10) Linear Programming Problems

10) LINEAR PROGRAMMING PROBLEMS

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Linear Programming

1. For the following shaded region, the linear in equations are:

a)

x+ y ≤ 5, x ≥2, y ≥1

b) x+ y ≥ 5, x ≥2, y ≥1 c) x+ y ≥ 5, x ≥2, y ≤1 d) x+ y ≤ 5, x ≥2, y ≤1 2. For the following shaded region, the linear in equations are:

a)

2 x +3 y ≥3,−5 x + 4 y ≤0, 3 x+ 4 y ≤18, x ≥ 0, y ≥ 0 b)

2 x +3 y ≥3,−5 x + 4 y ≤0, 3 x+ 4 y ≥18, x ≥ 0, y ≥ 0 c)

2 x +3 y ≥3,−5 x + 4 y ≥0, 3 x+ 4 y ≤18, x ≥ 0, y ≥ 0 d)

2 x +3 y ≤3,−5 x + 4 y ≤ 0, 3 x+ 4 y ≤18, x ≥ 0, y ≥ 0 5. For the following shaded region, the linear constraints are:

a) 2 x + y ≤2, y−x ≥−1, x+ 2 y ≥ 8, x ≥0, y ≥ 0 b) 2 x + y ≥ 2, y− x ≥−1, x+ 2 y ≤ 8, x ≥0, y ≥ 0 c)

2 x + y ≥ 2, y− x ≥−1, x+ 2 y ≤ 8, x ≥0, y ≥ 0

d) 2 x + y ≤2, y−x ≥−1, x+ 2 y ≥ 8, x ≥0, y ≥ 0 3. For the following shaded region, the linear in equations are:

a)

y−x ≥3, y−x ≥6, 0 ≤ x ≤ 4, y ≥ 0

b)

y−x ≥3, y−x ≤6, 0 ≤ x ≤ 4, y ≥ 0

c)

y−x ≤3, y−x ≥6, 0 ≤ x ≤ 4, y ≥ 0

d)

y−x ≥3, x− y ≤6, 0 ≤ x ≤ 4, y ≥ 0

6. For the following shaded region, the linear constraints are:

a)

x ≥ 2, y ≥ 3, x ≤6, y ≤5

b) x ≥ 2, y ≥ 3, x ≥6, y ≤5 c)

x ≥ 2, y ≥ 3, x ≤6, y ≤5

d) x ≥ 2, y ≥ 3, x ≤6, y ≥5 4. For the following shaded region, the linear constraints are:

a)

x+ y ≤ 0, 2 x + y ≤ 4, x ≥0, y ≤2

b) x+ y ≥ 0, 2 x + y ≥ 4, x ≥0, y ≤2 c)

x+ y ≤ 0, 2 x + y ≥ 4, x ≥0, y ≥ 0

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d) x+ y ≥ 0, 2 x + y ≤ 4, x ≥0, y ≤2 7. For the following shaded region, the linear constraints are:

a)

x+ y ≤ 4, 3 x+ y ≥ 3, x + 4 y ≥ 4, x ≥ 0, y ≥ 0

b) x+ y ≤ 4, 3 x+ y ≥ 3, x + 4 y ≤ 4, x ≥ 0, y ≥ 0 c)

x+ y ≤ 4, 3 x+ y ≤ 3, x +4 y ≥ 4, x ≥ 0, y ≥ 0

d) x+ y ≥ 4, 3 x+ y ≥ 3, x + 4 y ≥ 4, x ≥ 0, y ≥0 8. For the following shaded region, the linear constraints are:

a) 3 x+2 y ≥20, 3 x+ y ≥ 15, x ≥0, y ≥ 0 b) 3 x+2 y ≥20, 3 x+ y ≤ 15, x ≥0, y ≥ 0 c) 3 x+2 y ≤20, 3 x+ y ≤ 15, x ≥0, y ≥0 d) 3 x+2 y ≤20, 3 x+ y ≥ 15, x ≥0, y ≥ 0 9. For the following shaded region, the linear constraints are:

c)

2 x + y ≤12, x + y ≤ 7, x +2 y ≤10, x , y ≥0

d) 2 x + y ≤12, x + y ≤ 7, x +2 y ≥10, x , y ≥0 10. For the following shaded region, the linear constraints are:

a) 2 x +5 y ≥80, x + y ≤ 20, x ≥ 0, y ≥ 0 b) 2 x +5 y ≤80, x+ y ≤ 20, x ≥ 20, y ≥ 0 c)

2 x +5 y ≤80, x+ y ≥ 20, x ≥ 0, y ≥ 0

d) 2 x +5 y ≥80, x + y ≥ 20, x ≥ 0, y ≥ 0 11. For the following shaded region, the linear constraints are:

a)

x+ y ≥ 5,2 x + y ≥ 4,0 ≤ x ≤ 3,0 ≤ y ≤3

b) x+ y ≤ 5,2 x + y ≤ 4, 0 ≤ x ≤ 3,0 ≤ y ≤ 3 c)

x+ y ≥ 5,2 x + y ≤ 4, 0 ≤ x ≤ 3,0 ≤ y ≤3

d) x+ y ≤ 5,2 x + y ≥ 4, 0 ≤ x ≤ 3,0 ≤ y ≤3 12. For the following shaded region, the linear constraints are:

a) a) 2 x + y ≥12, x + y ≤ 7, x+2 y ≤10, x , y ≥0

x+ 2 y ≤16, x + y ≥ 12, 2 x + y ≥14, x ≥ 0, y ≥ 0

b) 2 x + y ≤12, x + y ≥ 7, x+2 y ≤10, x , y ≥0 P a g e |3

b)

x+ 2 y ≥16, x + y ≤ 12, 2 x + y ≥14, x ≥ 0, y ≥ 0 c)

x+ 2 y ≥16, x + y ≥ 12, 2 x + y ≥14, x ≥ 0, y ≥ 0

x+ y ≥ 5, 4 x + y ≥ 4, x +5 y ≥ 5, x ≤ 4, y ≤ 4

b) x+ y ≤ 5, 4 x + y ≤ 4, x +5 y ≥ 5, x ≤ 4, y ≤ 4 c)

x+ y ≤ 5, 4 x + y ≥ 4, x +5 y ≤ 5, x ≤ 4, y ≤ 4

d) x+ y ≤ 5, 4 x + y ≥ 4, x +5 y ≥ 5, x ≤ 4, y ≤ 4

d)

x+ 2 y ≥16, x + y ≥ 12, 2 x + y ≤14, x ≥ 0, y ≥ 0 13. For the following shaded region, the linear constraints are:

a)

a)

x+ y ≤ 5, x +2 y ≥ 4, 4 x+ y ≥ 12, x ≥0, y ≥ 0

b) x+ y ≤ 5, x +2 y ≥ 4, 4 x+ y ≤ 12, x ≥0, y ≥ 0 c) x+ y ≤ 5, x +2 y ≤ 4, 4 x+ y ≤ 12, x ≥0, y ≥0 d) x+ y ≥ 5, x +2 y ≥ 4, 4 x+ y ≤ 12, x ≥0, y ≥ 0 14. For the following shaded region, the linear constraints are:

16. For the following shaded region, the linear constraints are:

a)

x+ 2 y ≤ 8,2 x+ y ≤ 2, x − y ≤ 1, x ≥ 0, y ≥ 0

b) x+ 2 y ≤ 8,2 x+ y ≥ 2, x − y ≥ 1, x ≥ 0, y ≥ 0 c)

x+ 2 y ≤ 8,2 x+ y ≥ 2, x − y ≤ 1, x ≥ 0, y ≥ 0

d) x+ 2 y ≥ 8,2 x+ y ≥ 2, x − y ≤ 1, x ≥ 0, y ≥ 0 17. For the following shaded region, the linear constraints are: a)

2 x +3 y ≥3, 3 x+ 4 y ≤18,−7 x+ 4 y ≥14, x , y ≥ 0 b)

2 x +3 y ≥3, 3 x+ 4 y ≥18,−7 x+ 4 y ≤14, x , y ≥ 0 c)

2 x +3 y ≥3, 3 x+ 4 y ≥18,−7 x+ 4 y ≤14, x , y ≥ 0 a) 3 x+ 4 y ≤12, 4 x +3 y ≤12, x ≥ 0, y ≥ 0

d)

b) 3 x+ 4 y ≥12, 4 x +3 y ≤12, x ≥ 0, y ≥ 0

2 x +3 y ≥3, 3 x+ 4 y ≤18,−7 x+ 4 y ≤14, x , y ≥ 0

3 x+ 4 y ≤12, 4 x +3 y ≤12, x ≥ 0, y ≥ 0

18. For the following shaded region, the linear constraints are:

c)

d) 3 x+ 4 y ≥12, 4 x +3 y ≥12, x ≥ 0, y ≥ 0 15. For the following shaded region, the linear constraints are:

a)

2 x +3 y ≥6, 4 x+6 y ≤ 24,−3 x+ 2 y ≤ 3, x ≥ 0, y ≥0

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b)

2 x +3 y ≥6, 4 x+6 y ≤ 24,−3 x+ 2 y ≥ 3, x ≥0, y ≥0 c)

2 x +3 y ≥6, 4 x+6 y ≥ 24,−3 x+ 2 y ≤ 3, x ≥0, y ≥0 d) c) d) None of these 19. Find the linear inequalities for which the 21. For the following shaded region, the linear solution set is the shaded region given in constraints are: figure below:a) 2 x −3 y ≤ 6,2 x− y ≥1, x ≥ 1, y ≤ 0

2 x +3 y ≤6, 4 x+6 y ≤ 24,−3 x+ 2 y ≤3, x ≥ 0, y ≥0

b) 2 x −3 y ≤ 6,2 x− y ≥1, x ≤ 1, y ≥ 0 c)

2 x −3 y ≤ 6,2 x− y ≤1, x ≥ 1, y ≤ 0

d) 2 x −3 y ≥ 6,2 x− y ≥1, x ≥ 1, y ≤ 0 22. The minimum value of a)

x+ y ≤ 4, x+ 5 y ≥ 4, 6 x+ 2 y ≤ 8,0≤ x ≤ 3,0 ≤ y ≤ 3 b)

x+ y ≤ 4, x+ 5 y ≤ 4, 6 x+2 y ≥ 8,0≤ x ≤ 3,0 ≤ y ≤ 3 c)

z=10 x+ 8 ysubjectto 4 x+ y ≥ 4, x+ 3 y ≥ 6, x + y ≥ 3, x ≥ 0, y ≥ a) 60 b) 27 c) 74/3 d) 32 23. For the following shaded region, the linear constraints except

x ≥ 0, y ≥ 0

are :

x+ y ≥ 4, x+ 5 y ≥ 4, 6 x+ 2 y ≥ 8,0≤ x ≤ 3,0 ≤ y ≤ 3 d)

x+ y ≤ 4, x+ 5 y ≥ 4, 6 x+ 2 y ≥ 8,0≤ x ≤ 3,0 ≤ y ≤ 3 20. The solution set of the linear inequalities

2 x + y ≥ 8 andx+2 y ≥10 is

a) 2 x + y ≥ 2, x− y ≥ 1, x +2 y ≤ 8 b) 2 x + y ≥ 2, x− y ≤1, x +2 y ≤ 8 c)

a)

2 x + y ≤2, x− y ≤1, x +2 y ≤ 8

d) 2 x + y ≥ 2, x− y ≥ 1, x +2 y ≥ 8 24.

For the following feasible region, the

linear constraints except

x ≥ 0, y ≥ 0

are :

b)

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a)

x ≥ 20, y ≤ 10, x +2 y ≥ 30

b) x ≤ 20, y ≥ 10, x +2 y ≥ 30 c)

x ≤ 20, y ≤ 10, x +2 y ≤ 30

d) x ≥ 20, y ≥ 10, x +2 y ≤ 30 25. For the following shaded region, the linear constraints are: a)

x+ y ≥ 60,5 x + y ≥100, x ≥ 0, y ≥ 0

b) x+ y ≤ 60,5 x + y ≤100, x ≥ 0, y ≥ 0 c)

x+ y ≥ 60,5 x + y ≤100, x ≥ 0, y ≥ 0

d) x+ y ≥ 60,5 x + y ≥100, x ≥ 0, y ≥ 0

x1 x2 x + x ≥1, + ≤1, x 1 ≥ 0, x2 ≥ 0 1 2 a) 8 3

28. For the following shaded region in following figure, the linear constraints( except

x ≥ 0, y ≥ 0 )are :

x1 x2 x + x ≤1, + ≥1, x 1 ≥ 0, x2 ≥ 0 1 2 b) 8 3 x1 x2 x + x ≤1, + ≤1, x 1 ≥ 0, x2 ≥ 0 1 2 c) 8 3 x1 x2 x + x ≥1, + ≥1, x 1 ≥ 0, x2 ≥ 0 1 2 d) 8 3 26. For the following shaded region, the linear constraints are:

a) 2 x + y ≤2, x− y ≤1, x + y ≤ 8 b) 2 x + y ≥ 2, x− y ≤1, x +2 y ≤ 8 c)

2 x + y ≤2, x− y ≥1, x +2 y ≤ 8

d) 2 x + y ≤2, x− y ≤1, x +2 y ≥ 8 29. For the following shaded region in following figure, the linear constraints are :

a) 4 x +6 y ≥ 24,5 x+ 3 y ≥ 15,2 y ≤5, x ≥ 0 b) 4 x +6 y ≤ 24,5 x+ 3 y ≤ 15,2 y ≤5, x ≥ 0 c)

4 x +6 y ≤ 24,5 x+ 3 y ≥ 15,2 y ≤5, x ≥ 0

d) 4 x +6 y ≥ 24,5 x+ 3 y ≤ 15,2 y ≤5, x ≥ 0 27. For the following shaded region, the linear constraints are:

a)

x+ 2 y ≤ 4, 2 x + y ≤ 4,1 ≤ x ≤2, y ≥0

b) x+ 2 y ≤ 4, 2 x + y ≥ 4,1 ≤ x ≤2, y ≥0 c)

x+2 y ≥ 4, 2 x + y ≤ 4,1 ≤ x ≤2, y ≥0

d) x+ 2 y ≥ 4, 2 x + y ≥ 4,1 ≤ x ≤2, y ≥ 0 30. For the following shaded region in following figure, the linear constraints are : P a g e |6

a) 2 x + y ≤ 6, x + y ≤ 4,1 ≤ x ≥ 0, y ≥ 0 b) 2 x + y ≥ 6, x + y ≤ 4,1 ≤ x ≥ 0, y ≥ 0 c)

2 x + y ≥ 6, x + y ≥ 4,1 ≤ x ≥ 0, y ≥ 0

d) 2 x + y ≤ 6, x + y ≥ 4,1 ≤ x ≥ 0, y ≥ 0 31. For the following shaded region in following figure, the linear constraints are :

a)

x y x y + ≥ 1, x + y ≤35, + ≥1, x ≥ 0, y ≥ 0 50 25 20 40 b)

x y x y + ≥ 1, x + y ≥35, + ≥1, x ≥ 0, y ≥ 0 50 25 20 40 c)

x y x y + ≤ 1, x + y ≥35, + ≥1, x ≥ 0, y ≥ 0 50 25 20 40 d) a)

x+ y ≤ 4, 3 x+ 8 y ≤ 24,10 x +7 y ≤ 35, x ≥ 0, y ≥ 0 b)

x+ y ≤ 4, 3 x+ 8 y ≤ 24,10 x +7 y ≥ 35, x ≥ 0, y ≥ 0

x y x y + ≥ 1, x + y ≥35, + ≤1, x ≥ 0, y ≥ 0 50 25 20 40 34. For the following shaded region in following figure, the linear constraints are :

c)

x+ y ≤ 4, 3 x+ 8 y ≥ 24,10 x +7 y ≤ 35, x ≥ 0, y ≥ 0 d)

x+ y ≥ 4, 3 x+ 8 y ≤ 24,10 x +7 y ≤ 35, x ≥ 0, y ≥ 0 32. For the following shaded region in following figure, the linear constraints are :

a) 7 x+ 9 y ≤ 63, x + y ≥1.5, x ≤ 6, y ≤ 5 b) 7 x+ 9 y ≤ 63, x + y ≥1.5, x ≤ 6, y ≤ 5 c)

7 x+ 9 y ≥ 63, x + y ≥1.5, x ≤ 6, y ≤ 5

d) 7 x+ 9 y ≥ 63, x + y ≤1.5, x ≤ 6, y ≤ 5 35. For the following shaded region in following figure, the linear constraints are : a) 5 x+3 y ≤150, 3 x+ 4 y ≤120, x ≥ 0, y ≥ 0 b) 5 x+3 y ≥150, 3 x+ 4 y ≤120, x ≥ 0, y ≥ 0 c)

