(1.4) [Points: (a) 2; (b) 2; (c) 2] Expand and simplify to a single fraction:
(a)
1 1 − 2 3
(b)
6a a 3a − + 5 10 20
1 1 − 2 3 (c) 1 1 − 4 6
8
(1.5) [Points: (a) 2; (b) 2; (c) 2; (d) 2] Calculate/simplify: √ 3 (a) 251/2 (b) (x 1/2 y −1/4 )4 (c) 27a 6 (d) p 1/5 (p 4/5 − p −1/5 )
9
(2.1) [Points: (a) 2; (b) 2; (c) 2] Solve the following equations for the unknown x: √ 3 1 3 (a) x = −6 (b) = (c) 3 − x = 2 5 x−1 2x + 3
10
(1.6) [Points: (a) 2; (b) 3; (c) 3] Solve the following inequalities: x−1 (a) −3x + 2 < 5 (b) ≤0 (c) x 3 < x x+3
11 (2.3) [Points: (a) 3; (b) 3; (c) 3] Solve the following equations: (a) 3x − 9x 2 = 0
12
(b) x 2 − 2x − 15 = 0
(c) 2P 2 = 2 − 3P
(2.4) [Points: (a) 3; (b) 4; (c) 4] Solve the following systems of equations: 3 3 + =3 p q 1.5p − 0.5q = 14 2x − y = 5 (a) (b) (c) 3 1 x + 2y = 5 2.5p + 1.5q = 28 − =7 p q
© Knut Sydsæter and Peter Hammond 2010
72 TEST II (Elementary Mathematics) Students who now enter university or college courses in economics tend to have a wide range of mathematical backgrounds and aptitudes. At the low end, they may have no more than a shaky command of elementary algebra. Or, at the high end, they may already have a ready facility with calculus, though often it is some years since economics students took their last formal mathematics course. Experience suggests therefore that right from the start of the course, it is very important that the instructor, as well as each individual student, should get some impression of what the student knows well, what is vaguely familiar, and what seems to be more or less forgotten or perhaps never learned at all. The present test is meant to test the students’ actual knowledge of some elementary mathematical topics of interest to economists. The level is nevertheless more advanced than for Test I and the topics covered are discussed in the earlier chapters of the main text. The maximum total score is 100. 1
[Points: (a) 2; (b) 2; (c) 2; (d) 2] (b) 3−15 + 3−15 + 3−15 = 3y ⇒ y =
(a) 25 + 25 = 2x ⇒ x = (c) 132 − 52 = 2
(d)
226 − 223 z = ⇒ z = 226 + 223 9
[Points: (a) 1+1; (b) 2+2] (a) Find the slopes of the following straight lines: (i) y = − 23 x + 4
(ii) 6x − 3y = 5
(b) Find the equation of the straight line that: (i) passes through (−2, 3) and has slope −2. (ii) passes through both (a, 0) and (0, b). 3
[Points: 5] Fill in the following table and sketch the graph of y = −x 2 + 2x + 4. −2
x
−1
0
1
2
3
y = −x 2 + 2x + 4
4
[Points: 4+4] Determine the maximum/minimum points for: (a) y = x 2 − 4x + 8
5
(b) y = −2x 2 + 16x − 14
[Points: 4+4] Perform the following polynomial divisions: (a) (2x 3 − 3x + 10) ÷ (x + 2)
(b) (x 4 + x) ÷ (x 2 − 1)
6
[Points: 5] f (x) = 43 x 3 − 15 x 5 . For what values of x is f (x) = 0?
7
[Points: 3+3+3+3] Sketch the graph of a function f in each of the following cases: (a) f (x) > 0 and f (x) > 0,
(b) f (x) > 0 and f (x) < 0
(c) f (x) < 0 and f (x) > 0,
(d) f (x) < 0 and f (x) < 0
© Knut Sydsæter and Peter Hammond 2010
4
73 8
[Points: 3+3] (a) The cost in dollars of extracting T tons of a mineral ore is given by C = f (T ). Give an economic interpretation of the statement that f (1000) = 50. (b) A consumer wants to buy a certain item at the lowest possible price. Let P (t) denote the lowest price found after searching the market for t hours. What are the likely signs of P (t) and P (t)?
