Paolo Pin
[email protected] http://venus.unive.it/pin
August, 2005
Rangaraian K. Sundaran, 1996, “A First Course in Optimization Theory”, Cambridge University Press:
Solutions of (some) exercises Appendix C (pp. 330-347): Structures on Vector Spaces. 1.
= =
1 1 kx + yk2 − kx − yk2 = < x + y, x + y > − < x − y, x − y > 4 4 1 < x, x > +2 < x, y > + < y, y > −(< x, x > −2 < x, y > + < y, y >) 4 1 4 < x, y > = < x, y > 2 4
2.
kx + yk2 + kx − yk2 = kx + yk2 − kx − yk2 + 2kx − yk2
Polarization Identity =⇒ = 4 < x, y > +2kx − yk2
= 2 < x, x > +2 < y, y > = 2 kxk2 + kyk2 2
3. Let’s take e.g. R2 , ~x = (1, 0) and ~y = (0, 2): kx + yk21 + kx − yk21 = 32 + 32 = 18 6= 10 = 2 · (12 + 22 ) = 2 kxk21 + kyk21 2
1
4. a) {xn } → x in (V, d1 ) if and only if: lim d1 (xn , x) = 0 =⇒ lim d2 (xn , x) ≤ lim c1 · d1 (xn , x) = c1 · lim d1 (xn , x) = 0
n→∞
n→∞
n→∞
n→∞
since d2 (xn , x) ≥ 0, ∀ n: limn→∞ d2 (xn , x) = 0 ; b) A is open in (V, d1 ) if and only if: ∀ x ∈ A, ∃ r > 0 s.t. B1 (x, r) = {y ∈ V | d1 (x, y) < r} ⊂ A =⇒ ∀ x ∈ A, ∃
r r > 0 s.t. B2 (x, ) = {y ∈ V | c2 · d2 (x, y) ≤ r} ⊆ B1 (x, r) ⊂ A c2 c2
the reverse of (a) and (b) is equivalent.
2
5. Consider R with the usual metric d(x, y) ≡ |x − y| and with the other one defined as δ(x, y) ≡ max{d(x, y), 1}; δ is a metric, to prove triangle inequality consider that: δ(x, z) = max{d(x, z), 1} ≤ max{d(x, y) + d(x, z), 1}
≤ max{d(x, y) + d(x, z), 1 + d(x, y), 1 + d(y, z), 2}
= max{d(x, y), 1} + max{d(x, z), 1} = δ(x, y) + δ(y, z)
∀ Bδ (x, r) ⊂ R, Bd (x, r) ⊆ Bδ (x, r), similarly ∀ Bd (x, r) ⊂ R, Bδ (x, min{r, 12 }) ⊆ Bd (x, r), so they generate the same open sets; nevertheless 6 ∃ c ∈ R+ such that d(x, y) < c · δ(x, y), ∀ x, y ∈ R, suppose by absurd it exists, then d(0, c) = c < c · δ(x, y) is impossible since δ(x, y) ≤ 1, ∀ x, y ∈ R.
2
6. d∞ (~x, ~y ) = max{|x1 − y1 |, |x2 − y2 |} is equivalent to any 1 p p p , with p ∈ N, p ≤ 1, because: dp (~x, ~y ) = |x1 − y1 | + |x2 − y2 |
1 d∞ ≤ dp ∧ dp ≤ d∞ 2 all the metrics are equivalent by symmetry and transitivity (see exercise 7). 7. If d1 and d2 are equivalent in V , ∃ b1 , b2 ∈ R such that ∀ x, y ∈ V :
2
2
;
d2 (x, y) ≤ d1 (x, y) ≤ b2 · d2 (x, y) b1 similarly if d2 and d3 are equivalent in V , ∃ c1 , c2 ∈ R such that ∀ x, y ∈ V : d3 (x, y) ≤ d2 (x, y) ≤ c2 · d3 (x, y) c1 hence ∃ (b1 · c1 ), (b2 · c2 ) ∈ R such that ∀ x, y ∈ V : d3 (x, y) ≤ d1 (x, y) ≤ b2 · c2 · d3 (x, y) b1 · c1
=⇒ d1 and d3 are equivalent. 2 8.
ρ(x, y) =
1 x 6= y 0 x=y
Positivity and Simmetry come from the definition; for Triangle inequality consider the exhaustive cases: x = y = z =⇒ d(x, z) = 0 = 0 + 0 = d(x, y) + d(y, z) x = y 6= z =⇒ d(x, z) = 1 = 0 + 1 = d(x, y) + d(y, z) x 6= y = z =⇒ symmetric
x = z 6= y =⇒ d(x, z) = 0 ≤ 1 + 1 = d(x, y) + d(y, z)
all different =⇒ d(x, z) = 1 ≤ 1 + 1 = d(x, y) + d(y, z) 2 9. We generalize from R to any V ; every subset X in (V, ρ) is closed because X C is open (every subset is open in (V, ρ)); in (V, ρ) the only convergent sequences are those constant from a certain point on (not only constant sequences as stated at page 337), consider a finite subset A of V , every sequence in A have a constant (hence converging) subsequence, consider now an infinite subset B of V , (by the Axiom of choice) we can construct a sequence with all different elements, where all subsequence will have all different elements, =⇒ compact subsets in (V, ρ) are all and only the finite ones. 2 10. Since that every subset X in (V, ρ) is closed (exercise 9), then cl(X) = X, ∀ X ⊆ V , and ∀ X ⊆ V, 6 ∃A ⊂ X s.t. A 6= X and cl(A) = X
3
neither for V itself, so that if V is uncountable, there exists no countable subset of V whose closure is V . 2 11. The correct statement of Corollary C.23 (page 340) is: A function f : V1 → V2 is continuous if and only if, ∀ open set U2 ⊆ V2 , f −1 (U2 ) = {x ∈ V1 |f (x) ∈ U2 } is an open set in V1 . (The inverse function of every open set is an open set.) Since every subset of (R, ρ) is open, every function (R, ρ) → (R, d) and (R, ρ) → (R, ρ) is continuous, that is F = G = {f | f : R → R}; clearly not every function (R, d) → (R, d) is continuous (here continuity has its standard meaning), then H ⊂ F = G and H = 6 F. 2 12. See exercise 11 for the correct statement of Corollary C.23 (page 340); the open sets in (Rn , d∞ ) and (Rn , d2 ) coincide, because: d∞ ∀ Bx,r d2 ∃ Bx,r
= {y ∈ Rn | max{|yi − xi |}i∈{1,...n} ≤ r} n X 1 d∞ n = {y ∈ R | ( (yi − xi )2 ) 2 ≤ r} ⊆ Bx,r i=1
and
∀ ∃q≡
r
d2 Bx,r
r2 d∞ s.t. Bx,q n
n X 1 = {y ∈ R | ( (yi − xi )2 ) 2 ≤ r} n
i=1
d2 = {y ∈ Rn | max{|yi − xi |}i∈{1,...n} ≤ q} ⊆ Bx,r
;