5 x+3 y ≤150, 3 x+ 4 y ≥120, x ≥ 0, y ≥ 0

d) 5 x+3 y ≥150, 3 x+ 4 y ≥120, x ≥ 0, y ≥ 0 33. For the following shaded region in following figure, the linear constraints are :

a) 8 x+5 y ≤ 200,2 x+5 y ≤ 100, x ≥ 0, y ≥0 P a g e |7

b) 8 x+5 y ≤ 200,2 x+5 y ≥ 100, x ≥ 0, y ≥0 c)

8 x+5 y ≥ 200,2 x+ 5 y ≤ 100, x ≥ 0, y ≥0

d) 8 x+5 y ≥ 200,2 x+ 5 y ≥ 100, x ≥0, y ≥0 36. For the following shaded region in following figure, the linear constraintsare :

39. The common region represented by the in equalities

x+ y ≤ 3, y ≥ 2,−2 x+ y ≤ 1, x ≥ 0, y ≥ 0is a) Triangle b) Quadrilateral c) Pentagon d) None of these 40. The common region represented by the in equalities

0 ≤ x ≤6,0 ≤ y ≤ 4 is

a) Triangle b) rectangle c) square d) Pentagon 41. The common region represented by the in a) 3 x+5 y ≥15, x ≥ 1, y ≥ 1 b) 3 x+5 y ≥15, x ≤ 1, y ≥ 1 c)

3 x+5 y ≤15, x ≥ 1, y ≥ 1

d) 3 x+5 y ≤15, x ≤ 1, y ≤ 1 37. For the following shaded region in following figure, the linear constraints are :

equalities

x ≤ 5,2 y ≤9, 9 x +10 y ≥ 45, is

a) Triangle b) Quadrilateral c) Pentagon d) None of these 42. The graph of the in equalities

3 x−4 y ≤ 12, x ≤1, x ≥ 0, y ≤ 0

lies fully in

a) First quadrant b) second quadrant c) Third quadrant d) Fourth quadrant 43. For the following shaded region in following figure, the linear constraints are :

a) 2 x +3 y ≥6, x +2 y ≥ 8, 0≤ x ≤ 4, y ≥ 0 b) 2 x +3 y ≤6, x+ 2 y ≤ 8,0 ≤ x ≤ 4, y ≥0 c)

2 x +3 y ≥6, x +2 y ≤ 8, x ≤ 4, x ≥ 0, y ≥ 0

d) 2 x +3 y ≤6, x+ 2 y ≥ 8, 0≤ x ≤ 4, y ≥0 38. For the following shaded region in following figure, the linear constraints are :

a)

x− y ≤2, 2 x + y ≥ 4, x+2 y ≤10, x ≥ 0, y ≥ 0

b) x− y ≤2, 2 x + y ≥ 4, x+2 y ≥10, x ≥ 0, y ≥ 0 c)

x− y ≤2, 2 x + y ≤ 4, x+2 y ≤10, x ≥ 0, y ≥ 0

d) x− y ≥2, 2 x + y ≥ 4, x+2 y ≤10, x ≥ 0, y ≥ 0 44. For the following shaded region in following figure, the linear constraints are : a)

x+ 2 y ≥ 2,3 x+ 5 y ≥ 15, x ≥0, y ≥ 0

b) x+ 2 y ≥ 2,3 x+ 5 y ≤ 15, x ≥0, y ≥0 c)

x+ 2 y ≤2, 3 x+ 5 y ≥ 15, x ≥0, y ≥0

d) x+ 2 y ≤2, 3 x+ 5 y ≥ 15, x ≥0, y ≥0 P a g e |8

a)

7 x+ 2 y ≤14, x+ y ≤ 5, x +5 y ≥5, y ≤3, x ≥ 0, y ≥ 0

47. For the following shaded region in following figure, the linear constraints are :

b)

7 x+ 2 y ≤14, x+ y ≥ 5, x +5 y ≥5, y ≤3, x ≥ 0, y ≥ 0 c)

7 x+ 2 y ≥14, x+ y ≤ 5, x +5 y ≥5, y ≤3, x ≥ 0, y ≥ 0 d)

7 x+ 2 y ≥14, x+ y ≤ 5, x +5 y ≤5, y ≤3, x ≥ 0, y ≥ 0 45. For the following shaded region in following figure, the linear constraints are :

a)

x+ y ≤ 7.5,2 x+ y ≥ 4,2 x +9 y ≥ 18,0 ≤ x ≤5, 0 ≤ y ≤ 3 b)

x+ y ≤ 7.5,2 x+ y ≥ 4,2 x +9 y ≤ 18,0 ≤ x ≤5, 0 ≤ y ≤ 3 c)

x+ y ≤ 7.5,2 x+ y ≤ 4,2 x +9 y ≥ 18,0 ≤ x ≤5, 0 ≤ y ≤ 3 d) a)

7 x+ 2 y ≤16, 2 x+7 y ≥14, x + y ≤ 9, x ≥ 0, y ≥ 0

x+ y ≥ 7.5,2 x + y ≥ 4,2 x +9 y ≥ 18,0 ≤ x ≤5, 0 ≤ y ≤ 3 48. For the following shaded region in following figure, the linear constraints are :

b)

7 x+ 2 y ≥14, 2 x+7 y ≥ 14, x + y ≤ 9, x ≥ 0, y ≥ 0 c)

7 x+ 2 y ≥14, 2 x+7 y ≤14, x + y ≤ 9, x ≥ 0, y ≥ 0 d)

7 x+ 2 y ≥14, 2 x+7 y ≥ 14, x + y ≥ 9, x ≥ 0, y ≥ 0 46. For the following shaded region in following figure, the linear constraints are :

a)

x+ y ≥ 0, 3 x +2 y ≥ 6, x ≥ 0, y ≤ 2

b) x+ y ≤ 0, 3 x +2 y ≤ 6, x ≥ 0, y ≤ 2 c)

x+ y ≤ 0, 3 x +2 y ≥ 6, x ≥ 0, y ≤ 2

d) x+ y ≥ 0, 3 x +2 y ≤ 6, x ≥ 0, y ≤ 2 49. For the following shaded region in following figure, the linear constraints are :

a)

2 x + y ≤14, 6 x +8 y ≤ 48, x +2 y ≤ 10, x ≥0, y ≥0 b)

2 x + y ≤14, 6 x +8 y ≥ 48, x +2 y ≤ 10, x ≥0, y ≥0 c)

2 x + y ≤14, 6 x +8 y ≤ 48, x +2 y ≥ 10, x ≥0, y ≥0 d)

a)

6 x+ y ≥ 6, 3 x +7 y ≥21,4 x + y ≤ 8, x ≥0, y ≥0 b)

6 x+ y ≤ 6, 3 x +7 y ≥21,4 x + y ≤ 8, x ≥0, y ≥0

2 x + y ≥14, 6 x +8 y ≤ 48, x +2 y ≤ 10, x ≥0, y ≥0

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c)

53. Maximize value z=15x+10y subject to the

6 x+ y ≥ 6, 3 x +7 y ≤21,4 x + y ≤ 8, x ≥0, y ≥0 d)

constraints

3 x+2 y ≤12, 2 x +3 y ≤15, x ≥ 0, y ≥ 0

is

6 x+ y ≥ 6, 3 x +7 y ≥21,4 x + y ≥ 8, x ≥0, y ≥ 0 50. For the following shaded region in following figure, the linear constraints are :

a) 61.5 b) 50 c) 60 d) 62.5 54. The minimum value of p=10x+15y subject to the constraints

2 x +3 y ≥12, 2 x + y ≥ 6, x ≥0, y ≥0

is

a) 90 b) 70 c) 60 d) None of these 55. The minimum value of z=4x+5y subject to the constraints

x ≤ 30, y ≤ 40, andx ≥ 0, y ≥ 0 is

a)

2 x +3 y ≤6, 4 x+3 y ≤24,2 x− y ≤2, x ≥ 0, y ≥ 0 b)

2 x +3 y ≥6, 4 x+3 y ≤ 24,2 x− y ≤ 2, x ≥ 0, y ≥0

a) 320 b) 200 c) 120 d) 0 56. The maximize value of p=40x+50y subject to the constraints

3 x+ y ≤ 9, x +2 y ≤ 8 andx ≥0, y ≥ 0is

c)

2 x +3 y ≥6, 4 x+3 y ≥ 24,2 x− y ≤ 2, x ≥0, y ≥0 d)

2 x +3 y ≤6, 4 x+3 y ≥ 24,2 x− y ≤ 2, x ≥ 0, y ≥0 51. For the following shaded region in following figure, the linear constraints are :

is

a) 0 b) 120 c) 230 d) 200 57. The maximize value of z=48x+40y subject to the constraints

2 x + y ≤ 90, x +2 y ≤ 80, x+ y ≤ 50 andx ≥ 0, y ≥ 0 is is a) 0 b) 2320 c) 2160 d) 1600 58. The minimum value of c=3x+y subject to the constraints

2 x +3 y ≤6, x+ y ≥ 1, andx ≥ 0, y ≥ 0 is

a) 2 x + y ≥3, x + y ≥ 5,−x + y ≤1, x ≥ 0, y ≥ 0 b) 2 x + y ≥3, x + y ≤ 5,−x + y ≤1, x ≥ 0, y ≥ 0 c)

2 x + y ≤3, x + y ≤ 5,−x + y ≤1, x ≥ 0, y ≥ 0

d) 2 x + y ≥3, x + y ≤ 5,−x + y ≤1, x ≥ 0, y ≥ 0 52. The solution set of the inequalities

is

a) 0 b) 3 c) 2 d) 1 59. The maximize value of p=6x+3y subject to the constraints

4 x + y ≤12, 2 x+2 y ≥10, andx ≥ 0, y ≥0 is a) 18 b) 16 c) 23 d) 36 60. The minimum value of c=6x+7y subject to the constraints

5 x+8 y ≤ 40, 3 x + y ≤ 6,andx ≥ 0, y ≥2 is

a) 12 b) 14 c) 16 d) 20 a) Wholly xy plane 61. The maximize value of z=5x+7y subject b) Open half plane which does not contain to following in equations given by the origin. 3 x+2 y ≤12, x+ y ≤ 5, andx ≥ 0, y ≥ 0 is is c) Half plane those contain the origin. d) Open half planes which contain the a) 25 b) 30 origin. c) 35 d) 34

2 x +3 y ≤6 is

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62. The minimize value of

12 x 1 +42 x 2

20 x+10 y ≥200, x + y ≥ 35, x+2 y ≥50∧2 x + y ≥ 20is is

subject to the constraints

x 1+2 x 2 ≥ 3, x 1 +4 x2 ≥ 4, 3 x 1 + x 2 ≥ 3∧x 1 ≥ 0, x 2 ≥ 0 is a) 40 c) 50

b) 45 d) 55

63. The maximize value of

z=5 x 1+3 x 2

x+ y ≥ 2,7 x +8 y ≤56, x ≤ 6 andy ≤5 is

subject to

x 1 ≤ 4, x2 ≤ 8,∧x 1+ x 2 ≤8, x 1 ≥ 0, x2 ≥ 0is a) 20 b) 32 c) 40 d) 50 64. The maximize value of z=5x+10y subject to following in equations given by

x+ 2 y ≤10, 3 x+ y ≤ 12, andx ≥ 0, y ≥ 0 is a) 40 c) 55

b) 45 d) 50

65. The maximize value of

z=8 x 1+3 x 2

3 x1 + x 2 ≤ 45, x 1+ x 2 ≤25, 2 x1 + x 2 ≤ 40 is b) 115 d) 150

z=40 x 1 +60 x2

b) 350 d) 425

z=6 x 1−2 x2 such

2 x 1−x 2 ≤2, x 1 ≤ 3, x 1 ≥ 0, x 2 ≥ 0 thevaluesof x1 , x2 are a) (3,4) b) (2,3) c) (1,2) d) None of these 69. The maximize value of z=4x+8y subject

x− y ≥0, 3 x+ y ≥ 3, andx ≤ 4 is

then the

is

3 x+2 y ≤210, 2 x + 4 y ≤300, y ≤ 65, x ≥ 0, y ≥ 0 is

that

to constraints

is

a) 15 b) 16 c) 17 d) 17.5 76. The maximum value of P=20x+30y subject to the constraints given by following in equations

4 x 1 + x 2 ≥ 10,3 x 1 +2 x 2 ≥12, x 2 ≥ 5,is

68. The maximize value of

a) 110 b) 100 c) 105 d) 115 73. The minimum value of z=3x+4y subject to the following constraints

x+ y ≤ 4.5, 2 x + y ≤7, x +2 y ≤ 8, x ≥ 0, y ≥ 0is

subject to constraints

a) 300 c) 400

is

maximum value of z=3x+4y is a) 15 b) 18 c) 16 d) 17 75. The maximum value of z=3x+4y subject to the set of linear in equation given by

subject to constraints

67. The maximize value of

x+ y ≤ 40, x+ 2 y ≤ 60, x ≥ 0, y ≥0 is

x+ 2 y ≤ 8,3 x +2 y ≤ 12, x ≥0, y ≥0

b) 392 d) 395

a) 100 c) 125

a) 215 b) 217 c) 220 d) 218 72. The maximum value of z=2x+3y under the following constraints

a) 3 b) 3.5 c) 4 d) 5 74. The constraints in equation are given by

x 1+ x 2 ≤20, 3 x1 +2 x 2 ≤ 48, x1 ≥ 0, x 2 ≥ 0 is

66. The maximize value of

is

x 1−x 2 ≥ 1,−x 1+3 x 2 ≤ 15, x1 ≥ 0, x 2 ≥ 0 is

z=22 x1 +18 x 2

subject to

a) 390 c) 391

a) 250 b) 220 c) 200 d) 210 71. The minimum value of z=28x+28y subject to constraints

is

a) 46 b) 48 c) 47 d) 50 70. The minimum value of z=4x+8y under the following constraints

a) 2400 c) 2500

b) 2450 d) 2600

77. The minimum value of

z=40 x 1 +60 x2

subject to the following constraints

4 x 1 + x 2 ≥ 10,3 x 1 +2 x 2 ≥12, x 1 ≥ 0, x 2 ≥5 is

is

a) 300 b) 350 c) 400 d) 425 78. For the following shaded region in following figure, the linear constraints are : a)

x 1+2 x 2 ≤ 2,3 x 1 + x 2 ≥ 3, 4 x 1+3 x 2 ≥ 6, x 1 , x 2 ≥ 0

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b)

x 1+2 x 2 ≥ 2,3 x 1 + x 2 ≥ 3, 4 x 1+3 x 2 ≥6, x 1 , x 2 ≥ 0 c)

x 1+2 x 2 ≥ 2,3 x 1 + x 2 ≤ 3, 4 x 1+3 x 2 ≥ 6, x 1 , x 2 ≥ 0 d)

x 1+2 x 2 ≥ 2,3 x 1 + x 2 ≥ 3, 4 x 1+3 x 2 ≤6, x 1 , x 2 ≥ 0 79. The maximum value of z=4x+5y subject to the constraints

x+ y ≤ 20, x +2 y ≤ 35, x−3 y ≤ 12is a) 90 b) 93 c) 95 d) 100 80. The objective function z=4x+3y can be maximized subject to the constraints

3 x+ 4 y ≤24, 8 x+ 6 y ≤ 48, 0 ≤ x ≤5∧0 ≤ y ≤ 6 is

83. Two kind of food A and B are being considered to form a weekly diet. The minimum weekly requirement of fats, carbohydrates, and proteins are 16, 23 and 16 units respectively. One kg of food A has 5, 15 and 7 units of these ingredients respectively and one kg of food B has 12, 5 and 8 units of ingredients respectively. The price of food A is 4 per kg and food B is Rs. 3 per kg. Formulate this LLP to minimize the cost and set of solution. a) Minimize z=4x+3y subject to

5 x+12 y ≥16, 15 x +5 y ≥23, 7 x+ 8 y ≥ 16, x ≥ 0, y ≥ 0 b) Minimize z=4x+3y subject to

5 x+12 y ≤16, 15 x +5 y ≥23, 7 x+ 8 y ≥ 16, x ≥ 0, y ≥ 0 c) Minimize z=4x+3y subject to