9
[Points: 2+2+2+2] Write down the general rules for differentiating the following: (a) y = f (x) + g(x) ⇒ y =
(b) y = f (x)g(x) ⇒ y =
(c) y = f (x)/g(x) ⇒ y =
(d) y = f (g(x)) ⇒ y =
10 [Points: 2+2+2+2+2+2+2+2] Differentiate the following functions: x (a) y = x 2 (b) y = x 5 /5 (c) y = (d) y = (x 2 + 5)6 x+1 (e) y = ex (f) y = ln x (g) y = 2x (h) y = x x 11
[Points: 2+2+2+2+2] Which of the following statements are correct? (a) The rule which converts a temperature measured in degrees Fahrenheit into the same temperature measured in degrees Celsius is an invertible function. (b) A concave function always has a maximum. (c) A differentiable function can only have an interior maximum at a stationary point for the function. (d) If f (a) = 0, then a is either a local maximum point or a local minimum point. (e) The conditions f (a) = 0 and f (a) < 0 are necessary and sufficient for a to be a local maximum point for f .
12
[Points: 2+2+2+2] In each of the following cases, decide whether the given formula is correct or not: 1 (a) x 2 dx = x 3 + C 3 (b) [f (x) + g(x)] dx = f (x) dx + g(x) dx (c) f (x)g(x) dx = f (x) dx g(x) dx b (d) x dx = b − a a
© Knut Sydsæter and Peter Hammond 2010
74 Answers to TEST I 73+2 75 7 3 · 72 = 4 = 4 = 75−4 = 71 = 7 (b) (5.5 − 3.5)3 = 23 = 8 4 7 7 7 −2 −2 −8 219 − 217 3 217 (22 − 1) −2 = (= −0.064) (d) 19 = = (c) 5 5 5 125 2 + 217 217 (22 + 1) 5
1. (a)
2. (a) 4x 4 y 2 = (2x 2 y)2 = 25 (b) 11 % of 3500 is 3500 · 11/100 = 3500 · 0.11 = 385 √ √ √ √ √ (c)√ 132 − 122 √ = (13√ + 12)(13 − 12) = 25 = 5 (or 132 − 122 = 169 − 144, etc.) (“ 132 − 122 = 132 − 122 = 13 − 12 = 1” is a SERIOUS mistake.) √ √ √ √ √ √ √ √ √ 3+ 2 ( 3 + 2)( 3 + 2) 3+2 3 2+2 (d) √ =5+2 6 √ = √ √ √ √ = 3−2 3− 2 ( 3 − 2)( 3 + 2) 3. (a) (x + 2y)2 = x 2 + 4xy + 4y 2 (c) (a + b)(a − b) = a 2 − b2
(b) (2x − 3y)2 = 4x 2 − 12xy + 9y 2
4. (a) 5a − (3a + 2b) − 2(a − 3b) = 5a − 3a − 2b − 2a + 6b = 4b (b) (x + 2)2 + (x − 2)2 − 2(x + 2)(x − 2) = [(x + 2) − (x − 2)]2 = 42 = 16 (c) (1 − x)2 (1 + x)2 = [(1 − x)(1 + x)]2 = (1 − x 2 )2 = 1 − 2x 2 + x 4 (d) (2 − a)3 = (2 − a)2 (2 − a) = (4 − 4a + a 2 )(2 − a) = 8 − 4a − 8a + 4a 2 + 2a 2 − a 3 = −a 3 + 6a 2 − 12a + 8 5. 8(1.05)6 billion dollars. 6. (a) 5a 2 b + 15ab2 = 5ab(a + 3b) (b) 9 − z2 = (3 − z)(3 + z) (c) p 3 q − 4p 2 q 2 + 4pq 3 = pq(p 2 − 4pq + 4q 2 ) = pq(p − 2q)2 . (One point for the first equality.) 1 1 3 2 3 2 1 7. (a) − = − = − = (b) 20 is the common denominator, so 2 3 2·3 3·2 6 6 6 6a a 3a 4 · 6a 2a 3a 24a − 2a + 3a 25a 5a − + = − + = = = 5 10 20 20 20 20 20 20 4 1 1 6 4 2 − − 12 = 12 = 2 (or 21 − 13 = 2( 41 − 16 ), so the ratio is 2). (c) 21 13 = 12 3 2 1 − − 4 12 12 6 12 √ √ √ 3 3 1/2 1/2 8. (a) 25 = 5 (b) (x y −1/4 )4 = x (1/2)·4 y (−1/4)·4 = x 2 y −1 (c) 27a 6 = 3 27 a 6 = 3a 2 (d) p 1/5 (p 4/5 − p −1/5 ) = p1/5 p 4/5 − p 1/5 p −1/5 = p1/5+4/5 − p 1/5−1/5 = p − 1 9. (a) 3x = −30, so x = −10 (b) 2x + 3 = 3(x − 1), so x = 6. (Neither denominator is 0 when x = 6.) √ (c) If 3 − x = 2, then 3 − x = 4, so x = −1. This is indeed a solution, as is easily checked. 10. (a) −3x + 2 < 5 or −3x < 3, so that x > −1. (Recall that an inequality is reversed if multiplied by a negative number.) (b) −3 < x ≤ 1. (Use a sign diagram. Note that the fraction is undefined if x = −3.) (c) x < −1 or 0 < x < 1. (x 3 < x, or x 3 − x < 0, and so x(x 2 − 1) < 0, or x(x − 1)(x + 1) < 0. Then use a sign diagram.) 11. (a) 3x(1 − 3x) = 0, so x = 0 or x = 1/3