hence also the continuous functions in (Rn , d∞ ) → (R, d) and the continuous functions in (Rn , d2 ) → (R, d) coincide. 2 13. [The p-adic valuation here is not well defined: if r = 0, a = 0 and any n would do; if r = 1 there are no a, b, n such that 1 = p−n ab , a is not a multiple of p. For a survey: http://en.wikipedia.org/wiki/P-adic_number ] 14. Since V with the discrete topology is metrizable from (V, ρ), topological compactness is equivalent to compactness in the sequential sense (page 343), hence exercise 9 proves that the compact subsets in (V, ρ) are all and only the finite ones. 2
4
15. As seen in exercise 11, every such function is continuous. 2 16. 1 2 d[(x, y), (x′ , y ′ )] = d1 (x, y)2 + d2 (x′ , y ′ )2
positivity and symmetry are straightforward from positivity of d1 and d2 ; triangle inequality: 1 2 d1 (x, z)2 + d2 (x′ , z ′ )2 1 2 ≤ (d1 (x, y) + d1 (y, z))2 + d2 (x′ , z ′ )2 1 2 ≤ (d1 (x, y) + d1 (y, z))2 + (d2 (x′ , y ′ ) + d2 (y ′ , z ′ ))2
d[(x, z), (x′ , z ′ )] =
= = ≤
=
=
1 2 d1 (x, y)2 + 2d1 (x, y)d1 (y, z) + d1 (y, z)2 + d2 (x′ , y ′ )2 + 2d2 (x′ , y ′ )d2 (y ′ , z ′ ) + d2 (y ′ , z ′ )2 h i 1 2 d1 (x, y)2 + d2 (x′ , y ′ )2 + d1 (y, z)2 + d2 (y ′ , z ′ )2 + 2 d1 (x, y)d1 (y, z) + d2 (x′ , y ′ )d2 (y ′ , z ′ ) h d1 (x, y)2 + d2 (x′ , y ′ )2 + d1 (y, z)2 + d2 (y ′ , z ′ )2 + . . . i 1 2 . . . + 2 d1 (x, y)d1 (y, z) + d1 (x, y)d2 (y ′ , z ′ ) + d2 (x′ , y ′ )d1 (y, z) + d2 (x′ , y ′ )d2 (y ′ , z ′ ) h i 1 2 2 ′ ′ 2 2 ′ ′ 2 ′ ′ ′ ′ d1 (x, y) + d2 (x , y ) + d1 (y, z) + d2 (y , z ) + 2 d1 (x, y) + d2 (x , y ) d1 (y, z) + d2 (y , z ) q √ √ √ since a + b + 2 ab = a + b 1 1 2 2 d1 (x, y)2 + d2 (x′ , y ′ )2 + d1 (y, z)2 + d2 (y ′ , z ′ )2
= d[(x, y), (x′ , y ′ )] + d[(y, z), (y ′ , z ′ )]
. 2
17. The base of a metric space (X, d1 ) is the one of its balls B(x, r), 1 every ball in (X, d1 ) × (Y, d2 ) = (X × Y, (d21 + d22 ) 2 ) is of the form: 1
B((x∗ , y ∗ ), r) = {(x, y) ∈ X × Y | (d1 (x, x∗ )2 + d2 (y, y ∗ )2 ) 2 ≤ r} ⊆ {(x, y) ∈ X × Y | max{d1 (x, x∗ ), d2 (y, y ∗ )} ≤ r}
= B(x∗ , r) × B(y ∗ , r)
for every open set in (X, d1 ) × (Y, d2 ) there is a couple of open sets, one in (X, d1 ) and one in (Y, d2 ), whose product contains them, moreover
5
B(x∗ , r1 ) × B(y ∗ , r2 ) = {(x, y) ∈ X × Y | d1 (x, x∗ ) ≤ r1 ∧ d2 (y, y ∗ ) ≤ r2 } 1
⊆ {(x, y) ∈ X × Y | (d1 (x, x∗ )2 + d2 (y, y ∗ )2 ) 2 ≤ r1 + r2 } = B((x∗ , y ∗ ), r1 + r2 )
then for every couple of open sets in (X, d1 ) and (Y, d2 ), there is an open set in the product space containing their product, then the two topologies coincide. 2 18. [(a) is not hard only for f compact =⇒ f sequentially compact, the other way round and (b) are really non trivial; see e.g.: http://www-history.mcs.st-and.ac.uk/˜john/MT4522/Lectures/L22.html ] 19. f topological continuous =⇒ f sequentially continuous: consider xn → x ∈ V , and an open Of (x) containing f (x) ∈ V ′ , f −1 Of (x) contains x and all of the xn after a certain n0 , then all of the f (xn ), after the same n0 , are in Of (x) , f is sequelntially continuous; f sequentially continuous =⇒ f topological continuous: non trivial. . . 2 20. a) 1) Clearly ∅, [0, 1] ∈ τ ; 2) if O1 , O2 ∈ τ , O1C and O2C are finite, but then also their union O1C ∪ O2C = (O1 ∩ O2 )C is, =⇒ O1 ∩ O2 ∈ τ ; ∀ α ∈ A, 3) if {Oα }α∈A ⊆ τ , OαC is finite T S S but then also their intersection α∈A OαC = ( α∈A Oα )C is, =⇒ α∈A Oα ∈ τ ; b) suppose ∃ x ∈ [0, 1] S with a countable base {Bi }i∈N , since BiC is finite ∀ i ∈ N, i∈N BiC is countable, S since [0, 1] is uncountable, ∃ y 6= x, y ∈ [0, 1] such that y 6∈ i∈N BiC , =⇒ y ∈ Bi ∀ i ∈ N, {y}C contains x but is not contained in any element of its base. 2
6
Chapter 3 (pp. 90-99): Existence of Solutions. 1. As a counterexample consider f : (0, 1) → R such that f (x) = x1 : sup f (x) = limx→0 (1,0)
1 = +∞. 2 x
2. It can be shown (e.g. by construction) that in R (in any completely ordered set) sup and inf of any finite set coincide respectively with max and min. If D ⊂ Rn is finite also f (D) = {x ∈ R | ∃ ~x ∈ D s.t. f (~x) = x} is. The result is implied by the Weierstrass theorem because every finite set A is compact, i.e. every sequence in A have a constant (hence converging) subsequence. 2 3. a) Consider x ¯ = max{D}, which exists because D is compact subset of R, f (¯ x) is the desired maximum, because 6 ∃x ∈ D such that x ≥ x ¯, =⇒ 6 ∃x ∈ D such that f (x) ≥ f (¯ x); b) consider D ≡ {(x, y) ∈ R2+ | x + y = 1} ∈ R2 , D is the closed segment from (0, 1) to (1, 0), hence it is compact, there are no two points in D which are ordered by the ≥ partial ordering, hence every function D → R is nondecreasing, consider: 1 x if x 6= 0 f (x, y) = 0 if x = 0 max f on D is limx→0 f (x, y) = limx→0
1 x
= +∞.