5 x+12 y ≥16, 15 x +5 y ≤23, 7 x+ 8 y ≥ 16, x ≥ 0, y ≥ 0

d) Minimize z=4x+3y subject to a) At only one point. 5 x+12 y ≥16, 15 x +5 y ≥23, 7 x+ 8 y ≤ 16, x ≥ 0, y ≥ 0 b) Two points only c) At an infinite number of points 84. The maximum value of z=5x+4y subject d) None of these 81. The maximum value of z=5x+10y subject to constraints 2 x + y ≤2, x− y ≥ 3, x ≥ 0 is to the following set of in equations a) 6 b) 8 2 x +3 y ≤6, x+ 4 y ≤ 4, x ≥ 0, y ≥ 0 is c) -12 d) 12 85. The maximum value of z=3x+2y subject a) 15 to the following set of in equations b) 16 c) 16.5 d) 17 82. A firm manufactures headache pills of two sizes A & B. size A contains 2 units of aspirin , 5 units of bicarbonate and one unit of codeine, while size B contains 1 nit of aspirin, 8 units of bicarbonate and 6 unit of codeine. it is found that it requires 12 units of aspirin, 74 units of bicarbonate and 24 units of codeine for the relief of headache. It is require determining the last number of pills a patient must take to get relief. a)

2 x + y ≥12, 5 x+ 8 y ≥ 74, x +6 y ≥24, x , y ≥ 0 b)

2 x + y ≤12, 5 x+ 8 y ≥ 74, x +6 y ≥24, x , y ≥ 0 c)

2 x + y ≥12, 5 x+ 8 y ≤ 74, x +6 y ≥24, x , y ≥ 0 d)

x+ y ≥ 1, x− y ≥−1, y −5 x ≤ 0, x + y ≤6, 0 ≤ x ≤3, y ≥0 is

a) 10 b) 12 c) 15 d) 9 86. The minimum value of z=5(x)+7(y) subject to the following linear in equations representing the common feasible region

5 x+8 y ≤ 40, 3 x + y ≤ 6, x ≥ 0, y ≥2 is a) 16 b) 14 c) 15 d) 14.5 87. The maximum value of z=48x+40y subject to the following constraints

2 x + y ≤ 90, x +2 y ≤ 80, x+ y ≤ 50, x ≥0, y ≥0 is a) 2160 b) 2320 c) 2300 d) 2175 88. The minimum value of z=20x+30y subject to the following constraints

2 x +5 y ≤80, x+ y ≤ 20, x ≥ 0, y ≥ 0 is a) 300 c) 480

b) 400 d) None of these

2 x + y ≥12, 5 x+ 8 y ≥ 74, x +6 y ≤24, x , y ≥ 0 P a g e |12

89. For the following shaded region in following figure, the linear constraints( except

x ≥ 0, y ≥ 0 a)

)are :

x+ y ≥ 2, x − y ≥ 2

b) x+ y ≥ 2, x − y ≤ 2 c)

x+ y ≤ 2, x − y ≤ 2

d) x+ y ≤ 2, x − y ≥ 2 90. For the following shaded region in following figure, the linear constraints are : a)

x+ y ≥ 1,2 x− y ≥ 0, x ≥0, y ≥0

b) x+ y ≤ 1,2 x− y ≤ 0, x ≥0, y ≥0 c) x+ y−1 ≥0, 2 x− y ≤ 0, x ≥ 0, y ≥ 0 d) x+ y−1 ≤0, 2 x− y ≥ 0, x ≥ 0, y ≥ 0 91. The maximum value of z=4x+7y in the feasible region constraints by in equations

2 x + y ≤14, x +2 y ≥ 16, x ≥0, y ≥0 is

a) 3 x+ 4 y ≤12, x−2 y ≤ 2, x ≥ 0, y ≤ 0 b) 3 x+ 4 y ≤12, x−2 y ≥ 2, x ≥ 0, y ≤ 0 c)

3 x+ 4 y ≤12, x−2 y ≥ 2, x ≥ 0, y ≥ 0

d) 3 x+ 4 y ≥12, x−2 y ≥ 2, x ≥ 0, y ≤ 0 94. For the following shaded region in following figure, the linear constraints are :

a) 56 b) 58 c) 28 d) 42 92. For the following shaded region in following figure, the linear constraints are :

a) 4 x +2 y ≤ 40,2 x+5 y ≤ 90, x ≥ 0, y ≥ 0 b) 4 x +2 y ≥ 40,2 x+ 5 y ≤ 90, x ≥ 0, y ≥ 0 a) x+ 2 y ≥ 4, 2 x− y ≤ 6, y ≥ 0 b) x+ 2 y ≤ 4, 2 x− y ≤ 6, y ≥ 0 c) x+ 2 y ≥ 4, 2 x− y ≥ 6, y ≥ 0

c)

4 x +2 y ≤ 40,2 x+5 y ≥ 90, x ≥ 0, y ≥ 0

d) 4 x +2 y ≥ 40,2 x+ 5 y ≥ 90, x ≥ 0, y ≥ 0 95. For the following shaded region in following figure, the linear constraints are :

d) x+ 2 y ≥ 4, 2 x− y ≤ 6, x ≥0, y ≥0 93. For the following shaded region in following figure, the linear constraints are :

P a g e |13

a) 3 x+ 4 y ≤12, 2 x +5 y ≤10, x ≥ 0, y ≥ 0 b) 3 x+ 4 y ≥12, 2 x +5 y ≥10, x ≥ 0, y ≥ 0 c)

3 x+ 4 y ≤12, 2 x +5 y ≥10, x ≥ 0, y ≥ 0

d) 3 x+ 4 y ≥12, 2 x +5 y ≤10, x ≥ 0, y ≥ 0 96. For the following shaded region in following figure, the linear constraints are :

a) 2 x + y ≥ 2, x− y ≤1, x +2 y ≤ 8, x ≥ 0, y ≥ 0 b) 2 x + y ≤2, x− y ≤1, x +2 y ≤ 8, x ≥ 0, y ≥ 0 c)

2 x + y ≥ 2, x− y ≥ 1, x +2 y ≤ 8, x ≥ 0, y ≥ 0

d) 2 x + y ≥ 2, x− y ≤1, x +2 y ≥ 8, x ≥ 0, y ≥ 0 98. A manufacturer produces two items A and B. both are processes on two machines I and II.A need 1 and 2 hours on the machine and B need 3 hours on machine I and 1 hour on machine II. If machine I can run for maximum 12 hrs per day and II for 8 hours per day. Construct problem in form of in equalities and find feasible solution graphically. a) 2 x +3 y ≤12, 2 x + y ≥ 8, andx ≥ 0, y ≥ 0 b) 2 x +3 y ≥12, 2 x + y ≤ 8, andx ≥ 0, y ≥ 0 c)

a) x+ 2 y ≥10, x + y ≤ 6, 0 ≤ x ≥ 4, y ≥ 0 b) x+ 2 y ≤10, x + y ≥ 6, 0 ≤ x ≥ 4, y ≥ 0 c)

x+ 2 y ≤10, x + y ≤ 6, 0 ≤ x ≤ 4, y ≥ 0

d) x+ 2 y ≥10, x + y ≥ 6, 0 ≤ x ≤ 4, y ≥ 0 97. For the following shaded region in following figure, the linear constraints are :

2 x +3 y ≥12, 2 x + y ≥ 8, andx ≥ 0, y ≥ 0

d) 2 x +3 y ≤12, 2 x + y ≤ 8,andx ≥ 0, y ≥ 0 99. For the following shaded region in following figure, the linear constraints(except

x ≥ 0, y ≥ 0

)are :

a) 2 x +3 y ≤12, 2 x + y ≥ 8 b) 2 x +3 y ≥12, 2 x + y ≤ 8 c)

2 x +3 y ≥12, 2 x + y ≥ 8

d) 2 x +3 y ≤12, 2 x + y ≤ 8 100. The maximum value of z=6x+4y subject to

2 x +3 y ≤30, 3 x+2 y ≤24, x+ 4 y ≥ 3, x ≥0, y ≥0 is a) 50 c) 47

b) 48 d) 46 P a g e |14

101. The maximum value of z=3x+2y subject to

running feet of teak, plywood and rosewood respectively. If A would sell for Rs.48 per unit and B would sell for Rs.40 per unit. The 0 ≤ x ≤3, 0 ≤ y ≤ 3, x + y ≤5, 2 x+ y ≥ 4 is maximum income would be a) 12 b) 13 a) 2000 b) 2100 c) 14 d) 15 c) 2160 d) None of These 102. The minimum value of z=4x+5y subject 108. The minimum value of z=2x+3y subject to following in equations to linear constraints

5 x+ y ≥ 1,0, x + y ≥6, x+ 4 y ≥ 12, x ≥ 0, y ≥ 0 is a) 25 b) 25.5 c) 26 d) 27 103. The maximum value of z=40x+50y subject to

3 x+ y ≤ 9, x +2 y ≤ 8, x ≥ 0, y ≥ 0 is

a) 220 b) 230 c) 235 d) 240 104. The maximum value of z=2x+5y subject to

x+ 4 y ≤ 24,3 x+ y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0is a) 33 b) 32 c) 27 d) None of these 105. The maximum value of z=5x+4y

5 x+3 y ≥15, x +2 y ≥ 6, x ≥ 0, y ≥ 0is a) 69/7 b) 65/7 c) 64/7 d) None of these 109. The standard weight of a trick has to be at least 5 Kg and has to contain two basic ingredients

B 1∧B 2 . B 1 cost Rs. 5/Kg and

B 2 cost Rs. 8/Kg. strength consideration dictate that the trick should not contain more than 4 Kg of

B 1 and should contain a

minimum of 2 Kg of

B 2 .The minimum cost

of trick is a) 30 b) 32 c) 31 d) 31.5 a) 46 b) 48 110. A firm produces two products A and B. c) 44 d) None of these The profit on each unit of product A is Rs.40 106. A company manufactures two products and that of the B is Rs. 50.Both of the electric iron and ceiling fan. Each of these products are processed on three machines M1, undergoes assembly and finishing processes. M2, and M3.The time required in hours by The time in hours required for each unit of the each product and total time available in hours products for different processes and time per week on each machine is as follows. available for each process in day given as Product machines Product Assembly Process M1 M2 M3 Finishing Process A 3 4 2 Electric Iron 4 B 4 5 8 2 x=no. of product A Ceiling fan 2 Total available 36 48 70 6 y=no. of product B Hours available 16 Hrs 18 Formulate the problem as L.P.P. to maximize per day profit z=40x+50y subject to linear constraints. The profit on each unit electric iron and ceiling a) fan sold are Rs.40 and Rs.50 respectively. 3 x+ 4 y ≤36, 4 x +5 y ≤ 48,2 x +8 y ≤70, x ≥ 0, y ≥ 0 Then the maximum value of profit is a) 250 b) 240 b) c) 230 d) 220 3 x+ 4 y ≥36, 4 x+5 y ≤ 48,2 x +8 y ≤70, x ≥ 0, y ≥ 0 107. A carpenter has 90, 80, and 50 running feet of teak, plywood and rosewood c) respectively. The product A requires 2,1 and 1 3 x+ 4 y ≤36, 4 x +5 y ≥ 48,2 x +8 y ≤70, x ≥ 0, y ≥ 0 running feet and product B requires 1,2 and 1 subject to

x+ 2 y ≥ 6,5 x+ 3 y ≤ 15, x ≤6, y ≤ 4 is

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d)

3 x+ 4 y ≤36, 4 x +5 y ≤ 48,2 x +8 y ≥70, x ≥ 0, y ≥ 0 111. An agriculturist wishes to mix fertilizer that will provide minimum of 15 units of potash, 20 units of nitrates and 24 units of phosphates. One unit of A provides 3 units of potash, 1unit of nitrate and 3 unit of phosphates and cost Rs. 120 per unit. One unit of B provides 1 unit of potash, 5 unit of nitrates and 2 units of phosphate. It costs Rs.60 per unit. Formulate the problem as L.P.P. to minimize the cost a)

3 x+ y ≥ 15, x +5 y ≥ 20,3 x+ 2 y ≥ 24, x ≥ 0, y ≥0 b)

3 x+ y ≥ 15, x +5 y ≥ 20,3 x+ 2 y ≤24, x ≥ 0, y ≥ 0 c)

3 x+ y ≥ 15, x +5 y ≤ 20,3 x+ 2 y ≥24, x ≥ 0, y ≥ 0 d)

3 x+ y ≤ 15, x +5 y ≥ 20,3 x+ 2 y ≥24, x ≥ 0, y ≥ 0 112. A firm makes two types of furniture’s namely tables and chairs. The profit is Rs.20 per chair and Rs.30 per table. Both are processed on three machines M1, M2, M3.The time required in hours by each product and total available time in hours per week on each machines are as follows. Machines Chairs Tables Available time in hours M1 3 3 36 M2 5 2 50 M3 2 6 60 The maximum value of profit is a) 320 b) 325 c) 330 d) None of these 113. The maximum value of z=10x+15y subject to linear constraints

2 x +3 y ≤36, 5 x+2 y ≤50,2 x+ 6 y ≤ 60, x ≥ 0, y ≥ 0is a) 160 c) 180

b) 170 d) None of these

114. The maximum value of z=40x+100y subject to following linear in equation

12 x +6 y ≤3000, 4 x +10 y ≤2000,2 x+ 3 y ≤ 900, x ≥ 0, y ≥ 0is a) 18000 b) 20000 c) 21000 d) 20,500 115. A company produces cars of two types A and B. To stay in business it must produce at least 5o models of A per month. However it does not have the facilities to produce more than 200 of that model per month. Also it does not have the facilities to produce more than 150 of model B per month, while the total demand does not exceed 300 per month. If the profit is Rs. 400 on each model A and Rs. 300 on each model B. The maximum profit is a) 1,05,000 b) 1,10,000 c) 1,15,000 d) 1,20,000 116. Vijay want to invest Rs.15, 000 in saving certificate (SC) and fixed deposit (FD).He want to invest at least Rs.3000 in SC and at least Rs. 5000 in fixed deposit. The rates of interest on SC is 8% and that on FD is 10%.formulate this as LPP problem, taking Rs. X and y interest in SC and FD respectively. a)

x+ y ≤ 15000, x ≥3000, y ≤ 5000

b) x+ y ≥ 15000, x ≥3000, y ≥ 5000 c)

x+ y ≤ 15000, x ≥3000, y ≥ 5000

d) x+ y ≤ 15000, x ≤3000, y ≥ 5000 117. If Ajay drives a car at speed of 60 km/hr, he has to spend Rs.5 per km on petrol. If he drives at a greater speed of 90 km/hr. The petrol cost increases to Rs. 8 per km. he has Rs. 600 to spend on petrol and wishes to travel the maximum distance within an hour. Formulate the above problem as LPP .assuming X= distance travelled at speed of 60 km/hr y= distance travelled at speed of 90 km/hr a) maximum value of z=x+y subject constraints

x /60+ y /90 ≤1, 5 x+ 8 y ≤ 600, x ≥0, y ≥0 b) maximum value of z=x+y subject constraints

x /60+ y /90 ≤1, 5 x+ 8 y ≥ 600, x ≥0, y ≥ 0

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c) maximum value of z=x+y subject constraints

x /60+ y /90 ≥1, 5 x+ 8 y ≤ 600, x ≥0, y ≥ 0

Maximum storage volume z=8x+12y subject to following linear to constraints. a)

1000 x +2000 y ≤8000, 6 x +8 y ≤ 72, x ≥ 0, y ≥ 0

d) maximum value of z=x+y subject constraints

b)

x /60+ y /90 ≥1, 5 x+ 8 y ≥ 600, x ≥0, y ≥ 0 118. In order to ensure optimal health of rabbits, a lab technician nee to feed the rabbits a daily diet containing a minimum of 24 gm of fats, 36 gm of carbohydrates and 4 gm of protein. Food X contains 8 gm of fats, 12 gm of carbohydrates and 2 gm of protein per gram and cost Rs. 20 per gram. Food y contains 12 gm of fats 12 gm of carbohydrates and 1 gm of protein per gram at cost of Rs. 30 per gram. Formulate the problem as LPP in order to ensure optimal health of rabbits at minimum cost assuming. Minimum cost

z=20 x 1+30 x 2

subject to the

following constraints. Let

x 1 = quantity (in grams) of food x

Let

x 2 = quantity (in grams) of food y

1000 x +2000 y ≤8000, 6 x +8 y ≥ 72, x ≥ 0, y ≥ 0 c)

1000 x +2000 y ≥8000, 6 x +8 y ≥ 72, x ≥ 0, y ≥ 0 d)

1000 x +2000 y ≥8000, 6 x +8 y ≤ 72, x ≥ 0, y ≥ 0 120. A manager of a hotel plans an extension not more than 50 room. At least 5 must be executive single rooms. The number of executive double rooms should be at least 3 times the number of executive single room. He charges Rs.3000 for an executive double room and Rs. 1800 for an executive single room per day. Formulate the above problem as LPP to obtain maximum income. Let x= number of executive double rooms Let y= number of executive single room Maximum z=300x+1800y subject to constraints.