(b) x = −3, x = 5
(c) P = −2 or P = 1/2
12. (a) x = 3, y = 1 (b) p = 10, q = 2 (c) Put x = 1/p, y = 1/q. Then 3x + 3y = 3 and 3x − y = 7. The solution to this system is x = 2 and y = −1. But then p = 1/2 and q = −1. © Knut Sydsæter and Peter Hammond 2010
75 Answers to TEST II 1. (a) 25 + 25 = 2 · 25 = 26 , so x = 6. (b) 3−15 + 3−15 + 3−15 = 3 · 3−15 = 3−14 , so y = −14. √ √ √ (c) 132 − 52 = 169 − 25 = 144 = 12. 223 (23 − 1) 23 − 1 7 226 − 223 = = = , so z = 7. (d) 26 2 + 223 223 (23 + 1) 23 + 1 9 2. (a) (i) −3/2 (ii) 2 (b) (i) y = −2x − 1 (ii) y = (−b/a)x + b 3. x
−2
−1
0
1
2
3
4
y = −x 2 + 2x + 4
−4
1
4
5
4
1
−4
The graph of the function is shown in the figure below. y 5 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4
y = −x 2 + 2x + 4
1
2
3
4
x
4. (a) Minimum 4 for x = 2. (Follows from x 2 − 4x + 8 = x 2 − 4x + 22 + 8 − 22 = (x − 2)2 + 4, or by using calculus.) (b) Maximum 18 for x = 4. (Follows from −2x 2 + 16x − 14 = −2(x 2 − 8x + 7) = −2(x 2 − 8x + 42 + 7 − 42 ) = −2[(x − 4)2 − 9] = −2(x − 4)2 + 18, or by using calculus.) 2x 3 − 3x + 10 = 2x 2 − 4x + 5 x+2 x+1 1 x4 + x = x2 + 1 + 2 = x2 + 1 + (b) 2 x −1 x −1 x−1
5. (a)
6. f (x) = 4x 2 − x 4 = x 2 (4 − x 2 ) = x 2 (2 − x)(2 + x) = 0 for x = 0 and for x = ±2. 7.
y
y
y
y y = f (x)
y = f (x)
y = f (x)
x
(a) © Knut Sydsæter and Peter Hammond 2010
y = f (x)
x
(b)
x
(c)
x
(d)
76 8. (a) The marginal cost is 50 when output is 1000 tons. (Or: The cost of extracting one ton more than 1000 tons is approximately 50 dollars.) (b) P (t) ≤ 0, because more search never leads to a higher best price, and generally leads to a lower one. However, P (t) is likely to be positive, because the longer you search the smaller gain in price you will obtain. (P (t) is decreasing and convex.) 9. (a) y = f (x) + g(x) ⇒ y = f (x) + g (x) (b) y = f (x)g(x) ⇒ y = f (x)g(x) + f (x)g (x) f (x)g(x) − f (x)g (x) (c) y = f (x)/g(x) ⇒ y = (g(x))2 (d) y = f (g(x)) ⇒ y = f (g(x))g (x) 10. (a) y = x 2 ⇒ y = 2x
(b) y =
1 5 5x
⇒ y = x4
(d) y = (x 2 + 5)6 ⇒ y = 6(x 2 + 5)5 2x = 12x(x 2 + 5)5 (f) y = ln x ⇒ y = 1/x
(g) y = 2x ⇒ y = 2x ln 2
x 1 ⇒ y = x+1 (x + 1)2 (e) y = ex ⇒ y = ex
(c) y =
(h) y = x x ⇒ y = x x (ln x + 1)
11. (a) Correct. (b) Wrong. (Consider y = −ex .) (c) Correct. (d) Wrong. (a could be an inflection point.) (e) Wrong. (The conditions are sufficient, but not necessary. For example, f (x) = −x 4 has a maximum at x = 0 and yet f (0) = f (0) = 0.) d 12. (a) Correct. (b) Correct. (c) Wrong, because f (x) dx g(x) dx dx = f (x) g(x) dx + g(x) f (x) dx = f (x)g(x), except for very special functions f and g. b b (d) Wrong, because a x dx = a 21 x 2 = 21 (b2 − a 2 ), which equals b − a only when a + b = 2 or a = b.
© Knut Sydsæter and Peter Hammond 2010