2
4. A finite set is compact, because every sequence have a constant (hence converging) subsequence. Every function from a finite set is continuous, because we are dealing with the discrete topology, and every subset is open (see exercise 11 of Appendix C for the correct statement of Corollary C.23 (page 340)). We can take a subset A of cardinality k in Rn and construct a one-to-one function assigning to every element of A a different element in R. n A compact and convex subset P A ⊂a R cannot have a finite number k ≥ 2 of elements (because otherwise a∈A k is different from every a ∈ A but should be in A). Suppose A is convex and ∃ a, b ∈ A such that f (a) 6= f (b), then f[a,b] (λ) = f (λa + (1 − λ)b) is a restriction of f : A → R but can be also considered as a function f[a,b] : [0, 1] → R. By continuity every element of [f (a), f (b)] ⊂ R must be in the codomain of f[a,b]
7
(the rigorous proof is long, see Th. 1.69, page 60), which is then not finite, so that neither the codomain of f is. 2 5. Consider sup(f ), it cannot be less than 1, if it is 1, then max(f ) = f (0). Suppose sup(f ) > 1, then, by continuity, we can construct a sequence {xn ∈ x s.t. ∀ x > R+ | f (xn ) = sup(f ) − sup(fn )−1 , moreover, since limx→∞ f (x) = 0, ∃¯ x ¯, f (x) < 1. {xn } is then limited in the compact set [0, x¯], hence it must converge to x ˆ. f (ˆ x) is however sup(f ) and then f (ˆ x) = max(f ). f (x) = e−x , restricted to R+ , satisfies the conditions but has no minimum. 2 6. Continuity (see exercise 11 of Appendix C for the correct statement of Corollary C.23 (page 340)) is invariant under composition, i.e. the composition of continuous functions is still continuous. Suppose g : V1 → V2 and f : V2 → V3 are continuous, if A ⊂ V3 is open then, by continuity of f , also f −1 (A) ⊂ V2 is, and, by continuity of g, also g−1 (f −1 (A)) ⊂ V1 is. g−1 ◦ f −1 : V3 → V1 is the inverse function of f ◦ g : V1 → V3 , which is then also continuous. We are in the conditions of the Weierstrass theorem. 2 7. −1 x ≤ 0 g(x) = x x ∈ (0, 1) 1 x≥1
,
f (x) = e−|x|
arg max(f ) = 0, max(f ) = e0 = 1 and also sup(f ◦ g) = 1, which is however never attained. 2 8. min p~ · ~x subject to ~x ∈ D ≡ {~y ∈ Rn+ | u(~y ) ≥ u ¯} s.t. u : Rn+ → R is continuous, p~ ≫ ~0 a) p~ · ~x is a linear, and hence continuous function of ~x, there are two possibilities: (i) u is nondecreasing in ~x, (ii) u is not; (i) D is not compact, we can however consider a utility U and define WU ≡ {~y ∈ Rn+ | u(~y ) ≤ U }, by continuity of u, WU is compact, for U large enough D ∩ WU is not empty and the minimum can be found in it;
8
(ii) for the components of ~x where u is decreasing, maximal utility may be at 0, so D may be compact, if not we can proceed as in (i); b) if u is not continuous we could have no maximum even if it is nondecreasing (see exercise 3); if ∃ pi < 0, and u is nondecreasing in xi , the problem minimizes for xi → +∞. 2 9. max p · g(~x) − w ~ ′ · ~x s.t. ~x ∈ Rn+ with p > 0, g continuous, w ~ ≫ ~0
a solution is not guaranteed because Rn+ is not compact.
2
10. | p~′ ~x ≤ w(H − l), l ≤ H} F (~ p, w) = {(~x, l) ∈ Rn+1 + F (~ p, w) compact =⇒ ~ p ≫ 0 (by absurd): if ∃ pi ≤ 0, the constrain could be satisfied also at the limit xi → +∞, F would not be compact; ~p ≫ 0 =⇒ F (~ p, w) compact: F is closed because inequalities are not strict, l is limited in [0, H], ∀ i ∈ {1, . . . n}, xi is surely limited in [0, w(H−l) ]. pi
2
11. ~ subject to φ ~ ∈ {ψ ~ ∈ RN | p~ · ψ ~ ≤ 0, ~y(φ) ~ ≥ 0} max U (~y (φ)) P ~ = ωs + N φi zis , U : RS → R is continuous and strictly increasing where ys (φ) + i=1 (~y ≫ ~y ′ in RS+ if yi ≥ yi′ ∀i ∈ {1, . . . , S} and at least one inequality is strict). ~ ∈ RN such that p ~ ≤ 0 and Z ′ φ ≫ ~0. Arbitrage: ∃ φ ~·φ A solution exists ⇐⇒ no arbitrage (=⇒) (by absurd: (A =⇒ B) ⇔ (¬B =⇒ ¬A) ) ~ ∈ RN such that p~ · ψ ~ ≤ 0 and Z t ψ ~ ≫ ~0, then any multiple k · ψ ~ of ψ ~ suppose ∃ ψ has the same property,
9
~ increases linearly with k and then also U (~y (k · ψ)) ~ increases with k, y~(k · ψ) ~ max U (~y (k · ψ)) is not bounded above; N ~ ~ (⇐=) PN∀ ψ ∈ R such that p~ · ψ ≤ 0 there are two possibilities: either i=1 ψi zis = 0 ∀ s ∈ {1, . . . S}, P or ∃ s ∈ {1, . . . S} such that N i=1 ψi zis < 0; ~ : R → R is constant, in the first case the function of k U (~y (k · φ)) then a maximum exists, ~ ≥ 0 impose a convex constraint on the feasible set, in the second case ~y (φ) by Weierstrass theorem a solution exists. 2
12. max W u1 (x1 , h(x)), . . . un (xn , h(x)) n+1
with h(x) ≥ 0, (~x, x) ∈ [0, w]
, w−x=
n X
xi
i=1