a)

8 x 1+12 x 2 ≥ 24,12 x 1 +12 x 2 ≥36, 2 x 1 + x 2 ≤ 4, x 1 ≥ 0, x 2 ≥0 b)

8 x 1+12 x 2 ≥ 24,12 x 1 +12 x 2 ≥36, 2 x 1 + x 2 ≥ 4, x 1 ≥ 0, x 2 ≥0 c)

a)

x+ y ≤ 50, y ≤ 3 x , 0 ≤ x ≤5, y ≥ 0

b) x+ y ≤ 50, y ≥ 3 x , 0 ≤ x ≤5, y ≥ 0 c)

x+ y ≥ 50, y ≥ 3 x , 0 ≤ x ≤5, y ≥0

d) x+ y ≥ 50, y ≤ 3 x , 0 ≤ x ≤5, y ≥ 0

121. Two products x and y when 8 x 1+12 x 2 ≥ 24,12 x 1 +12 x 2 ≤36, 2 x 1 + x 2 ≤ 4, x 1 ≥ 0, x 2 ≥0

manufactured must pass through machine operations I, II, III, are 46,100 and 300 Hrs respectively. Product A gives a profit of Rs. 6 8 x 1+12 x 2 ≤ 24,12 x 1 +12 x 2 ≥36, 2 x 1 + x 2 ≤ 4, x 1 ≥ 0, x 2 ≥0 per unit and product B gives a profit of Rs. 4 119. Vijay wants to buy some filling cabinets per unit. Formulate this problem as LPP for his office, the cost of cabinet x is Rs. 1000 problem to maximize the profit assuming per unit, requires 6 sq. feet of floor space and Let x= number of product A Let y= number of product B hold eight cubic feet of files. Cabinet y casts Rs.2000 per unit, requires eight square feet of Maximum z=6x+4y subject to constraints. a) floor space and hold 12 cubic feet of files. He has Rs. 8000 for this purchase through he 1 2 x + y ≤ 46, 4 x+ 2 y ≤ 100,3 x+ y ≤ 300, x ≥ 0, y ≥ 0 does not have to spend that much. The office 2 has room for no more than 72 square feet of b) cabinets. Formulate this problem as LLP in order to maximize storage volume. 1 2 x + y ≤ 46, 4 x+ 2 y ≤ 100,3 x+ y ≥ 300, x ≥ 0, y ≥ 0 Let x= no of cabinets x 2 Let y= no of cabinets y d)

P a g e |17

c)

N1 36 6 108 1 2 x + y ≤ 46, 4 x+ 2 y ≥ 100,3 x+ y ≤ 300, x ≥ 0, y ≥ 0 N2 3 12 2 36 d) N3 20 10 If the product A and B cost Rs. 20 and Rs.40 1 2 x + y ≥ 46, 4 x +2 y ≤ 100,3 x+ y ≤ 300, x ≥ 0, y ≥ 0 per unit respectively. Find how many each of 2 these two products should be bought to 122. A company produces two type of goods minimize cost assuming A and B, that requires gold and silver. Each x 1 = number of product A Let unit of type A requires 3 gm of silver and 1 gm of gold while that B requires 2 gm of silver and Let x 2 = number of product B 8 gm of gold. If each unit of type A bring out a a) 155 b) 150 profit of Rs.35 and that of type B Rs. 50, c) 160 d) 162 formulate this problem as LPP to get the 125. The maximum value of z=48x+40y maximum profit given subject to linear in equations Let x= number of good of type A 2 x + y ≤ 90, x +2 y ≤ 80, x+ y =50, x ≥0, y ≥ 0is y= number of good of type B a) a) 2300 b) 2320 maximizez=35 x +50 ysubjectto 3 x +2 y ≥ 12, x +2 y ≤8, x ≥c) 0, 2350 y≥0 d) 2360 126. The maximum value of z=16x+15y b) subject to linear in equations

maximizez=35 x +50 ysubjectto 3 x +2 y ≤ 12, x +2 y ≤8, x ≥ 0, y ≥ 0 2 x +3 y ≥12, 2 x + y ≥ 6, x ≥0, y ≥0 is c)

a) 50 b) 55 c) 60 d) 65 127. The maximum value of z=15x+10y d) subject maximizez=35 x +50 ysubjectto 3 x +2 y ≥ 12, x +2 y ≥8, x ≥ 0, y to ≥ 0linear in equations

maximizez=35 x +50 ysubjectto 3 x +2 y ≤ 12, x +2 y ≥8, x ≥ 0, y ≥ 0

123. A company produces bicycle and tricycles. Each of which must be produces through two types of machines A and B. machine A has a maximum of 120 hrs available and machine B has maximum of 180 Hrs available. Manufacturing tricycle requires 6 Hrs on machine A and 3 Hrs on machine B. manufacturing bicycle requires 4 hours of machine A and 10 hours on machine B. If profits are Rs. 50 for tricycle and 70 for bicycle. solve graphically and show that the maximum profit is a) 1500 b) 1550 c) 1600 d) 1650 124. At form horses are fed on various products having certain nutrient constituent necessary. The contents of the product (A and B) per unit in nutritional constituents are given below in the table. Nutrients Nutrient content in product Minimum quantity required A B

3 x+2 y ≤12, 2 x +3 y ≤15 is

a) 61 b) 60 c) 61.5 d) 60.5 128. For the following shaded region the linear constraints (except

x ≥ 0, y ≥ 0 ) are

a) 3 x+3 y ≤50, 3 x+ 5 y ≥ 80 b) 3 x+3 y ≤50, 3 x+ 5 y ≤ 80 c)

3 x+3 y ≥50, 3 x+ 5 y ≤ 80

d) 3 x+3 y ≥50, 3 x+ 5 y ≥ 80 129.

The vertex of common graph of in

equalities

2 x + y ≥ 2andx− y ≤ 3 is

a) (5/3,4/3) b) (5/3,-4/3) c) (-5/3,4/3) d) (-5/3,-4/3) 130. For the following shaded region, the linear constraints are: P a g e |18

a)

x+ 2 y ≤11, 3 x +4 y ≤ 30,2 x +5 y ≤ 30, x ≥ 0, y ≥ 0 b)

x+ 2 y ≥11, 3 x +4 y ≤ 30, 2 x +5 y ≤ 30, x ≥ 0, y ≥ 0 c)

x+ 2 y ≥11, 3 x +4 y ≥ 30, 2 x +5 y ≤ 30, x ≥ 0, y ≥ 0 d)

x+ 2 y ≥11, 3 x +4 y ≤ 30, 2 x +5 y ≥ 30, x ≥ 0, y ≥ 0 131. The true statement for the graph of in equations

3 x+2 y ≤6, 6 x+ 4 y ≥ 20, x ≥0, y ≥0 a) Both graph are disjoint b) Both do not contain origin c) Both contain point d) None of these 132. The minimum value of z=12x+8y subject to constraints

4 x + y ≥ 4, x+3 y ≥6, x + y ≥ 3, x ≥ 0, y ≥ 0 is

a) 25 b) 78/3 c) 76/3 d) 74/3 138. The minimum value of z=2x+3y subject to linear in equations

2 x +7 y ≥22, x + y ≤ 6,5 x+ y ≥ 10 is a) 13 b) 14 c) 15 d) 16 139. Solve graphically z=15x+12y subject to linear in equations

2 x + y ≥ 2, x− y ≤ 3, x ≥ 0

then vertex of in equalities is a) (5/3,4/3) b) (-5/3,4/3) c) (-5/3,-4/3) d) (5/3,-4/3) 140. The maximum value of z=40x+50y subject to linear constraints is

3 x+ y ≤ 9, x +2 y ≤ 8, x ≥ 0, y ≥ 0 is a) 230 b) 220 c) 215 d) 225 141. The maximum value of z=60x+15y subject to linear constraints is

x+ y ≤ 50,3 x + y ≤90, x ≥ 0, y ≥ 0 is

a) 76/6 b) 76/5 a) 1850 b) 1800 c) 76/3 d) 74/3 c) 1700 d) 1750 133. The minimum value of z=x+y subject to 142. Two types of food packets A and B are constraints available. Each contains vitamins A1 and B1.A x+ y ≤ 8, 6 x+ 4 y ≥12,5 x+ 8 y ≥ 20, x ≥ 0, y ≥ 0 is person needs 4 decigrams of A1 and 12 decigrams of B1 every day. Food packets A a) 16/7 b) 3 contains 2 decigram of vitamin A1 and 4 c) 19/7 d) 18/7 decigram of vitamin B1.Food packets A and B 134. The maximum value of z=x+y subject cost Rs.15 and Rs 10 for A and B respectively. to following in equations solve LLP graphically to minimize the cost. x− y ≥0, 3 x+ y ≥ 3, x ≥ 0, y ≥ 0is a) 30 b) 40 a) 8 b) 8.5 c) 35 d) 32 c) 7 d) 7.5 143. If a man rides his motor cycle at 135. The maximum value of z=5x+5y 25km/hr, he has to spend Rs.2 per km on subject to constraints petrol. If he rides it a faster speed of 40 km/hr, the petrol cost increases to Rs. 5 per km. He 10 x+7 y ≤ 35,3 x+ 8 y ≤ 24, x + y ≤ 4, x ≥ 0, y ≥ 0 is has Rs. 100 to spend on petrol and wishes to a) 15 b) 20 find what is the maximum distance he travel c) 18 d) 16 within 1 hour. 136. The maximum value of z=3x+5y a) 20 b) 30 subject to constraints c) 25 d) None of these x+ y ≤ 4, 2 x +3 y ≤12,0 ≤ x ≤ 2,0 ≤ y ≤3 is 144. A machine can produces by using 2 units of chemicals and 1 units of a compound a) 16 b) 18 and can produce product B by using unit of c) 17 d) 15 chemicals and 2 units of compound. Only 800 137. The minimum value of z=12x+8y units of chemical and 1000 units of compound subject to linear in equations are available. The profit per unit of A and B 4 x + y ≥ 4, x+ y ≥ 3, x +3 y ≥ 6, x ≥ 0, y ≥ 0 is are Rs.30 and Rs.20 respectively. Formulate a P a g e |19

LLP and solve graphically to maximize profit.max. profit is a) 12000 b) 13000 c) 14000 d) 15000 145. A diet for the seek person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 units of calories. Two foods A and B are available at a cost Rs. 4 and Rs.3 per unit respectively. If one unit of A contains 200 units of vitamins, 1 units minerals and 40 units of calories. One unit of food B contains 100 units of vitamins, 2 units of minerals and 40 units of calories. Minimum value of z=4x+3y is a) 120 b) 110 c) 1005 d) 115 146. A has Rs.1500 for purchase of rice and wheat. A bag of rice and wheat cost Rs.180 and Rs.120 respectively. He has storage capacity of 10 bags only. He earns a profit of Rs. 11 and Rs. 9 per bag of rice and wheat. Maximum value of profit is a) 150 b) 100 c) 120 d) 110 147. A firm manufactures two products A and B on which profit earned per unit Rs.3 and Rs. 4 respectively. each product is processed on two machines M1 and M2.the product A requires one minute of processing time on M1 and 2 minute on M2.B requires one minute of M1 and one minute of M2.M1 is available for not more than 600 minutes during any day. The value of maximum profit is a) 1600 b) 1620 c) 1650 d) None 148. If one gram of wheat provides 0.1 gm of proteins and 0.25 gm of carbohydrates and the corresponding values of rice are 0.05 gm and 0.5 gm of carbohydrates. Wheat costs Rs.2 per Kg and rice Rs. 8 per Kg. The minimum requirements of proteins and carbohydrates for an average child are 50 g and 200 gm respectively. Find the quantities in which wheat and rice should be mixed in the daily diet to provide minimum daily requirements of proteins and carbohydrates. The value of minimum cost is a) 2.4 b) 2.5 c) 2.6 d) 2.3 149. A carpenter has 20 and 15 sq. meters of plywood and sun mica respectively. He

produces A and B. Product A requires 2 and 1 sq. meters and product B requires 1 and 3 sq. meters of plywood and sun mica respectively, if the profit on one piece of product A is Rs.30 and that on the piece of product B is Rs.20,The no. of pieces of product A and B to be made to maximize his profit are a) (9,2) b) (2,9) c) (2,8) d) (8,2) 150. The extreme values of the function

f ( x , y ) =2 x + y subject to

over the convex polygon x

x ≤ 2, y ≥−2, x +2 y ≤ 4, is

a) 4,-8 b) 4,8 c) 8,4 d) -8,4 151. Which of the following is the common region for

x+ y ≥ 2, x +3 y ≥ 9, x ≤2, x ≥ 0, y ≥ 0

a)

b)

c)

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a) 15 b) 16 c) 17 d) 14 155. Solve the following LPP graphically, Maximize

7 z=3 x+ 4 ysubjectto 0 ≤ x ≤ 4, y ≥ , x+ y ≤ 5,2 x + y ≥ 4, x ≥ 0, 2

d) 152. A business firm produces two types of product I and II. The profit for product I Rs.100 per ton and that for product II is Rs. 70 per ton. The plant of three production department A, B and C. The equipment in each department can be used for 8 h0urs a day. Production I requires 2 hours in department C per ton. Product II requires 1 hour in department B and 1 hour in department C per ton. Formulate the problem as LPP to maximize the profit is a) 690 b) 680 c) 700 d) None of these 153. A company manufacturer two type of toy A and B, each type of toy requires 2 minutes for cutting and 1 minute for assembling. Each toy of type B requires 3 minute for cutting and 4 minutes of assembling. There are 3 hours available for cutting and 2 hours 40 minutes available for assembling. On selling a toy of type A , he get a profit Rs. 10 and that on selling a toy of type B , he gets profit of Rs.20The value of maximum profit is a) 17 1 c) 173

.the maximum value of Z is a) 18.5 b) 18 c) 17.5 d) 17 156. The maximum value of

z=5 x+ 3 ysubjectto 2 x+3 y ≤ 30,3 x+ 2 y ≤ 24, x +4 y ≥ 3, 0≤ is a) 40 c) 39 157.

The maximum value of

z=x + y

subject to linear constraints

3 x+ 4 y ≤18, x−6 y ≤3, 2 x +3 y ≥3, x , y ≥ 0 is a) 155/22 b) 125/22 c) 130/22 d) 128/22 158. The maximum value of

z=6 x +9 ysubjecttolinearconstraintsx+ y ≤ 20,, 2 x +5 y ≤ 80, is a) 155 b) 160 c) 150 d) 165 159. The extreme value of linear function z(x,y)=2x+5y over the convex polygon given by

x+ 3 y ≤ 3, x− y ≤ 3, y ≥−1, x ≥ 1 are

a) 3,6 b) -6,3 c) -3,6 d) 6,-3 160. The maximum value of

z=2 x+3 ysubjecttolinearequationsx + y ≥5, 5 x+ 3 y ≥ 15,2 x

1

b) 172 d) 18

b) 41 d) 34.5

is a) 10 b) 12 c) 11 d) 10.5 161. The minimum value of

154. A company produces two types of vehicles car and jeeps. Each of these are z=2 x+ ysubjecttolinearequations 2 x + y ≥ 4, x + y ≤3, 2 x−3 produced on three machine I,II and III.A car needs 4 hours of machine I, 3 hours of is machine II and 2 hours of machine III and jeep a) 3 b) 4 needs 3 hours, 2 hours and 3 hours of c) 4.5 d) 3.5 machine I, II and III. machine I, II, and III are to 162. The minimum value of be used at least 16 hrs, 12 hrs and 10 hrs a z=3 x+ 3 ysubjecttolinearequations 7 x+ 4 y ≥28, x+ y ≤ 6, x ≥ day respectively to minimize the cost .the production cost of a car and jeep are Rs. 4 is lacs and Rs.3 Lacs respectively. construct the a) 12 b) 13 problem as LPP .The minimum value of cost is c) 14 d) None P a g e |21

163.