sufficient condition for continuity is that W , ~u and h are continuous, the problem is moreover well defined only if ∃ x ∈ [0, w] such that h(x) ≥ 0; P the simplex ∆w ≡ {(~x, x) ∈ [0, w]n+1 | x + ni=1 xi = w} is closed and bounded, if h(x) is continuous H = {(~x, x) ∈ [0, w]n+1 | h(x) ≥ 0} is closed because the inequality is not strict, then the feasible set ∆w ∩ H is closed and bounded because intersection of closed and bounded sets. 2 13. max π(x) = max xp(x) − c(x) x∈R+
with p : R+ → R+ and c : R+ → R+ continuous, c(0) = 0 and p(·) decreasing; a) ∃ x∗ > 0 such that p(x∗ ) = 0, then, ∀ x > x∗ , π(x) ≤ π(0) = 0, the solution can then be found in the compact set [0, x∗ ] ∈ R+ ; b) now ∀ x > x∗ , π(x) = π(0) = 0, as before; c) consider c(x) = 2p¯ · x, now π(x) = not bounded above. 2
p¯ 2
· x, so that the maximization problem is
14. a) p~(2) max v(c(1), c(2)) subject to ~c(1) ≫ ~0, ~c(2) ≫ ~0, p~(1) · ~c(1) + · ~c(2) ≤ W0 1+r
10
(2) b) we can consider, in R2n , the arrays ~c = (~c(1), ~c(2)) and ~p = (~ p(1), p~1+r ), the conditions
~(2) p · ~c(2) ≤ W0 ~c(1) ≫ ~0, ~c(2) ≫ ~0, p~(1) · ~c(1) + 1+r become ~c ≫ ~0, p~ · ~c ≤ W0
we are in the conditions of example 3.6 (pages 92-93), and ~p ≫ ~0 if and only if p~(1) ≫ ~0 and p~(2) ≫ ~0. 2 15. 0 ≤ x1 ≤ y 1 0 ≤ x2 ≤ f (y1 − x1 ) max π(x1 ) + π(x2 ) + π(x3 ) subject to 0 ≤ x3 ≤ f (f (y1 − x1 ) − x2 )
let’s call A the subset of R3+ that satisfies the conditions, A is compact if it is closed and bounded (Th. 1.21, page 23). It is bounded because A ⊆ [0, y1 ] × [0, max f ([0, y1 ])] × [0, max f ([0, max f ([0, y1 ])])] ⊂ R3+ and maxima exist because of continuity. Consider (ˆ x1 , x ˆ2 , x ˆ3 ) ∈ Ac in R3+ , then (if we reasonably suppose that f (0) = 0), becasue of continuity, at least one of the following holds: x ˆ1 > y1 , x ˆ2 > f (y1 − x ˆ1 ) or x ˆ3 > f (f (y1 − x ˆ1 ) − x ˆ2 ). If we take r = max{ˆ x1 − y 1 , x ˆ2 − f (y1 − x ˆ1 ), xˆ3 − f (f (y1 − x ˆ1 ) + x ˆ2 )}, r > 0 and c B((ˆ x1 , x ˆ2 , x ˆ3 ), r) ⊂ A . Ac is open =⇒ A is closed. 2
11
Chapter 4 (pp. 100-111): Unconstrained Optima. 1. Consider D = [0, 1] and f : D → R, f (x) = x: arg maxx∈D f (x) = 1, but Df (1) = 1 6= 0. 2 2. ′
2
f (x) = 1 − 2x − 3x = 0 for x = ′′
f (x) = −2 − 6x ,
1±
√
1+3 = −3
−1 1 3
f ′′ (−1) = 4 = −4 f ′′ ( 31 )
−1 is a local minimum and 13 a local maximum, they are not global because limx→−∞ = +∞ and limx→+∞ = −∞. 2 3. The proof is analogous to the proof at page 107, reverting the inequalities.
2
4. a) fx = 6x2 + y 2 + 10x =⇒ y = 0 ∧ x = 0 fy = 2xy + 2y 12x + 10 2y 2 D f (x, y) = 2y 2x + 2 (0, 0) is a minimum; b) fx = e2x + 2e2x (x + y 2 + 2y) = e2x (1 + 2(x + y 2 + 2y)) fy = e2x (2y + 2)
=⇒ y = −1
1 1 1 =⇒ f ( , −1) = − e 2 2 2 since limx→−∞ f (x, −1) = 0 and limx→+∞ f (x, −1) = +∞, the only critical point ( 21 , −1) is a global minimum; 1 + 2(x + 1 − 2) = 0 =⇒ x =
c) the function is symmetric, limits for x or y to ±∞ are +∞, ∀ a ∈ R: fx = ay − 2xy − y 2 fy = ax − 2xy − x2 2
1 1 2 2 =⇒ (0, 0) ∨ (0, a) ∨ (a, 0) ∨ (− a, − a) ∨ (− a, − a) 5 5 5 5
D f (x, y) =
2y a − 2x − 2y a − 2x − 2y 2x
12
(− 25 a, − 51 a) and (− 15 a, − 52 a) are local minima ∀ a ∈ R; d) when sin y 6= 0, limits for x ±∞ are ±∞, fx = sin y =⇒ (0, kπy) ∀ k ∈ Z = {. . . , −2, −1, 0, 1, . . .} fy = x cos y
f is null for critical points, adding any (ǫ, ǫ) to them (ǫ < π2 ), f is positive, adding (ǫ, −ǫ) to them, f is negative, critical points are saddles; e) limits for x or y to ±∞ are +∞, ∀ a ∈ R, fx = 4x3 + 2xy 2 fy = 2x2 y − 1
=⇒ y =
1 1 =⇒ 4x3 = − =⇒ impossible 2 2x x
there are no critical points; f) limits for x or y to ±∞ are +∞, fx = 4x3 − 3x2 fy = 4y 3
3 =⇒ (0, 0) ∨ ( , 0) 4
since f (0, 0) = 0 and f ( 43 , 0) < 0, ( 43 , 0) is a minimum and (0, 0) a saddle; g) limits for x or y to ±∞ are 0, fx = fy = since f (1, 0) = minimum;
1 2
1−x2 +y 2 (1+x2 +y 2 )2 −2xy (1+x2 +y 2 )2
)
=⇒ (1, 0) ∨ (−1, 0)
and f (−1, 0) = − 21 , the previous is a maximum, the latter a
h) limits for x to ±∞ are +∞, limits for y to ±∞ depends on the sign of (x2 − 1),
fx fy
y = 0 =⇒ x = 2 ∨ 3 √ = x8 + 2xy 2 − 1 =⇒ x = 1 =⇒ y = ± 47 2 = 2y(x − 1) ∨ x = −1 =⇒ impossible 3 2 x + 2y 2 4xy 2 8 D f (x, y) = 4xy 2y(x2 − 1)
13
5 √ √ 7 0 7 2 √4 =⇒ saddle, = =⇒ saddle, D f (1, 4 ) = 0 0 7 0 √ 5 √ − 7 4 √ D2 f (1, − 47 ) = =⇒ saddle. 2 − 7 0
D2 f (2, 0)
3 2
5. The unconstrained function is given by the substitution y = 9 − x: f (x) = 2 + 2x + 2(9 − x) − x2 − (9 − x)2 = −2x2 + 18x − 61 which is a parabola with a global maximum in x =
9 2
=⇒ y = 29 . 2
6. a) x∗ is a local maximum of f if ∃ǫ ∈ R+ such that ∀y ∈ B(x∗ , ǫ), f (y) ≤ f (x∗ ), considering then that: f (x∗ ) − f (y) = lim inf ∗ y→x
lim inf
y∈B(x∗ ,ǫ)→x∗
f (x∗ ) − f (y) ≥ 0
∀ y < x∗ , x∗ − y > 0 ∧ ∀ y > x∗ , x∗ − y < 0 −1 · lim inf = − lim sup we have the result; b) a limit may not exist, while lim inf and lim sup always exist if the function is defined on all R; c) if x∗ is a strict local maximum the inequality for lim inf y→x∗ is strict, so also the two inequalities to prove are. 2 7. Consider f : R → R: f (x) =
1 x∈Q 0 otherwise
where Q ⊂ R are the rational number; x = 0 is a local not strict maximum, f is not constant in x = 0. 2 8. f is not constant null, otherwise also f ′ would be constant null, since limy→∞ f (y) = 0, f has a global maximum x ¯, with f (¯ x) > 0 and f ′ (¯ x) ′ x is the only point for which f (x) = 0 =⇒ x = x ¯. 2 9. a) φ′ > 0: df dφ ◦ f = φ′ (f ) , dxi dxi
df dφ ◦ f = 0 =⇒ =0 dxi dxi
Df (~x∗ ) = ~0 =⇒ Dφ ◦ f (~x∗ ) = ~0
14
b) d2 φ ◦ f df i df df d h ′ d2 f φ (f ) = φ′′ (f ) = + φ′ (f ) dxi dxj dxj dxi dxi dxj dxi dxj d2 φ ◦ f d2 f df = 0 =⇒ = φ′ (f ) dxi dxi dxj dxi dxj Df (~x∗ ) = ~0 =⇒ D 2 φ ◦ f (~x∗ ) = φ′ (f (~x∗ )) · D 2 f (~x∗ ) a negative definite matrix is still so if multiplied by a constant.
2
10. Let us check conditions on principal minors (see e.g. Hal R. Varian, 1992, “Microeconomic analysis”, Norton, pp.500-501).
f x1 x1 . . . f x1 xn .. .. D2 f (~x) = ... . . f x1 xn . . . f xn xn
is positive definite if: 1) fx1 x1 > 0, 2) fx1 x1 · fx2 x2 − fx21 x2 > 0, ... n) |D 2 f (~x)| > 0;
−fx1 x1 . . . −fx1 xn gx1 x1 . . . gx1 xn .. .. .. = D 2 − f (~x) = .. .. D2 g(~x) = ... . . . . . −fx1 xn . . . −fxn xn gx1 xn . . . gxn xn
is negative definite if: 1) −fx1x1 < 0, 2) (−fx1 x1 ) · (−fx2 x2 ) − (−fx1 x2 )2 > 0, ... n) |D 2 − f (~x)| < 0 if n is odd, |D 2 − f (~x)| > 0 if n is even;
the ith principal minor is a polinimial where all the elements have order i, it is easy to check that odd principal minors mantain the sign of its arguments, while even ones are always positive, hence D 2 f (~x) positive definite =⇒ D 2 f (~x) negative definite. 2
15
Chapter 5 (pp. 112-144): Equality Constraints. 1. a) L(x, y, λ) = x2 − y 2 + λ(x2 + y 2 − 1)
=0
Lx = 2x + 2λx =⇒ λ = −1 ∨ x = 0 =0
Ly = −2y + 2λy =⇒ λ = 1 ∨ y = 0 x = 0 ⇒ y = ±1 λ = 1 =0 2 2 Lλ = x + y − 1 =⇒ y = 0 ⇒ x = ±1 λ = −1
when λ = 1 we have a minimum, when λ = −1 a maximum; b) 2
2
2
f (x) = x − (1 − x ) = 2x − 1 =⇒
min for x = 0 max for x = ±∞
the solution is different from (a) because the right substitution is y ≡ admissible only for x ∈ [−1, 1]. 2
√
1 − x2 ,
2. a) Substituting y ≡ 1 − x: f (x) = x3 − (1 − x)3 = 3x2 − 3x + 1 =⇒ max for x = ±∞
b)
L(x, y, λ) = x3 + y 3 + λ(x + y − 1) ) =0 Lx = 3x2 + λ =⇒ λ = −3x2 =0
Ly = 3y 2 + λ =⇒ λ = −3y 2
=⇒ x = ±y x=y
Lλ = x + y − 1 =⇒ x = y = x=y=
1 2
is the unique local minimum, as can be checked in (a).
1 2 2
3. a) L(x, y, λ) = xy + λ(x2 + y 2 − 2a2 )
=0
Lx = y + 2λx =⇒ (y = 0 ∧ λ = 0) ∨ y = −2λx =0
Ly = x + 2λy =⇒ (x = 0 ∧ λ = 0) ∨ x = −2λy √ x = 0 ⇒ y = ± √2a =0 Lλ = x2 + y 2 − 2a2 =⇒ y = 0 ⇒ x = ± 2a y = −2λx ∧ x = −2λy
16
√ √ (0, ± 2a) and (± 2a, 0) are saddle points, because f (x, y) can be positive or negative for any ball around them; y = −2λx ∧ x = −2λy imply: λ = ± 12 , x = ±y, and x = ±a, the sign of x does not matter, when x = y we have a maximum, when x = −y a minimum. b) substitute x ˆ ≡ x1 , yˆ ≡
1 y
and a ˆ≡
1 a
L(ˆ x, y, λ) = x ˆ + yˆ + λ(ˆ x2 + yˆ2 − a ˆ2 )
1 =0 Lxˆ = 1 + 2λˆ x =⇒ x ˆ=− λ 2 1 =0 Lyˆ = 1 + 2λˆ y =⇒ yˆ = − λ = x ˆ 2 √ 2 =0 ˆ = yˆ = ± a Lλ = x ˆ2 + yˆ2 − a ˆ2 =⇒ x 2
for x ˆ and yˆ negative we have a minimum, otherwise a maximum. c) L(x, y, x, λ) = x + y + z + λ(x−1 + y −1 + z −1 − 1)
√ =0 Lx = 1 − λx−2 =⇒ x = ± λ √ =0 . . . =⇒ y = ± λ √ =0 . . . =⇒ z = ± λ
we have a minimum when they are all negative (x = y = z = −3) and a maximum when all positive (x = y = z = −3). d) substitute xy = 8 − (x + y)z = 8 − (5 − z)z: f (z) = z(8 − (5 − z)z) = z 3 − 5z 2 + 8z 2
=0
fz = 3z − 10z + 8 =⇒ z =
5±
√ 25 − 24 2 = 4 3 3
as z → ±∞, f (z) → ±∞, fzz = 6z − 10 is negative for z = 43 (local maximum) and positive for z = 2 (local minimum),
17
when z = 34 , x + y = 5 − z = 11 3 and xy = 8 − (x + y)z = =⇒ one is also 34 and the other 73 ,
28 9
when z = 2, x + y = 5 − z = 3 and xy = 8 − (x + y)z = 2 =⇒ one is also 2 and the other is 1; e) substituting y ≡
16 x:
=0
f (x) = x + 16 =⇒ fx = 1 − x162 =⇒ x = ±4 x when x = ±4 and y = ± 14 we have a minimum, maxima are unbounded for limx→0 and limx→±∞ ; f) substituting z ≡ 6 − x and y ≡ 2x: f (x) = x2 + 4x − (6 − x)2 = 16x − 36 , f (x) is a linear function with no maxima nor minima.