The minimum value of

c) 4 d) 6 171. z=2 x+3 ysubjecttolinearequationsx + y ≥1, y ≥2 x , x +2 y ≤ 6,A person wants to invest up to an amount of Rs. 30,000 in fixed income is securities. His friend recommended investing a) 8/3 b) 7/3 in two bonds, bond A yielding 7% per annum c) 5/3 d) None of these and bond B yielding 10 % per annum. he 164. The minimum value of decides to invest at the most Rs.1200 in bond z=3 x+ 6 ysubjecttolinearequations 8 x+ 9 y ≤ 72, x + yB≥6,4 y ≥ 28,Rs.6000 x ≥0, y ≥0in bond A. he also want andx+7 at least that the amount invested in bond A should be is more than the amount invested in bond B. a) 20 b) 21 maximum value of profit is c) 22 d) None of these a) 2360 b) 2460 165. The minimum value of c) 2820 d) None z=5 x+10 ysubjecttolinearequationsx −2 y ≤ 4, x+3 y 172. ≥ 3,2 x +Solve y ≥ 2, xgraphically ≥ 0, y ≥ 0 following linear in equations and minimize z=5x+8y subject to is x+ y ≥ 5,3 x +3 y ≥ 12, x ≤ 4, y ≥ 2, x , y ≥ 0 The a) 10 b) 10.5 c) 11 d) 11.5 minimum value of z is 166. The minimum value of a) 30 b) 30.5

z=2 x+3 ysubjecttolinearequationsx −2 y ≤−1, 2 x − y ≥ 1,c) x +31 4 y ≤ 4, x ≥0, y ≥0

is a) 2 b) 3 c) 5 d) 29/3 167. The maximum value of

d) 32 173. Solve graphically following linear in equations and maximize z=4x+y subject to

3 x+2 y ≤30, x+ y ≥ 10,2 x + y ≥ 10, x , y ≥ 0

and

shade the common feasible region. The

z=3 x+ 2 ysubjecttolinearconstraints 3 x +4 y ≥ 12, x−2 y ≤3, x ≥ 0,value y ≥ 0 of z is maximum is a) 16 b) 17 c) 15 d) None of these 168. Solve graphically linear in equations and shade common feasible region.

a) 40 b) 45 c) 30 d) None of these. 174. Solve graphically following linear in equations and minimize z=40x+100y subject to

12 x +6 y ≤3000, 4 x +10 y ≤2000, 2 x +3 y ≤ 900, x , y ≥ 0 2 x +3 y ≥3, 3 x+ 4 y ≤18,−7 x+ 4 y ≤14, x−6 y ≤3, x ≥ 0, y ≥ 0

The maximum value of z=x+y is a) 129/11 b) 129/22 c) 131/11 d) 133/22 169. The shaded common feasible region is given by linear constraints other than

x ≥ 0, y ≥ 0

are

x+ y ≤ 8, x+2 y ≥ 4,6 x +4 y ≥ 12, x , y ≥ 0

a) 3 x−2 y ≥−3, 4 x+ 6 y ≤ 24,2 x +3 y ≤ 6 b) 3 x−2 y ≤−3, 4 x+ 6 y ≥ 24,2 x +3 y ≥ 6 c)

3 x−2 y ≥−3, 4 x+ 6 y ≤ 24,2 x +3 y ≥ 6

d) 3 x−2 y ≤−3, 4 x+ 6 y ≤ 24,2 x +3 y ≥ 6 170. Solve graphically following linear in equations and minimize z=2x+3y subject to

2 x −3 y ≤ 6,2 x + y ≥ 4, x + y ≤ 3 a) 7

and shade the common feasible region. The minimum value of z is a) 16000 b) 17000 c) 18000 d) None of these. 175. Solve graphically following linear in equations and minimize z=30x+20y subject to and

shade the common feasible region. The minimum value of z is a) 60 b) 50 c) 40 d) None of these. 176. Maximize z=2x+5y subject to constraints

3 x+ y ≤ 21, x +4 y ≤ 24, x + y ≥ 9, x , y ≥ 0 a) 31 c) 32

is

b) 33 d) 35

b) 5 P a g e |22

177.

For the following shaded region the

linear constraints except x , y ≥0 a)

are

x+ y ≤ 4, 3 x+ 8 y ≤ 24,5 x +3 y ≥ 18

b) x+ y ≤ 4, 3 x+ 8 y ≥ 24,5 x +3 y ≤ 18 c)

x+ y ≤ 4, 3 x+ 8 y ≤ 24,5 x +3 y ≤ 18

d) x+ y ≥ 4, 3 x+ 8 y ≤ 24,5 x +3 y ≤ 18 178.

The profit on each toaster and mixture are Rs.50 and Rs.100.The basic feasible solution are a)

4 x +2 y +s 1 ≤ 16, 2 x +6 y + s2 ≤18, x1 x 2 , s 1 , s2 ≥ 0, b)

4 x +2 y +s 1 ≤ 16, 2 x +6 y + s2 ≥18, x1 x 2 , s 1 , s2 ≥ 0, c)

The maximum value of

4 x +2 y +s 1 ≥ 16, 2 x +6 y + s2 ≤18, x1 x 2 , s 1 , s2 ≥ 0, z=3 x+ 4 ysubjecttolinearequationsx + y ≤ 40, x +2 y ≤60, x+ 2 y ≤70, x ≥ 0, y ≥ 0

is a) 120 b) 130 c) 135 d) 140 179. The maximum value of

d)

4 x +2 y +s 1 ≥ 16, 2 x +6 y + s2 ≥18, x1 x 2 , s1 , s2 ≥ 0,

183. To formulate the problem for solution by the simplex method, we must add artificial z=3 x+ 4 ysubjecttolinearequationsx + y ≤ 300, x +2 y ≤variables 400, x+2 yto≤500, x ≥ 0, y ≥ 0 a) Only equality constraints is b) Only greater than constraints. a) 1010 b) 1000 c) Both (a) and (b) c) 1020 d) 1015 d) None of these. 180. The maximum value of 184. If for a given solution, a slack variable is z=3 x+ 4 ysubjecttolinearequationsx +2 y ≤8, 2 x + y ≤7, x + y ≤ 4.5, x ≥ 0, y ≥1 equal to zero then a) The solution is optimal. is b) The solution is infeasible. a) 16 b) 18 c) The entire amount of resources with the c) 17 d) 20 constraints in which the slack variable 181. Minimize has assumed. z=21 x1 +15 x 2 subjectto x 1 +2 x 2 ≤6, 4 x 1 +3 x 2 ≤ 12, x 1 ,appears x2 ≥ 0 d) All the above. Express above LPP in standard form. All basic 185. For maximization LP model, the simplex model is terminated when all values feasible solutions are a) 2 b) 3 c) 4 d) 6 182. A company manufacture two products namely toaster and mixture. Each of them are first assembled and then finished. The time required for each unit of product for assembly and finishing processes in a day are given below Product assembly finishing Toaster 4 3 Electric mixture 2 6 Hrs available 16 18 per day

a)

C j−z j ≥ 0

b)

C j−z j ≤ 0

c)

C j−z j =0

d)

z j ≤0

186.

The maximum value of

z=4 x+5 ysubjecttolinearequations 2 x +3 y ≤ 12,2 x + y ≤ 8, x is a) 20 b) 22 c) 21 d) 24 187. A company manufactures two types of cricket bats B1 and B2.Both the products pass through machines I and II. The time required for processing each product on the machine I and II and availability of these machines is given below. Product machine I II 4 5 3 7 P a g e |23

Available 1800 1500 Material is enough to produce 300 types of bats B1 and 400 types of B2.each unit of bat gives a profit of Rs.50 and Rs.100.Formulate the above A LPP, The constraints are a)

4 x +3 y ≤1800, 5 x+7 y ≥ 1500,0 ≤ x ≤ 300,0≤ y ≤ 400 b)

4 x +3 y ≤1800, 5 x+7 y ≤1500, 0 ≤ x ≤ 300,0≤ y ≤ 400 c)

4 x +3 y ≥1800, 5 x+7 y ≤ 1500,0 ≤ x ≤ 300,0≤ y ≤ 400 d)

4 x +3 y ≥1800, 5 x+7 y ≥ 1500,0 ≤ x ≤ 300,0≤ y ≤ 400 188. The maximum value of the linear objective function z in given feasible region is

contains 1 units of potash, 8 units of phosphate and 3 units of nitrate. Formulate the problem to maximize profit, given that type A gives Rs.50 and type B gives Rs.70 profit per unit. The maximum value of profit z is a) 250 b) 249 c) 248 d) 252 193. A firm produces two types of gadgets A and B. They are produced at foundry and then sent to finishing to the machine shop. The no of man hours of labor required in each shop for production of A and B, the available man hours are as follows. Product foundry machining A 10 5 B 6 4 Time available 60 35 Formulate the above problem as an LLP.

z=4 x+5 ysubjecttox+ y ≤ 8, x ≥ 2 y , x+ y ≤ 8, x ≥ 0, y ≥ 0 a) 17 c)

b) 17.5 1 3

17

b) 10 x+6 y ≤ 60,5 x +4 y ≥ 35, x ≥0, y ≥ 0

d) 18

189. For the following shaded region, the linear in equations are: a) x−2 y ≥ 0, 4 x +3 y ≥24, 0 ≤ x ≤5, y ≥0 b) x−2 y ≥ 0, 4 x +3 y ≤24, 0 ≤ x ≤5, y ≥0 c)

x−2 y ≤ 0, 4 x +3 y ≤ 24, 0 ≤ x ≤5, y ≥0

d) x−2 y ≤ 0, 4 x +3 y ≥ 24, 0 ≤ x ≤5, y ≥0 190.

The maximum value of

z=3 x+ 6 ysubjecttolinearequationsy≥ is

a) 10 x+6 y ≤ 60,5 x +4 y ≥ 35, x ≥0, y ≥ 0

c)

10 x+6 y ≥ 60,5 x +4 y ≤ 35, x ≥0, y ≥ 0

d) 10 x+6 y ≥ 60,5 x +4 y ≥ 35, x ≥ 0, y ≥ 0 194. A shopkeeper buys 2 types of food for his shop .tea at Rs. 4o a packet and coffee at Rs. 60 a tin. He has Rs. 1500 available and decides that at least 30 tins should be bought and one third of the tins should be of coffee. He makes a profit of Rs.10 on tins of tea and Rs 20 on a tin of coffee. Formulate the problem as LPP for maximum profit.

x , y ≤ 2 x , x + y ≤ 8, x ≥0,2yx≥+3 0 y ≥75, x+ y ≥ 30, 2 y ≥ x , x ≥ 0, y ≥ 0 2 a) b) 2 x +3 y ≤75, x+ y ≤ 30,2 y ≥ x , x ≥ 0, y ≥ 0

a) 50 b) 60 c) 2 x +3 y ≤75, x+ y ≥ 30,2 y ≤ x , x ≥ 0, y ≥ 0 c) 40 d) 55 191. Laura is to buy some orange and d) 2 x +3 y ≤75, x+ y ≥ 30,2 y ≥ x , x ≥ 0, y ≥ 0 lemons. Lemans are Rs. 50 and orange are for 195. A firm produces two types of health pills Rs. 25.she must not buy more than 2 kg of A and B.A contains 2 units of proteins 5 units lemons and she must buy at least 4 kg of of vitamin D and one unit of vitamin B1.B orange. She is told to at least 6 kg of fruits contains 1,8 and 6 units of protein, vitamin D together. Minimum value of cost z is and vitamin B1.it is found that it requires 12 a) 140 b) 150 unit of proteins ,26 units of vitamin D and 24 c) 160 d) 170 units of vitamin D. formulate this problem as 192. A agriculturalist wishes to mix two LLP . brands of fertilizer type A and B. Type A a) contains 1 unit of potash , 3 units of 2 x + y ≥12, 5 x+ 8 y ≥ 26, x +6 y ≥ 24, x , y ≥ 0 phosphates and 5 units of nitrate. Type B P a g e |24

b)

2 x + y ≤12, 5 x+ 8 y ≥ 26, x +6 y ≥24, x , y ≥ 0 c)

Rs.350 and a sewing machine Rs.250.He sells a fan at a profit of Rs.20 and a machine at the profit of Rs.15.formulate this problem as LPP to maximize profit solution.

2 x + y ≥12, 5 x+ 8 y ≥ 26, x +6 y ≤24, x , y ≥ 0 d) None of these 196. A farm is engaged in feeding pigs, who are fed on various farm products .in view of need to ensure certain nutrient constituent it is necessary to buy 2 product A and B. The content of products A and B are given below. Nutrient nutrient content in minimum amount A B M1 36 6 108 M2 3 12 36 M3 20 10 100 Formulate this problem as LLP a)

a) 350 x+250 y ≤ 4000, x + y ≤25, x , y ≥ 0 b) 350 x+250 y ≤ 4000, x + y ≥25, x , y ≥ 0

350 x+250 y ≥ 4000, x + y ≤25, x , y ≥ 0

c)

d) 350 x+250 y ≥ 4000, x + y ≥25, x , y ≥ 0

199. A foundry produces two kinds of steel casting P and Q. The product is to be fettled, machined and then finishing process. The time in hrs required is as follows. P Q Fettling 2 ½ Machining 4 2 Finishing 3 1 Total hrs available for fettling. Machining and 36 x+ 6 y ≥ 108,3 x +12 y ≥ 36,20 x +10 y ≥ 100, x , y ≥ 0finishing are 45, 150 and 250 respectively. Profit P costs Rs.5 and B gives a profit of Rs.4 b) per unit. Formulate this problem as LPP to 36 x+ 6 y ≥ 108,3 x +12 y ≥ 36,20 x +10 y ≤ 100, x , y ≥ 0maximize profit. a) c)

36 x+ 6 y ≥ 108,3 x +12 y ≤ 36,20 x +10 y ≥ 100, x , y ≥ 0 2 x 1 + 1 x 2 ≤ 45, 4 x 1 +2 x 2 ≤150, 3 x1 + x 2 ≥ 250 x 1 , x 2 ≥ 0 2 d)

36 x+ 6 y ≤ 108,3 x +12 y ≥ 36,20 x +10 y ≥ 100, x , y ≥ 0

b)

1

2 x 1 + x 2 ≤ 45, 4 x 1 +2 x 2 ≤150, 3 x1 + x 2 ≤ 250 x 1 , x 2 ≥ 0 197. A former want to make sure that his 2 herd gets the minimum daily requirement of c) three basic nutrients A, B and C. daily requirements are 15 units of A , 20 units of B 1 and one unit of C. one gram of product P has 2 2 x 1 + 2 x 2 ≤ 45, 4 x 1 +2 x 2 ≥150, 3 x1 + x 2 ≤ 250 x 1 , x 2 ≥ 0 units of A , one unit of B and one unit of C. d) The cost of P is Rs.12 per gram and the cost of Q is Rs.18 per gram. Formulate this problem 1 2 x 1 + x 2 ≥ 45, 4 x 1 +2 x 2 ≤150, 3 x1 + x 2 ≤ 250 x 1 , x 2 ≥ 0 as LPP to determine units of p and Q should 2 the farmer buy so that the cost is minimum. 200. A company produces two types of food Then the system of in equations are stuffs F1 and F2 which contains vitamins V1, a) 2 x + y ≤15, x + y ≥ 20, x+3 y ≥ 0, x , y ≥ 0 V2 and V3.F1 contains 1 mg of V1 , and 2 mg 2 x + y ≥15, x + y ≤ 20, x+3 y ≥ 0, x , y ≥ 0 of V2, and 1 mg of V3,whereas F2 contains b) 1mg of V1 , 1 mg of V2 and 4 mg of c) 2 x + y ≥15, x + y ≥ 20, x+3 y ≥ 0, x , y ≥ 0 V3.minimum daily requirements of these vitamins are 6 mg of V1, 7 mg of V2 and 8 mg d) 2 x + y ≥15, x + y ≥ 20, x+3 y ≤ 0, x , y ≥ 0 of V3.The cost of 1unit of F1 is Rs.2 and that 198. A dealer wishes to purchase a no. of of F2 is Rs.3.write the linear problem to find fans and sewing machines. He has Rs. 4000 to the least expensive diet that would supply to invest and has space of 25 items. A fan costs body . P a g e |25

x+ y ≥ 6, 2 x + y ≥7, x+ 4 y ≤ 8, x , y ≥ 0

a)

b) x+ y ≥ 6, 2 x + y ≤7, x+ 4 y ≥ 8, x , y ≥ 0

x+ y ≥ 6, 2 x + y ≥7, x+ 4 y ≥ 8, x , y ≥ 0

c)

d) x+ y ≤ 6, 2 x + y ≥ 7, x+ 4 y ≥ 8, x , y ≥ 0 201.

The maximize value of

z=5 x+10 y

subject to the constraints

2 x +3 y ≤18, x+ 4 y ≤16, x ≤ 6, x ≥ 0, y ≥ 0

at the

c) (-4,0) (10,4) (9,4) d) (4,0) (4,10) (9,4) 206. The convex polygon given by

x ≤ 4, x− y ≥0, 3 x+ y ≥ 3 the vertices of convex polygon a) (4,9)(4,-4)(-1,0) b) (4,9)(4,-4)(-1,0) c) (4,-9)(4,4)(1,0) d) (4,9)(4,4)(-1,0) 207. The common region represented by the inequalities.