2
4. Actually a lemniscate is (x2 +y 2 )2 = x2 −y 2 , while (x2 −y 2 )2 = x2 +y 2 identifies only the point (0, 0). (x2+y2)2−x2+y2 = 0 0.5
y
A lemniscate 0
−0.5 −1
−0.8
−0.6
−0.4
−0.2
0 x
0.2
0.4
0.6
0.8
1
In the lemniscate x + y maximizes, by simmetry, in the positive quadrant, in the point where the tangent of the explicit function y = f (x) is −1; for x and y positive the explicit function becomes: (x2 + y 2 )2 = x2 − y 2
x4 + 2x2 y 2 + y 4 − x2 + y 2 = 0
y 4 + (2x2 + 1)y 2 + x4 − x2 = 0
−2x2 − 1 +
√
4x4 + 4x2 + 1 − 4x4 + 4x2 2 s √ −2x2 + 1 + 8x2 + 1 y = f (x) = 2 y
2
=
computing f ′ (x) and finding where it is −1 we find the arg max = (x∗ , y ∗ ), (−x∗ , −y ∗ ) will be the arg min. 2
18
5. a) (x − 1)3 = y 2 implies x ≥ 1, f (x) = x3 − 2x2 + 3x − 1 =⇒ fx = 3x2 − 4x + 3 which is always positive for x ≥ 1, since f (x) is increasing, min f (x) = f (1) = 1 (y = 0); b) the derivative of the constraint D((x − 1)3 − y 2 ) is (3x2 − 6x + 3, −2y)′ , which is (0, 0)′ in (1, 0) and in any other point satisfying the constraint, hence the rank condition in the Theorem of Lagrange is violated. 2 6. a) 1 L(~x, ~λ) = ~c′ ~x + ~x′ D~x + ~λ′ (A~x − ~b) 2 n n m n n X X X 1 XX ci xi + = Aij xj − bi λi xj Dji xi + 2 i=1
i=1
Lxi = ci + Dii xi + = ci +
n X
1 2
=
n X j=1
xj Dji +
j=1,j6=i m X
1 2
i=1 n X
j=1
xj Dij +
j=1,j6=i
m X
λj Aji
j=1
λj Aji
xj Dij +
j=1
j=1
Lλi
j=1 n X
Aij xj − bi
b). . . 7. The constraint is the normalixation |~x| = 1, the system is: f (~x) = ~x′ A~x n n X X xj Aji xi = i=1 j=1 n X
xj Aji
f xi = 2
j=1 Pn ∗ j=1,j6=i xj Aji
x∗i = −
Aii
x~∗ = −
0
A21 A11
.. .
0 .. .
... ... .. .
A1n Ann
A2n Ann
...
A12 A22
such that |x~∗ | = 1
19
An1 A11 An2 A22
.. . 0
~∗ · x ≡ B · x~∗
for any eigenvectors of the square matrix B, its two normalization (one opposite of the other) are critical points of the problem; A11 . . . An1 .. , ∀ ~x is constant, .. since D 2 f (~x) = 2 ... . .
A1n . . . Ann all critical points are maxima, minima or saddles, according wether A is positivedefinite, negative-definite or neither. 2 8. a) 1
0.9
0.8
0.7
stock
0.6
0.5
0.4
0.3
0.2
0.1
0
0
50
100
150
200 days
250
300
350
400
dt , The function s(t) quantifying the stock is periodic of period T = x dI RT
s(t)
x in this period the average stock is 0 T = x·T 2T = 2 , in the long run this will be the total average;
b)
Lx Ln
x L(x, n, λ) = Ch + C0 n + λ(nx − A) 2 Ch C0 Ch = 2 + λn =0 =⇒ n = − ∧x=− = C0 + λx 2λ λ r Ch C0 Ch C0 =0 Lλ = nx − A =⇒ = A =⇒ λ = ± 2λ2 2A
x and n negative have no meaning so λ must be negative.
2
9. The condition for equality constraint to suffice is that u(x1 , x2 ) is nondecreasing, this happens for α ≥ 1 ∧ β ≥ 1; if otherwise one of them , say α is less than 1, the problem is unbounded at limx1 →0 xα1 + xβ2 = +∞. 2 10.
20
min w1 x1 + w2 x2 s.t. (x1 , x2 ) ∈ X ≡ {(x1 , x2 ) ∈ R2+ | x21 + x22 ≥ 1} a) 2
1.8
1.6
1.4
X=x21+x22 ≥ 1
x
2
1.2
1
0.8
0.6
−w /w 1
2
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1 x1
1.2
1.4
1.6
1.8
2
if w1 < w2 it is clear from the graph that (1, 0) is the cheapest point in X, similarly, if w2 < w1 , the cheapest point is 0, 1, if w1 = w2 they both cost w1 = w2 ; b) if nonnegativity constraints are ignored: min w1 x1 + w2 x2 =
x21 +x22 ≥1
lim
(x1 ,x2 )→(−∞,−∞)
w1 x1 + w2 x2 = −∞
similarly for (x1 , x2 ) → (+∞, +∞) if w1 and w2 are positive.
2
1
11. [x 2 is not defined if x < 0. . . ] 1
1
max x 2 + y 2 s.t. px + y = 1 1
Lx = Ln =
1
L(x, y, λ) = x 2 + y 2 + λ(px + y − 1) ) 1 − 12 x + λp =0 2 =⇒ x = (−2λp)−2 ∧ y = (−2λ)−2 1 − 12 + λ y 2 =0
Lλ = px + y − 1 =⇒ p(−2λp)−2 + (−2λ)−2 = 1 p p2 + p 1 p2 + p p + = = 1 =⇒ λ = ± 4λ2 p2 4λ2 4λ2 p2 2p 2
the sign of λ does not matter to identify x∗ = p21+p and y ∗ = p2p+p , the unique critical point, with both positive components, q 1 1 1 1 1+p 2 2 = x∗ 2 + y ∗ 2 = √1+p p is greater e.g. than 0 + 1 = 1, 2 p +p
being the unique critical point it is a maximum.
21
2
Chapter 6 (pp. 145-171): Inequality Constraints. 1. Since the function is increasing in both variables, we can search maxima on the boundary, √ substituting y = 1 − x2 (which means x ∈ [0, 1] =⇒ y ∈ [0, 1]): f (x) = ln x+ln
p 1 1 1 2x 1 − 2x2 1 − x2 = ln x+ ln(1−x2 ) =⇒ fx = − = 2 x 2 1 − x2 (1 − x2 )x
√ 2 2 =⇒ y = =⇒ x = 2 2 which is the argument of the maximum. 2 √
=0
2. The function is increasing, we search maxima on the boundary, √ √ we can substitute to polar coordinates y ≡ x sin ρ, z ≡ x cos ρ: f (ρ) =
√
x(py sin ρ + pz cos ρ) =⇒ fρ = =0
=⇒
py sin ρ = cos ρ pz
=⇒
√ x(py cos ρ − pz sin ρ)
y py = z pz
economically speaking, marginal rate of substitution equals the price ratio, sin ρ = ±
p2y
py + p2z
∧ cos ρ = ±
p2y
pz + p2z
√ y=± x
p2y
py + p2z
√ ∧ z=± x
p2y
pz + p2z
√ py √ pz √ py √ pz x p2 +p2 ) is a maximum and (− x p2 +p x p2 +p2 ) is a minimum. 2 ( x p2 +p 2, 2,− y
z
y
z
y
z
y
z
3. a) We can search maxima on the boundary determined by I: max x1 x2 x3 s.t. x1 ∈ [0, 1], x2 ∈ [2, 4], x3 = 4 − x1 − x2 max f (x1 , x2 ) = 4x1 x2 − x21 x2 − x1 x22 s.t. x1 ∈ [0, 1], x2 ∈ [2, 4] consider only x1 , 4x2 −x22 f (x1 ) = 4x1 x2 − x21 x2 − x1 x22 has a maximum for x1 = 2x = 2 − 21 x2 , 2 by simmetry a maximum is for x1 = x2 = 43 , which is however not feasible, nevertheless, for x1 ∈ [0, 1] the maximum value for x2 ∈ [2, 4] is 2, we get then x1 = 1 and x3 = 1; b)
22
max x1 x2 x3 s.t. x1 ∈ [0, 1], x2 ∈ [2, 4], x3 =
6 − x1 − 2x2 3
1 2 max f (x1 , x2 ) = 2x1 x2 − x21 x2 − x1 x22 s.t. x1 ∈ [0, 1], x2 ∈ [2, 4] 3 3 2x2 − 32 x22 2 x 3 2
f (x1 ) has a maximum when x1 = f (x2 ) when x2 =
2x1 − 13 x22 4 x 3 2
3 2
=
= 3 − x2 ,
− 14 x1 ,
the two together give x1 = 3 − 32 + 41 x1 =⇒ x1 = 2 and x2 = 1, not feasible, nevertheless, for x1 ∈ [0, 1] the maximum value for x2 ∈ [2, 4] is again 2, hence x1 is again 1 and x3 = 1. 2 4. The argument of square-root must be positive, the function is increasing in all variables, so we can use Lagrangean method:
L(~x, λ) =
T X t=1
Lx i =
T X √ xt − 1 2−t xt + λ t=1
−1 2−i−1 xi 2
+λ
if λ = 0 all xi are 0 (minimum); otherwise, for i < j, i, j ∈ {1, . . . T }: − 12
2−i−1 xi
− 12
= 2−j−1 xj
=⇒
x 1 i
2
xj
=
xi 2j+1 = 2j−i =⇒ = 22(j−i) 2i+1 xj
the system becomes x1 = x1 , x2 = 14 x1 , . . . xT = since they must sum to 1: x1 = 1/
T −1 X i=0
1 x , 22(T −1) 1
T −1
T −1
X 1 X −i 1 4−i 4−i , x2 = / 4 , . . . xT = (T −1) / 4 4 i=0 i=0
as T increases the sum quickly converges to
1 1− 14
= 43 .
2
5. a) min w1 x1 + w2 x2 s.t. x1 x2 = y¯2 , x1 ≥ 1, x2 > 0 the feasible set is closed but unbounded as x1 → +∞ b) substituting x2 =
y¯2 x1
we get:
23
2
f (x1 ) = w1 x1 + w2 xy¯1 2
=0
f ′ (x1 ) = w1 − w2 xy¯2
=⇒ x1 = +
1
2
f ′′ (x1 ) = 2w2 xy¯3 > 0 1
if
q
w2 w1 y
r
w2 y w1
since x1 > 1
since x1 > 1
≥ 1, this is the solution, and x2 =
q
w1 w2 y,
otherwise 1 is, and x2 = y 2 , in both cases ∃ ~λ satisfying Kuhn-Tucker conditions: L(x1 , x2 , λ1 , λ2 ) for
r w
2
w1
y,
r
w1 y w2
w1 x1 + w2 x2 + λ1 (¯ y 2 − x1 x2 ) + λ2 (x1 − 1)
=
w2 y
=0
Lx2 = w2 − λ1 x1 =⇒ λ1 = =0
r
w1 w2
Lx1 = w1 − λ1 x2 + λ2 =⇒ λ2 = 0 for (1, y 2 ) =0
Lx2 = w2 − λ1 x1 =⇒ λ1 = w2 =0
Lx1 = w1 − λ1 x2 + λ2 =⇒ λ2 = w2 y 2 − w1 . 2 6. 1
max
(x1 ,x2 )∈R2+
f (x1 , x2 ) =
max
(x1 ,x2 )∈R2+
1
1
p1 x12 + p2 x12 x23 − w1 x1 − w2 x2
1
1
substitute x ≡ x12 and y ≡ x23 , the problem becomes max f (x, y) = max x(p1 + p2 y) − w1 x2 − w2 y 3
(x,y)∈R2+
(x,y)∈R2+
=0
fx = p1 + p2 y − 2w1 x =⇒ x = p2 p1 p2 y =0 fy = xp2 − 3w2 y = + 2 − 3w2 y 2 =⇒ y = 2w1 2w1 2
only the positive value will be admissible; for p1 = p2 = 1 and w1 = w2 = 2 we have:
24
p1 + p2 y 2w1 r p2
− 2w21 ±
p22 2w1
2
3w2
+ 6 w2wp12 p1
√ 97 − 1 97 = =⇒ x = y= 6 24 96 4 1 is negative semidefinite for it is a maximum, because D 2 f (x, y) = 1 −12y every y ≥ 0, it is a global one for positive values because it is the only critical point. 2 − 14 +
q
1 16
+6
√
1
7. a) Substituting x ≡ x12 and y ≡ x2 , considering that utility is increasing in both variables, the problem is: max f (x, y) = x + x2 y such that 4x2 + 5y = 100
(x,y)∈R2+