5 x+ 4 y ≥20, x ≤ 6 andy ≤ 4, x ≥ 0, y ≥ 0 is point a) (6,0) b) (6,2) a) (1,4) (-4,0) (6,0) c) (4.8,2.8) d) (0,4) b) (4,0)(-6,0)(1,-4) 202. A printing company prints two types of magazines c) (4,0) (6,0) (6,4)(1,4) A and B. The company earns Rs. 10 and Rs.15 on each d) (4,0)(-6,0)(6,4) magazine A and B respectively. These are processed on 208. Maximize z=x + y subject to three machines I, II, and III and total times in hour x+ 2 y ≤ 8 and 3 x+2 y ≤12, x ≥ 0, y ≥ 0 available per week on each machine is as follows

is

a) Zmax=7 b) Zmax=5 c)

Zmax=6

d) Zmax=9 209.

Minimize

z=x + y ,

subject to

5 x+2 y ≥10,2 x+ 3 y ≥ 6, x ≥1, x ≥ 0 andy ≥0 a) 5 c) 4 203.

Minimum value of

b) 3 d) 6

z=4 x+ 8 y

is

a) Z min ¿ 20/11 under

the constraints

b) Z min ¿ 27/11 c)

Z min ¿ 28/11

200 x+100 y ≥2000, x+ 2 y ≥ 50,40 x + 40 y ≥ 1400, x ≥ 0, y ≥0

d) Z min ¿ 25/11

is a) 225 b) 175 c) 200 d) 280 204. If the constrains in an linear programming problems are changed: a) Solution is not defined. b) The objective function has to be modified. c) The problem is to be revaluated. d) The change in constraints is ignored. 205. The common region determined by

15 y+ 12 x ≤ 80, 3 x−4 y ≤ 12, andx ≥ 4, x ≥ 0, y ≥ 0 is

210. The following inequalities and indicate the solution set

x+ 2 y ≥ 2,3 x+ y ≥ 3, 4 x +3 y ≥ 6, x ≥ 0, y ≥ 0 a) (2,0)(0,3) b) (2,1)(2,3) c) (-2,0)(0,3) d) (2,6)(0,-3) 211. The Maximum value of the objective function

z=x + y

subject to constraints

x+ 2 y ≤ 8,3 x +2 y ≤ 12andx ≥ 0, y ≥ 0 a) 10 c) 5

b) 20 d) 15

a) (4,0) (10,-4) (-9,-4) b) (-4,0) (10,-4) (9,-4) P a g e |26

212.

z=6 x +3 y

The maximum value

2 x +3 y ≤13, x+ y ≤ 5, andx ≥ 0 andy ≥ 0 is

213.

b) 20 d) 21

The minimum value of

p=7 x +4 y

subject to constraints

x+ y ≤ 8, 6 x+ 4 y ≥12, 5 x+8 y ≥ 20 andx ≥ 0, y ≥ 0 is a) 56 c) 58 214.

b) 32 d) 12

The minimum value of

p=12 x+ 8 y

subject to constraints

4 x + y ≥ 4, x+3 y ≥6, x + y ≥ 3 andx ≥0, y ≥0 is a) 36 c) 24 215.

b) 32 d) 25.33

The maximum value of

subject to

216.

z=20 x+ 50 y

x+ y ≤ 5, 0≤ x ≤ 3∧0 ≤ y ≤3 is

a) 180 c) 190 The lines

The point which the minimum value of

p=30 x +20 y

subject to constraints a) 30 c) 0

220.

b) 90 d) 195

5 x+ 4 y ≥20, x ≤ 6, y ≤ 4 form

[M.H.C.E.T. 2005] a) A square b) A rhombus c) A triangle d) A quadrilateral 217. The point which the minimum value of

subject to constraints

x+ y ≤ 8, x+2 y ≥ 4,6 x +4 y ≥ 12 andx ≥ 0, y ≥ 0 is a) (8,0) b) (1,1.5) c) (4,0) d) (1.5,2) 221. Which of the following statements is correct? a) If a L.P.P. admits two optimal solutions, then it has an infinite number of optimal solutions. b) Every L.P.P. admits an optimal solution c) A L.P.P. admits a unique solution d) All L.P.P. admits an optimal solution 222. If the constraints in an linear programming problem are changed a) Solution is not defined. b) The objective function has to be modified. c) The problem is to be revaluated. d) The change in constraints is ignored 223. The maximum and minimum values of P ,

p=20 x +50 y subject to

x+ y ≤ 5, 0≤ y ≤3,

are

respectively a) 150,60 b) 160,0 c) 60,0 d) 190,0 224. The following inequalities and indicate the solution set.

5 x+ y ≥ 10, x + y ≥6, x+ 4 y ≥ 8, x ≥0 andy ≥ 0 is

a) (8,0)(5,-1)(-2,5) b) (8,0)(5,1)(2,5)(0,10) c) (-8,0)(5,-1)(2,-5) p=4 x+ 5 y subject to constraints d) (8,0)(-5,-1)(2,5) 225. The solution set of the following system of 5 x+ y ≥ 10,2 x+ 2 y ≥12, 5 x+ 8 y ≥ 20 andx ≥ 0, y ≥ 0 is inequalities Also find the co-ordinates of the vertices of the common region. a) (2,4) b) (1,5) x+ 2 y ≥ 6,5 x+ 3 y ≥ 15, x ≥ 0 andy ≥ 0 is c) (4,2) d) (0,10) 218. The corner points of common region are a) (-6,0)(2,2.5)(0,-5)

2 x + y ≥ 9, x +2 y ≥ 9, x+ y ≥ 7 andx ≥ 0, y ≥ 0 is

b) (6,0)(2,-2.5) c) (6,0)(-2,2.5)(0,5) a) (0,9)(2,5)(0,9) b) (0,9)(2,5)(2,5) d) (6,0)(2,2.5)(0,5) c) (0,9)(2,5)(0,9) d) (9,0)(5,2)(2,5)(0,9) 226. The solution set of the following inequalities. Write the co-ordinates of the vertices of the feasible 219. Which of the following is not a convex region. set? if

a) {( x , y ) :2 x+2 y ≤7 }

b)

{( x , y ) : x 2+ y❑2 ≤ 4 } c)

{x : x =5 }

d)

3 x+ 4 y ≥12, 6 x+ 3 y ≥ 18, x ≥0 andy ≥ 0 is a) 4,0)(0,-6) b) (-4,0)(0,-6) c) (-4,0)(0,6) d) (4,0)(0,6)

2 2 {( x , y ) :3 x +2 y ❑ ≤6 }

P a g e |27

227.

234.

The solution set of inequalities.

2 x +3 y ≥6, x +4 y ≤ 4, x ≥ 0 andy ≥ 0

contains

a) (3, 0) and (0, 4) b) (0, 4) and (3, 0) c) (0, 3) and (4, 0) d) (3, 0) and (4, 0) 228. The solution set of inequalities.

x+ y ≥ 1,7 x +9 y ≤63, y ≤5, x ≤ 6, x ≥ 0 andy ≥ 0

is

x+ 2 y ≥ 6,5 x+ 3 y ≥ 15, x ≤7, y ≤ 4, x ≥0, y ≥0

is

a) (7,0)(7,4)(2,2.5) b) (7,0)(7,-4)(2,2.5) c) (7,0)(7,-4)(2,2.5) d) (7,0)(7,4)(2,2.5)

Lies in. a) 1st quadrant including the origin b) 1st and 2nd quadrants c) 2nd quadrant including the origin. d) 1st quadrant not including origin 229. The solution set of inequalities.

236.

is

a) (3.8,5)(0,2)(3,0) b) (3,8,0.5)(0,-2)(3,0) c) (3.8,0.5)(0,2)(3,0) d) (3.8,0.5)(0,-2)(-3,0) 230. The solution set for following inequalities.

2 x + y ≤ 9, x +2 y ≥ 9, x+ y ≥ 7 x ≥ 0 andy ≥ 0

is

a) (-9,0)(2.5,-5)(0,9) b) (-9,0)(-2.5,5)(-9,0) c) (-9,0)(2.5,5)(0,9) d) (9,0)(2.5,5)(0,9) 231. The solution set of inequalities.

x− y ≤ 4, x−2 y ≥ 0, y ≥−3

inequalities. 3 x−4 y ≥ 12, x ≥5, y ≥−2

a) (5,3/4)(5,0) b) (-6,-2)(6,1) c) (6,-2)(-6,1) d) (-6,2)(6,-1) 235. The following inequalities and indicate the solution set

the points.

2 x +5 y ≤10, 2 x−3 y ≥ 6, x ≥ 0 andy ≥ 0

The solution set of the following

5 X 1+ 4 X 1 ≥ 9, X 1+ X 2 ≥ 3, X 1 ≥ 0, X 2 ≥ 0

Which of the following point lies inside the solution set? a) (1, 3) b) (1, 4) c) (1, 1) d) (1, 2) 237. The solution set of the following inequalities.

4 x +6 y ≤ 24,5 x+ 3 y ≤ 15,2 x ≤5, x ≤ 0, y ≤ 0

is

a) (-2.5,0)(0,4) b) (2.5,0)(0,-4) c) (2.5,0)(0,4) d) (-2.5,0)(0,-4) 238. Which of the following is shaded region of if

z=2 x+5 y

subject to

2 x + 4 y ≤ 8, 3 x+ y ≤ 0, x + y ≥ 0, y ≥ 0

is

a) (1,-3)(4,0)(-6,15)(-6,3) b) (1,-3)(-4,0)(6,1.5)(-6,3) c) (1,3)(-4,0)(6,1.5)(-6,3) d) (1,-3)(4,0)(6,1.5)(-6,-3) 232. The solution set of the system of the inequalities nd find the extreme points of the solution set

x− y ≤3, 2 x + y ≤5, y ≤2, x ≥ 0 andy ≥0 is

a)

a) (0,3)(2.8,0)(2,2)(0,2) b) (0,-3)(2.8,0)(-2,-2)(0,-2) c) (0,-3)(2.5,0)(2,2)(0,2) d) (0,4)(2.8,0)(-2,2)(0,2) 233. Minimize n

m

m

Z =∑ ❑ ∑ CijXijsubjectto ∑ Xij=bj , j=1, … n i j =1 i=1 j=1 s a L.P.P. with number of constraints. a) m-n b) mn c) m/n d) M+n

b)

P a g e |28

a)

c)

b) d) 239. The corner points of the common region are if.

x ≤ 5,2 y ≤9∧9 x +10 y ≥ 45 is

a) (0,5), (5,4.5), (4.5,0) b) (5,0),(5,4.5), (0,4.5) c) (5,0), (5,4.5) d) (0,5), (5,4.5) 240. The corner points of the common region are if.

x− y ≤3, 2 x + y ≤5, y ≤2, x ≥ 0 is

a) A(0,-3),C(3/2,2),D(0,2) b) A(0,-3),C(8/3,-1/2),C(3/2,2) c) A(0,-3),B(8/3,-1/2), C(3/2,2),D(0,2) d) B(8/3,-1/2), C(3/2,2),D(0,2) 241. The corner points of the common region are if.

c)

2 x + y ≥ 9, x +2 y ≤ 9, x+ y ≥ 7, x ≥ 0, y ≥ 0 is

a) A(0,9) B(2,5) C(2,5) D(0,9) b) A(9,0) B(5,2) C(2,5) c) A(9,0) C(2,5) D(0,9) d) A(9,0) B(5,2) C(2,5) D(9,0) 242. Which of the following is shaded region of if

z=x + y

subject to

5 x+2 y ≥10, 2 x +3 y ≥6, x ≥ 1, x ≥ 0, andy ≥ 0

d) 243. Which of the following is shaded region of if

z=30 x+ 20 y

subject to

x+ y ≤ 8, x+2 y ≥ 4,6 x +4 y ≥ 12, x ≥ 0, andy ≥ 0

P a g e |29

245.

Which of the following is shaded region

of if

z=6 x +3 y

subject to

x+ y ≤ 5, x +2 y ≥ 4, 4 x+ y ≤ 12, y ≥ 0, andx ≥ 0

a)

a)

b)

b)

c)

c)

d) 244. The number of linear constrains for which the shaded area in the figure below is the solution set is a) 6 b) 4 c) 5 d) 3

d)

P a g e |30

246.

The corner points of the common region

are if.

2 x + y ≥ 9, x +2 y ≤ 9, x+ y ≥ 7, x ≥ 0, y ≥ 0 is a)

A(0,9) B(2,5) C(2,5) D(0,9)

b) A(9,0) B(5,2) C(2,5) c) A(9,0) C(2,5) D(0,9) d) A(9,0) B(5,2) C(2,5) D(0,9) 247. Which of the following is shaded region

p=2 x+ 3 y

of if

subject to

x ≤ 4, x+ 2 y ≥7, x ≥ 0, y ≥0 a) 2 x +5 y ≥80, x + y ≤ 20, x ≥ 0, y ≤ 0 b) 2 x +5 y ≥80, x + y ≥ 20, x ≥ 0, y ≥ 0 c)

2 x +5 y ≤80, x+ y ≤ 20, x ≥ 0, y ≥ 0

d) 2 x +5 y ≤80, x+ y ≤ 20, x ≤ 0, y ≤ 0 a)

249.

For the following Shaded region, the

linear constraints (except

≥ 0 andy ≥ 0

) are

b)

a) 2 x + y ≤2, x− y ≤1, x + y ≤ 8 b) 2 x + y ≤2, x− y ≥1, x +2 y ≤ 8 c)

c)

2 x + y ≤2, x− y ≥1, x +2 y ≥ 8

d) 2 x + y ≥ 2, x− y ≤1, x +2 y ≤ 8 250.

Shaded region represent by

d) 248. Shaded region represent by

P a g e |31

c) a) 4 x −2 y 3 b) 4 x −2 y 3

4 x −2 y −3

c)

d) 4 x −2 y −3 251.

For the maximum value of

Z =6 x 1−2 x 2 suchthat 2 x 1−x 2 ≤2, x 1 ≤ 3 andx 1, x 2 ≥0 the value of x1 and x2 are a) 2,3 b) 1,2 c) 2,4 d) 3,4 252. Which of the following is shaded region

z=5 x+7 y

of if

d) 253. Which of the following is shaded region of

C=12 x + 42 y

subject to

5 x+8 y ≤ 40, 3 x + y ≤ 6, x ≥ 0, y ≥2

a) a)

b) b)

P a g e |32

c) c)

d) 254. Which of the following is shaded region

z=3 x+ 2 y

of if

subject to

0 ≤ x , y ≤ 3, x + y ≤5, 3 x+ y ≥ 4

d) 255. The corner points of the common region are if.

5 x+ y ≥ 10, x + y ≥6, x+ 4 y ≥ 8, x ≥0, andy ≥ 0 is a) (0,10) (1,5) (2/3,16/3) b) (0,10) (2/3,16/3) c) (0,10) (1,5) (2/3,16/3) (8,0) d) (0,10) (1,5) (2/3,16/3) (0,7) 256. Consider the inequalities

x 1+ x 2≤ 3,2 x 1+5 x 2 ≥ 10, x 1, x 2≥ 0

a)

which of

the following points does not lie in the feasible region? a) (2,2) b) (1,2) c) (4,2) d) (2,1) 257. Which of the following is shaded region of if

C=12 x + 42 y

subject to

x+ 2 y ≥3, x +4 y ≥ 4, 3 x+ y ≥ 3, x ≥0, y ≥ 0

b)

P a g e |33

a)

x a) line segment IG b) area DHF c) area AHC d) line segment EG 259. Which of the following is shaded region

z=30 x+ 25 y

of if

subject to

5 x+ y ≥ 10, x + y ≥6, x+ 4 y ≥ 12, x ≥ 0, y ≥0

b)

a) c)

b)

d) 258. The feasible region for the following constraints

L1 ≤ 0 L 2 ≥0 ; L 3=0, x ≥0, y ≥0

P a g e |34

c)

b)

c)

d) 260. Which of the following is shaded region

p=50 x +100 y

of if

subject to

2 x +6 y ≤ 60,5 x+ 2 y ≤50, x + y ≤ 12, x ≥ 0, y ≥ 0

d) 261. Which of the following is shaded region

p=5 x +10 y

of if

subject to

x+ 2 y ≤10, 3 x+ y ≤ 12, x ≥ 0, y ≥ 0

a)

a) P a g e |35

a) b)

b)

c)

c) d) 262. The corner points of the feasible region are if.

3 x+ 4 y ≥12, 6 x+ 3 y ≥ 18, x ≥0, andy ≥ 0 is

a) (0,4) (12/5,6/5) (6,0) b) (4,0) (12/5,6/5) c) (4,0) (12/5,6/5) (0,6) d) 12/5,6/5) (0,6) 263. Which of the following is shaded region of if

z=5 x 1+7 x 2 ,

d) 264. The vertex of common graph of

x 1+ x 2≤ 4, 3 x 1+8 x 2 ≤ 24,5 x 1+3 x 2 ≤ 18, x 1 ≥ 0, x 2 ≥0

inequalities.

2 x + y ≥ 2,andx− y ≤ 3,is

a) (0,0) b) (5/3,4/3) c) (-4/3,5/3) d) (5/3,-4/3)

P a g e |36

265. are if.

The corner points of the common region

3 x−4 y ≥ 12, x ≥6, andy ≥−2 is

a) A(-2,6), B(3/2,6) b) A(-2,6), B(6,3/2) c) A(6,-2), B(6,2/3) d) A(6,-2), B(6,3/2) 266. The maximum of

z=5 x+ 2 y , subjecttothex + y ≤7, x+ 2 y ≤10, x , y ≥ 0 is a) 10 b) 35 c) 26 d) 70 267. The corner points of the common region are if.

c)

6 x+ 5 y −30 ≤0, 3 x−4 y+ 12≥ 0, x ≥ 0, andy ≥ 0 is a) O(0,0),A(5,0), B(20/13,54/13) b) O(0,0), B(20/13,54/13),C(0,3) c) O(0,0), B(54/13,20/13),C(0,3) d) O(0,0), A(5,0), B(20/13,54/13), C(0,3) 268. Which of the following is shaded region of

c=12 x+ 8 y

subject to

4 x + y ≥ 4, x+3 y ≥6, x + y ≥ 3, x ≥ 0, y ≥ 0

d) 269. The corner points of the common region are if.

x ≤ 4, 3 x+ y ≥ 3 is

a) (4,0) (0,1) (3,0) b) (0,4) (1,0) (3,0) c) (4,0) (1,0) (0,2) d) (4,0) (1,0) (0,3) 270. The corner points of the feasible region are if. a)

4 x +2 y ≥ 80, 2 x +5 y ≤180, x ≥ 0, y ≥ 0 is

a) (40,0) b) (90,0) c) (0,20) d) (0,36) 271. Which of the following is shaded region of

z=12 x −5 y

subject to

x−2 y ≥ 3, x − y ≥ 4, x ≥ 1, x ≥ 0, y ≥ 0

b)

a) P a g e |37

a) b)

b)

c)

c)

d) 272. Which of the following is shaded region of

z=2 x+5 y

subject to

x+ 4 y ≥ 24,3 x+ y ≤ 21, x ≤ 9, x ≥ 0, y ≥ 0

d) 273.

The maximum value of

subject to a) 100 c) 160

z=3 x+ 4 y

x+ y ≤ 40, x+ 2 y ≤ 60, x ≥ 0, y ≥0

is

b) 110 d) 140 P a g e |38

274.

Maximize graphically

p=3 x +5 ysubjectto 4 x +3 y ≤ 12, y −x ≤ 2, x ≥ 0 andy ≥ 0 a)

Pmax =121/7

b)

Pmax =119/7

c)

Pmax =118/7

d)

Pmax =116 /7

275.

Minimize graphically

z=2 x+3 ysubjecttoconstraintsx + 2 y ≥ 4, 5 x +3 y ≥12, x ≥ 0 andy ≥0 a)

Using graphical method. a)

Z min =6

5 7

b)

Z min =6

4 7

c)

Z min =7

7 4

d)

Z min =6

6 7

276.

The corner points of the feasible region

are if.

4 x +2 y ≥ 80, 2 x +5 y ≤180, x ≥ 0, y ≥ 0 is b)

a) B(90,0), C(35,5/2) b) A(20,0), B(90,0) c) A(20,0), B(90,0), C(5/2,35) d) B(90,0), C(5/2,35) 277. Maximize graphically

Z =3 x +2 ysubjecttox+2 y ≤2, x ≤ 1, x ≥ 0 andy ≥ 0 a)

Z max =3

b)

Z max =5

c)

Z max =4

d)

Z max =1

278. are if.

c)

The corner points of the common region

2 x +5 y ≤10, 2 x−3 y ≤ 6, x ≥0, y ≥ 0 is

a) (0,2) (0,2) (2,0) b) (0,2) (0,2) (0,2) c) (0,2) (0,2) (1,2) d) None of These 279. Which of the following is shaded region of

z=3 x+ 2 y

subject to

2 x +3 y ≥6, 2 x+ y ≥ 4, 0 ≤ x ≤ 4, 0≤ y ≤6

d) 280. The vertices of the feasible region

{

}

(x , y) ≤ 4, x− y ≥ 0,3 x + y ≥3 are x

a) (1, 0), (0 ,4), (4, 4) and (3, 3) b) (0, 1), (4 ,0), (0, 4) and (3/4, 3/4) c) (1, 0), (4,0), (4, 4) and (3/4, 3/4) d) (0, 1), (1 ,4), (4, 4) and (3, 3) 281. The corner points of the common region are if.

3 x−4 y ≤ 12, y ≤−1 P a g e |39

a) (-1,8/3) b) (8/3,-1) c) (5/2,0) d) (0,5/2) 282. Which of the following is shaded region of

z=30 x+ 20 y

subject to

x+ y ≤ 8, 6 x+ 4 y ≥12, 5 x+8 y ≥ 20 x >0, y >0

a) O(0,0), B(20/19,45/19), C(0,3) b) O(0,0), A(2,0),B(45/19, 20/19), C(0,3) c) O(0,0), A(2,0), B(20/19,45/19), C(0,3) d) A(2,0), B(20/19,45/19), C(0,3) 284. The points which provides the solution to the linear programming problems: max

( 2 x +3 y ) subject to constraints x ≥ 0, y ≥ 0, 2 x +2 y ≤ 9, 2 x + y ≤7, x+ 2 y ≤ 8 i s a) (1, 3:5) b) (3, 2:5) c) (2, 3:5) d) (2, 2:5) 285. Which of the following is shaded region of

z=3 x+ 5 y

subject to

x ≤ 4, y ≥ 2, 2 x +3 y ≥12, 6 x−5 y ≥−6, x ≥ 0, y ≥ 0 a)

a)

b)

b) c)

c) d) 283. The corner points of the common region are if.

3 x+5 y ≤15, 5 x+ 2 y ≤10, x ≥ 0, y ≥ 0 is P a g e |40

292.

Minimize

z=3 x+ y , subjectto2 x+ 3 y ≥ 6,2 x + y ≥ 4, 0 ≤ x ≤ 4, 0 ≤ y ≤ 6 is

d) 286. To maximize the objective function

z=x +2 y

under the constraints

x− y ≤2, x+ y ≤ 4 andx , y ≥0 is a) X=1, y=2, b) X=1, y=4, c) X=0, y=4, d) X=0, y=3, 287. Minimize

z=5 z=9 z=8 z=6

a)

Z min =4

b)

Z min =3.5

c)

Z min =3

d)

Z min =4.5

293.

The extreme values of the function

f ( x , y ) =x+2 y . over the convex polygon subject to,

x ≥ 0, y ≥ 0, x+ 4 y ≥ 4,2 x +3 y ≤ 6

are

at

a) (1,1) b) (5,3) c) (2,3) d) (12/5,2/5) 294. Minimize value of z=5 x+7 ysubjectto2 x+ y ≥ 8, x+2 y ≥10, x ≥ 0 andy ≥0

z=12 x +8 y , subjectto 4 x + y ≥ 4, x+3 y ≥ 6, x + y ≥6, x ≥ 0, y ≥

a) 44 b) 38 c) 40 d) 42 288. The Maximize value of

is

accurs at a) (1/2,8/3) b) (1/3,7/3) c) (1/2,7/3) d) (1/3,8/3) z=5 x+5 ysubjecttox + y ≤ 4,3 x+ 8 y ≤ 24, 10 x +7 y ≤35, x , y ≥0 295. The Minimum value of

a) 22 b) 26.4 c) 24.8 d) 25 289. The Maximize value of

z=2 x1 +3 x 2 subjecttotheconstraints 2 x 1 +7 x 2 ≥ 22, x 1 + x 2 ≥ 6, is a) 20 b) 10 c) 14 d) 16 296. Number of optimum solutions for the LLP, maximum

p=4 x+ 2 ysubjectto 4 x+2 y ≤ 46, x +3 y ≤24, x ≥ 0, y ≥ 0 is a) 42 b) 46 c) 48 d) 52 290. The Maximize value of

p=3 x +5 ysubjecttox ≤2, y ≤3, x+ y ≤ 4, x ≥ 0, y ≥ 0 is a) 15 c) 16 291. Minimize

b) 18 d) 20

z=6 x +3 y , subjecttoconstraints , x+ y ≤ 5, x +2 y ≥ 4, 4 x+ y ≤ is a) More than one but finite b) Infinite c) No solution d) 1 297. The extreme values of the function

f ( x , y ) =x+2 y . over the convex polygon x z=25 x+ 30 y , subjectto 4 x+3 y ≥ 60,2 x+ 3 y ≥ 36, x ≥0 andy ≥ 0 a)

Z min=420

Z min=423 b) c)

Z min =424

d)

Z min =421

subject to,

x ≥ 0, y ≥ 0, x+ 4 y ≥ 4,2 x +3 y ≤ 6

are

at a)

Z min =3

b)

Z min =4

c)

Z min =2 P a g e |41

d) 298. of

Z min =5 Which of the following is shaded region

z=x + ysubjecttotheconstraintsx+ 2 y ≤ 8,3 x +2 y ≤ 12, x , y ≥ is a) 10 c) 5

C=12 x + 42 y

300.

b) 20 d) 15

The maximum value

z=6 x +3 y

subject to constraints

2 x +3 y ≤13, x+ y ≤ 5, andx ≥ 0 andy ≥ 0 is

a)

a) 30 b) 20 c) 0 d) 21 301. Objective function of L.P.P. is a) A constant b) A function to be optimized c) A relation between the variable d) None of these 302.

The minimum value of

p=7 x +4 y

subject to constraints

x+ y ≤ 8, 6 x+ 4 y ≥12, 5 x+8 y ≥ 20 andx ≥ 0, y ≥ 0 is a) 56 c) 58 303. b)

b) 32 d) 12

The minimum value of

p=12 x+ 8 y

subject to constraints

4 x + y ≥ 4, x+3 y ≥6, x + y ≥ 3 andx ≥0, y ≥0 is a) 36 c) 24 304.

b) 32 d) 25.33

The maximum value of

subject to

z=20 x+ 50 y

x+ y ≤ 5, 0≤ x ≤ 3∧0 ≤ y ≤3 is

a) 180 b) 90 c) 190 d) 195 305. A vertex of the linear inequalities

2 x +3 y ≤6, x+ 4 y ≤ 4, x ≥ 0, y ≥ 0 c)

a) (1,0) c) (12/5,2/5) 306.

The lines

is

b) (1,1) d) (2/5,12/5)

5 x+ 4 y ≥20, x ≤ 6, y ≤ 4 form

[M.H.C.E.T. 2005] a) A square b) A rhombus c) A triangle d) A quadrilateral 307. The points which provides the solution to the solution to the L.P.P. maximize

z=2 x+3 ysubjecttoconstraintsx+ 2 y ≤ 8,2 x +2 y ≤ 9, 2 x + y ≤ d) 299. The Maximum value of objective function

is a) (3,2.5) c) (2,2.5)

b) (2,3.5) d) (1,3.5) P a g e |42

308.

The point which the minimum value of

p=4 x+ 5 y

subject to constraints

5 x+ y ≥ 10,2 x+ 2 y ≥12, 5 x+ 8 y ≥ 20 andx ≥ 0, y ≥ 0 is a) (2,4) b) (1,5) c) (4,2) d) (0,10) 309. The corner points of common region are if

2 x + y ≥ 9, x +2 y ≥ 9, x+ y ≥ 7 andx ≥ 0, y ≥ 0 is

a) (0,9)(2,5)(0,9) b) (0,9)(2,5)(2,5) c) (0,9)(2,5)(0,9) d) (9,0)(5,2)(2,5)(0,9) 310. Which of the following is not a convex set?

c)

a) {( x , y ) :2 x+2 y ≤7 } 2

2

b) {( x , y ) : x + y❑ ≤ 4 } c)

{x : x =5 } 2

d) 312. The Shaded region of

2 ❑

d) {( x , y ) :3 x +2 y ≤6 } 311.

The Shaded region of

x+ y ≤ 5,0 ≤ x ≤ 4,0 ≤ y ≤ 0 is

x+ y ≤ 2, 0≤ x ≤ 1, y ≥ 0 is

given by

given by

a) a)

b)

b)

P a g e |43

c)

c)

d) 314. The Shaded region of

d) 313. The Shaded region of

5 x+8 y ≤ 40, 3 x + y ≤12, x ≥ 0, y ≥ 0 is

a)

given by

3 x+2 y ≤6, x+ y ≥ 2, x ≥ 0, y ≥ 0 is

given by

a)

b) b)

P a g e |44

c) c)

d) 316. The Shaded region of

x+ y ≤ 3, 0≤ x ≤ 2,0 ≤ y ≤ 2is

d) 315. The Shaded region of

4 x +3 y ≤12,−x+ y ≤ 2, x ≥0, y ≥0 is

given by

given by

a)

a)

b)

b)

P a g e |45

c)

d) 317. The Shaded region of

x+ y ≥ 5,2 x +3 y ≤ 21, 4 x +3 y ≤ 24, x ≥ 0, y ≥ 0 is given by

c)

d) 318. The Shaded region of

10 x+2 y ≥20, 5 x+5 y ≥30, x ≥ 0, y ≥ 0 is

given by

a) a)

b)

b)

P a g e |46

c)

c)

d) 320. The Shaded region of

d) 319. The Shaded region of

3 x+ y ≤ 9, 4 x +3 y ≤ 17, x ≥ 0, y ≥ 0is

a)

given by

x+ y ≥ 5, x ≥ 4, y ≤ 2, x ≥ 0, y ≥ 0is

given by

a)

b) b)

P a g e |47

c) c)

d) 322. The Shaded region of

x+ y ≥ 5, 0≤ x ≤ 4, 0 ≤ y ≤ 2 is d) 321. The Shaded region of

x ≤ 4, y ≤ 8, andx + y ≤ 8, x ≥ 0, y ≥ 0 is

given by

given by

a)

a)

b)

b) P a g e |48

c) c)

d) 323. The Shaded region of

d) 324. Shaded region is represented by

4 x + y ≥ 4, x+3 y ≥6, x + y ≥ 3, andx ≥ 0, y ≥ 0 is given by

a) 2 x +5 y ≥80, x + y ≤ 20, andx ≥ 0, y ≥ 0 a)

b) 2 x +5 y ≥80, x + y ≥ 20, andx ≥ 0, y ≥ 0

2 x +5 y ≤80, x+ y ≤ 20, andx ≤ 0, y ≤ 0

c)

d) 2 x +5 y ≤80, x+ y ≤ 20, andx ≥ 0, y ≥ 0 325.

If

3 x+2 y ≥15, 2 x + y ≥ 6 andx ≥ 0, y ≥ 0,

The corner points of the common region are a) (0,6),(5,0),(15/7,12/7) b) (6,0),(0,5),(15/7,0) c) (6,0),(15/7,12/7),(0,5) d) (6,0),(12/7,15/7),(50),(12/7,15/7) b)

P a g e |49

326.