substituting y = 20 − 54 x2 the problem becomes: 4 max f (x) = x + 20x2 − x4 5
x∈R+
16 3 x 5 with numerical methods it is possible to calculate that the only positive x for which fx = 0 is x∗ ≃ 3.5480, where f (x) ≃ 128.5418, 48 2 since fxx = 40 − 48 5 x < 40 − 5 · 5 < 0, it is a maximum, we obtain x∗1 ≃ 12.5883 and x∗2 ≃ 9.9294; fx = 1 + 40x −
b) buying the coupon the problem is: max fa (x, y) = x + x2 y such that 3x2 + 5y = 100 − a
(x,y)∈R2+
which becomes: max fa (x) = x +
x∈R+
100 − a 2 3 4 x − x 5 5
12 3 x 5 the value of a that makes the choice indifferent is when fa (x) ≃ 128.5418 (as without coupon) and fa′ = 0, for lower values of a the coupon is clearly desirable. 2 fa′ = 1 + (200 − 2a)x −
8. a) max u(f, e, l) s.t. l ∈ [0, H], uf > 0, ue > 0, ul > 0
25
if α is the amount of income spent in food, f = max u
αwl p
and e =
(1−α)wl : q
αwl (1 − α)wl , , l s.t. l ∈ [0, H], α ∈ [0, 1], uf > 0, ue > 0, ul < 0 ; p q
b) αwl (1 − α)wl , , l + λ1 l + λ2 (H − l) + λ3 α + λ4 (1 − α) L(l, α, ~λ) = u p q du Ll = + λ1 − λ2 dl du Lα = + λ3 − λ4 dα and the constraints; c) the problem is max
l∈[0,16], α∈[0,1]
1
1
f (l, α) = (α3l) 3 ((1 − α)3l) 3 − l2 1
1
2
= (α) 3 (1 − α) 3 (3l) 3 − l2 we can decompose the two variables function: 1 1 2 f (l, α) = g(α)(3l) 3 − l2 , where g(α) ≡ (α) 3 (1 − α) 3 2 since (3l) 3 is always positive, for l ∈ [0, 16], and g(α) is always positive, for α ∈ [0, 1], g(α) maximizes alone for α = 12 , where f ( 12 ) = 64; now we have: 1
2
f (l) ≡ 64(3l) 3 − l2 =⇒ fl = 128 · (3l)− 3 − 2l 1
=0
1
=⇒ l∗ = 64 · 3− 3 · l∗ − 3 4
4
fll = −128 · (3l)− 3 − 2
=⇒ l∗ 3 ≃ 64 · 0.6934 =⇒ l∗ ≃ 17.1938 ←
always negative
l∗ is a maximum but is not feasible, however, since the function is concave, f (l) is increasing in [0, 16], the agent maximizes working 16 hours (the model does not include the time for leisure in the utility) and splitting the resources on the two commodities. 2 9. 1
max x13 + min{x2 , x3 } s.t. p1 x1 + p2 x2 + p3 x3 ≤ I
26
in principle Weierstrass theorem applies but not Kuhn-Tucker because min is continuous but not differentiable. The cheapest way to maximize min{x2 , x3 } is however when x2 = x3 , the problem becomes: 1
max x13 + x2 s.t. p1 x1 + (p2 + p3 )x2 ≤ I now also Kuhn-Tucker applies.
2
10. a) max p · f (L∗ + l) − w1 L∗ − w2 l s.t. l ≥ 0, f ∈ C 1 is concave =⇒ fl < 0 b) L(l, λ) = p · f (L∗ + l) − w1 L∗ − w2 l + λl Ll = p · fl (L∗ + l) − w2 + λ
Lλ = l
l = 0 and λ = w2 − p · fl (L∗ + l); c) p·f (L∗ +l)−w1 L∗ −w2 l maximizes once (by concavity) for p·fl (L∗ +l) = w2 , when this happens for l ≤ 0, the maximum is (by chance) the Lagrangean point, when instead this happens for l > 0, the maximum is not on the boundary. 2 11. a) 1
1
max py x14 x24 − p1 (x1 − K1 ) − p2 (x2 − K2 ) s.t. x1 ≥ −K1 , x2 ≥ −K2 1
1
L(x1 , x2 , λ1 , λ2 ) = py x14 x24 − p1 (x1 − K1 ) − p2 (x2 − K2 ) + λ1 (K1 + x1 ) + λ2 (K2 + x2 ) 1 −3 1 py x1 4 x24 − p1 x1 + λ1 x1 Lx 1 = 4 1 1 −3 py x14 x2 4 − p2 x2 + λ2 x2 Lx 2 = 4 Lλ1 = K1 + x1 Lλ2
= K2 + x2 1
1
b) the unbounded maximum of f (x1 , x2 ) = x14 x24 − x1 − x2 + 4 is for:
27
1
3
1
− 34
3 4
1 4
−3
fx1 = 41 x1 4 x24 − 1 = 0 =⇒ x14 = 14 x24 1 4
fx2 = 41 x1 x2
− 1 = 0 =⇒ x2 = 14 x1
bounds are respected, the firm sells most of x1 and buys only
1 16
)
=⇒ x1 = x2 =
units of x2 to produce
c) is analogous to (b) since the problem is symmetric.
1 4
1 2 4
=
1 16
units of y;
2
12. a) max py (x1 (x2 + x3 )) − w ~ ′ ~x L(~x, ~λ) = py (x1 (x2 + x3 )) − w ~ ′ ~x + ~λ′ ~x Lx i
Lx1 = py (x2 + x3 ) − w1 + λ1
i={2,3}
= py x1 − wi + λi
Lλi = xi
x1 =
w3 − λ3 w2 − λ2 = py py x2 + x3
=⇒ λ3 − λ2 = w3 − w2 =
w1 − λ1 py
b) [ w4 ??? ] ~x ∈ R3+ and ~λ ∈ R3 are not defined by the equations; c) the problem can be solved considering the cheapest between x2 and x3 , suppose it is x2 , the problem becomes: max py x1 x2 − w1 x1 − w2 x2 0 py which has critical point ( wpy2 , wpy1 ) but its Hessian is not negative semidefpy 0 inite. The problem maximizes at (+∞, +∞) for any choice of py ∈ R+ and w ~ ∈ R3+ . 2
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