The objective function

z=4 x+3 y

can

be maximized subject to the constraints

3 x+ 4 y ≤24, 8 x+ 6 y ≤ 48, x ≤ 5, y ≤ 6 andx ≥ 0, y ≥ 0 a) At only one point b) At two points only c) At an infinite number of points d) none 327. Solution set of equations

a)

x−2 y ≥ 0, 2 x− y+2 ≤ 0, x ≥ 0, y ≥ 0 a) empty c) infinite 328. The constraints

b) Close half plane d) First quadrant

– x + y ≤1,−x+3 y ≤ 9,andx ≥ 0, y ≥ 0 define a) A bounded feasible b) Unbounded feasible region c) Both bounded and unbounded region d) none 329. The point which the maximum

z=3 x+ 2 y

b)

subject to constraints

x+ y ≤ 2, andx ≥ 0, y ≥ 0

is

a) (0,0) b) (1.5,1.5) c) (2,0) d) (0,2) 330. The point which the maximum

z=3 x+ 2 y

subject to constraints

x+ y ≤ 3, andx ≤ 2,−2 x+ y ≤ 1, x ≥0, y ≥0

c)

is

a) (2,3) b) (2/3,7/3) c) (2,1) d) (2,2) 331. The point which the maximum value of

z=9 x +13 y

subject to constraints

2 x +3 y ≤18, 2 x + y ≤10, andx ≥0, y ≥0

is

a) (4,3) b) (0,6) c) (5,0) d) (3,4) 332. The point which the minimum value of

p=30 x +20 y

subject to constraints

x+ y ≤ 8, x+2 y ≥ 4,6 x +4 y ≥ 12 andx ≥ 0, y ≥ 0

d) 334. Which of the following is the common region for :

x ≤ 3 y , x + y ≥ 8, x ≥ 6, x ≥ 0, y ≥ 0 ?

is

a) (8,0) b) (1,1.5) c) (4,0) d) (1.5,2) 333. The graph of inequalities

x+ y ≥ 2, 2 x + y ≤ 8 ; x ≥ 0, y ≥0

is

P a g e |50

a) a)

b) b)

c) c)

d) 335. Which of the following is the common region for :

3 x+5 y ≥15, x ≤ 4, y ≤2, x ≥ 0, y ≥ 0 :

d) 336. A company recruits male workers (x) and female workers (y) under the condition that the company must have at least 2 female workers and total number of workers should not be more than 9, then x and y can be related as a)

y ≥ 2, x+ y ≤ 9

b)

y ≤ 2, x+ y ≤ 9 P a g e |51

c)

y >2, x + y ≤ 9

d) None of these 337. Which of the following is the feasible region for :

x+ y ≤ 4, x ≤ 2, y ≤ 1, x + y ≥1, x ≥ 0, y ≥ 0 :

a)

a)

b)

b)

c)

c)

d) 339.

If the graph

y=f ( x)

passes through

the points (0,3),(1,6),(2,11) and (3,18), then

f ( x) d) 338. Which of the following is the common region for :

equals 2

b) 2 x + x +3

2

d) x −2 x −3

a)

x +2 x +3

c)

x −2 x +3

2

2

x− y ≥5,2 x+ 3 y ≥ 12, y ≤ 9, x ≥0, y ≥ 0: P a g e |52

340.

Minimum value of

z=4 x+ 8 y

under

the constraints

345.

The Maximum value of

subject to

x ≤ 2, y ≤ 3, x + y ≤ 4, x ≥ 0, y ≥ 0is

200 x+100 y ≥2000, x+ 2 y ≥ 50, 40 x + 40 y ≥ 1400 andx ≥ 0, y ≥ 0 is a) 200 b) 225 c) 175 d) 280 341. The solution set of the inequations

2x−y≥1

is

a) Half plane that contains the origin b) Open half plane not containing the origin c) Whole xy- plane except the points lying on the line

2 x − y=1

d) None of these 342. The solution set of the linear inequations

a) 15 c) 18

346.

b) 16 d) 20

The Maximum value of

subject to

b) 3 d) 8/3

The Minimum value of

subject to

p=x +3 y

2 x + y ≥ 6, x + y ≥ 4, x ≤ y , x ≥ 0, y ≥ 0 is

a) 8 c) 6 348.

p=x + y

2 x + y ≤ 4, x +2 y ≤ 4, x ≥0, y ≥0 is

a) 2 c) 4/3 347.

p=3 x +5 y

b) 7 d) 5

The Maximum value of

z=5 x+5 y

x+ y ≥ 1,7 x +9 y ≤63, y ≤5, x ≤ 6 andx ≥ 0, y ≥ 0 liesin subject to x+ y ≤ 4,3 x +8 y ≤ 24, 10 x +7 y ≤35, x , y ≥ 0is ST

a) 1 quadrant including the origin b) 1ST quadrant not including the origin c) 1ST and 2nd quadrant d) None of these 343. If a young man rides his motor cycle at 25 km per hour he has to send Rs.2 per km on petrol, if he rides it at a faster speed of 40 km per hour, the petrol costs increases to Rs. 5 per km. He has Rs. 100 to spend on petrol ,using linear programming the maximum distance, he can travel within one hour is a) 20 km b) 30 km c) 40 km d) 45 km 344. Suppose every gram of wheat provides 0.1 gm of proteins and 0.25 gm of carbohydrates and the corresponding values of the rice are 0.05 gm and 0.5 gm respectively. Wheat costs Rs. 2 per kg and rice Rs. 8 .Thus minimum daily requirements of protein and carbohydrates for an average child are 50 gm and 200 gm respectively. The quantities in which wheat and rice should be mixed in the daily diet to provide the minimum daily requirements of proteins and carbohydrates at minimum cost, is respectively. a) 400 gms,300 gms b) 400 gms,200 gms c) 500 gms,100 gms d) None of these

a) 22 c) 26.4 349.

b) 20 d) 25

The Maximum value of

z=3 x+ 5 y

subject to

x+ 2 y ≤2000, x + y ≤ 1500, x ≤ 600, x , y ≤ 0 is a) 5000 c) 6000 350.

b) 5300 d) 6500

The Maximum value of

p=4 x+ 2 y

subject to

4 x +2 y ≤ 46, x +3 y ≤ 24, x ≤ 600, x ≥0, y ≥0 is a) 42 c) 52 351.

b) 48 d) 46

The minimum value of

30 x+20 y

subject to the constraints

x+ y ≤ 8, 6 x+ 4 y ≥12, 5 x+8 y ≥ 20, x ≥0, y ≥0 is a) 50 b) 30 c) 60 d) 120 352. The number of optimum solutions for the LPP, The Maximize value of

z=6 x +3 y

subject to constraints

x+ y ≤ 5, x +2 y ≥ 4, 4 x+ y ≤ 12, x ≥0, y ≥ 0is a) 1 b) More than one but finite c) Infinite P a g e |53

d) No solution 353.

z=3 x+ 2 y

The Maximum value of

subject to constraints

0 ≤ x ≤3, 0 ≤ y ≤ 3, x + y ≤5, 2 x+ y ≥ 4, is a) 12 c) 14 354.

b) 13 d) 24

The Maximum value of

z=3 x+ 2 y

subject to

2 x +3 y ≥6,2 x+ y ≥ 4, 0 ≤ x ≤ 4, 0 ≤ y ≤ 6 is a) 13/4 b) 12/5 c) 13/2 d) 12/7 355. The Maximum value of

z=6 x 1+11 x 2 s . t .

subject to

2 x 1 + x 2 ≤ 104, x 1 +2 x 2 ≤76, x 1 ≥ 0, x 2 ≥0, is a) 240 b) 540 c) 440 d) 340 356. Shaded region is represented by a) 2 x +5 y ≥80, x + y ≤ 20, x ≥ 0, y ≥ 0 b) 2 x +5 y ≥80, x + y ≥ 20, x ≥ 0, y ≥ 0 c)

2 x +5 y ≤80, x+ y ≤ 20, x ≥ 0, y ≥ 0

d) 2 x +5 y ≤80, x+ y ≤ 20, x ≤ 0, y ≤ 0 357.

In equation

y−x ≤0

represents

3 per unit, respectively. if one unit of A contains 200 units of vitamins, one unit of minerals and 40 calories and one unit of food B contains 100 units of vitamins, two units of minerals and 40 calories. The combination of food A and B used will be least cost (respectively) at a) (0,35) b) (20,15) c) (50,0) d) (0,0) 360. A carpenter has 20 and 15 Sq. meters of plywood and sun mica respectively. He produces product A and B. Product A requires 2 and 1 sq. meters and product B requires 1 and 3 sq. meters of ply wood and sun mica respectively. if the profit on one piece of product A is Rs.30 and that on one piece of product B is Rs.20, how many piece of products A and B should be made to maximize his profit? a) (0,35) b) (20,15) c) (50,0) d) (0,0) 361. Find the extreme values of the function

f ( x , y )=2 x + y subject to

x ≥ 0, y ≥ 0, x+ 4 y ≥ 4,2 x +3 y ≤ 6

a) 4,2 c) -4,2 362.

over the convex polygon X

Maximize

b) 2,4 d) 2,-4

z=5 x+7 y

subject to

x+ y ≤ 4, 3 x+ 4 y ≤12, x ≥ 0, y ≥ 0 is

a) The half plane that contains the positive a) 30 b) 20 X-axis. c) 31 d) 21 b) Closed half plane above the line y=x 363. Find the extreme values of the function which contains positive Y-axis. f ( x , y )=2 x + y over the convex polygon X c) Half plane that contains the negative Xaxis. d) None of these. subject to x ≤ 2, y ≥−2, x− y ≥−2, x +2 y ≤ 4 358. For The Maximum value of a) 4,8 b) 4,-8 z=5 x+ 2 ys .t . subject to c) 8,4 d) -8,4 364. A candidate appearing for the 2 x +3 y ≥6, x−2 y ≤ 2, 6 x+ 4 y ≤24,−3 x+2 y ≤3, x ≥ 0, y ≥ 0 competitive test finds the following information on seeing the question paper : the values of x and y are respectively, 1. The total time is 3 hours a) 18/7,2/7 2. The paper has two section-section A b) 7/2,3/4 and section B c) 3/2,15/4 3. To answer at least two questions from d) 3/2,3/4 section A and at least three question 359. A diet for a sick person must contain at from section B. least 1400 units of vitamins, 50 units of minerals and 1400 units of calories. Two foods A and B are available at a cost of Rs.4 and Rs. P a g e |54

4.

A question of first section carries 10 marks and that of section B carries 15 marks. 5. Time to answer question of section A is 15 minutes and that of section B is 25 minutes. 6. The maximum number of question to be answered is 10. The maximum marks he can secure is a) 115 b) 125 c) 100 d) 90 365. The Minimum value of

doubtful. He solve the questions by one, then number of questions solved by him, are a) 67 b) 90 c) 79 d) 80 369. Minimize n

m

n

n

j=1

j=1

z=∑ ∑ cijxijsubjectto ∑ xij ≤ ai ,i =1,… , mand ∑ xij=b j j=1 i=1

a) M+n b) m-n c) Mn d) m/n 370. The feasible region for the following

L1 ≤ 0, L2 ≥ 0, L3=0, x ≥0, y ≥ 0 in z=2 x1 +3 x 2 subjectedtothecontraints 2 x 1 +7 x2 ≥ 22, x 1constraints + 2 x 2 ≥ 10, x1 ≥ 0, x 2 ≥ 0, is

a) 14 b) 20 c) 10 d) 16 366. The manager of an oil refinery wants to decide on the optimal mix of two possible blending processes I and II of which the inputs per production run are as follows Process Crude A Crude B gasoline X gasoline Y I 5 3 5 8 II 6 5 4 4 The maximum amounts available of crude A and B are 250 and 200 units respectively. At least 150 units of gasoline X and 8 units of Y are required. The profit per production run from processes I and II Rs. 400 and Rs. 500 respectively. The number of constraints is a) 3 b) 4 c) 5 d) 6 367. The vertex of common graph of in equalities

2 x + y ≥ 2andx− y ≤ 3,is

the diagram shown is a) Area DHF b) Area AHC c) Line segment EG d) Line segment IG 371. Which of the following is not true on linear programming problem? a) A slack variable is a variable added to the left hand side of a less than or equal to constraints to convert it into an equality. b) A surplus variable is a variable subtracted from the left hand side of a greater than or equal to constraints to convert it into an equality. c) A basic solution which is also in the feasible region is called a basic feasible region. d) A column in the simplex table that contains all the variables in the solution is called pivot or key column. 372. Which of the following does not represent a convex polygon? a)

x 2+ y 2 ≤ 9

b) x+ y ≤ 1

a) (0,0) b) (5/3,-4/3) 2 2 2 2 c) x + y ≥1 d) 3 x + 4 y ≤ 1 c) (5/3,4/3) d) (-4/3,5/3) 368. In the examination of the C.E.T. the 373. Shaded region represented by total marks of the mathematics are 300. If the answer is right, marks provided 3 and if the answer is wrong, marks provided -1. A student knows the correct answer of 67 questions and remaining questions are doubtful for him. He takes the time

1

12 min. To give correct

answer and 3 minute. That for doubtful. Total time is 3 hours. In the question paper after every two simple questions, one question is

P a g e |55

a) (4/3,2/3) b) (1,1) c) (0,2/3) d) (0,2) 378. Shaded region represented by

a) 4 x −2 y ≤ 3

b) 4 x −2 y ≤−3

4 x −2 y ≥ 3

d) 4 x −2 y ≥−3

c) 374.

For the following shaded region, the

linear constraints

exceptx ≥ 0 andy ≥0 ¿

are

a) 2 x +5 y ≥80, x + y ≤ 20, x ≥ 0, y ≤ 0 b) 2 x +5 y ≥80, x + y ≥ 20, x ≥ 0, y ≥ 0 c)

2 x +5 y ≤80, x+ y ≤ 20, x ≥ 0, y ≥ 0

d) 2 x +5 y ≤80, x+ y ≤ 20, x ≤ 0, y ≤ 0 379.

Which of the following is the feasible

region for:

x− y ≤3, 2 x + y ≤ 4, y ≥ 3, x ≥ 0, y ≥ 0

a) 2 x + y ≤2, x− y ≤1, x + y ≤ 8 b) 2 x + y ≤2, x− y ≤1, x +2 y ≤ 8 c)

2 x + y ≤2, x− y ≥1, x + y ≤ 8

a)

d) 2 x + y ≤2, x− y ≤1, x +2 y ≥ 8 375.

The Maximum value of

z=6 x 1−2 x2 suchthat 2 x 1−x 2 ≤ 2, x 1 ≤ 3,∧x 1 , x 2 ≥0, thevaluesof x 1∧x 2 are a) 3,4 b) 2,3 c) 1,2 d) None of these 376. If the constraints in the linear programming problem are changed a) The problem is to be revaluated. b) Solution is not defined c) The objective function has to be modified. d) The change in constraints is ignored. 377. The solution set of linear constraints

x−2 y ≥ 0, 2 x− y ≤ 2 andx , y ≥ 0

b)

is P a g e |56

384. set:

Which of the following is not a convex

a) {x : x =5 } 2

2

b) {( x , y ) :3 x +2 y ≤ 6 } c)

{( x , y ) : x 2+ y 2 ≤ 4 }

d) {( x , y ) :2 x+5 y y , x ≥ 6000, y ≥12000 Maximize

z=

7 10 x+ y=0.07 x +0.1 y 100 100

z atA ( 6000,24000 )=0.07 (6000)+ 0.1(24000)=2820 z atB ( 6000,12000 )=0.07( 6000)+0.1(12000)=¿ z atC (18000,12000 )=0.07(18000)+0.1(12000)=2460 Maximum value of z is 2820 at A (6000, 24000).

P a g e |109

172

(c)

z atA ( 0,5 )=5(0)+8(5)=40 z atB ( 3,2)=5( 3)+8 (2)=31 z atC ( 4,2)=5 (4)+8 (2)=36 Minimum value of z is 31 at B (3,0).

173

(a)

z atA ( 0,15)=4(0)+1( 15)=15 z atC (10,0 )=4 ( 10 ) +1 ( 10 ) =40 Maximum value of z is 40 .

174

(c)

P a g e |110

z atA ( 0,200)=40(0)+100(200)=20000 z

atB

=40 ( 2500 /11 ) +100 (1200 /11 )=20000

1200 , ( 2500 11 11 )

z atC ( 450,0)=40 (450)+100(0)=18000 Maximum value of z is 18000 at C(450,0) .

175

(a)

z atD ( 1,3/ 2)=30 (1)+20(3/2)=60 z atC ( 4,0) =30 (4)+20 (0)=120 z atB ( 8,0)=30 (8)+ 20(0)=240 z atA ( 0,8 ) =30(0)+20(8)=160 Minimum value of z is 60 at D(1,3/2).

176

(d)

z atA ( 4,5 )=2(4 )+ 5(5)=33 z

atB

=2 ( 180/11 ) +5 ( 5/11 ) =385/11

5 , ( 180 11 11 )

z atC (6,3 )=2(6)+5 (3)=27 Maximum value of z is 35 at B(180/11,5/11). P a g e |111

178

(d)

z atA ( 0,30)=3 (0)+4 (30)=120 z atB ( 20,20)=3(20)+ 4(20)=140 z atB ( 20,20)=3(20)+ 4(20)=140 z atC ( 40,0)=3(40)+4 (0)=120 Maximum value of z is 140.

179

(b)

z atA ( 0,200)=3( 0)+ 4 (200)=800 z atB ( 200,100)=3(200)+ 4 (100)=1000 z atC (300,0 )=3(300)+4 (0)=900 Maximum value of z is 1000 at B(200,100).

P a g e |112

180

(c)

z atA ( 3,1)=3 (3)+ 4 (1)=13 z atB ( 2.5,2)=3 (2.5)+ 4 (2)=15.5 z atC (1,3.5 )=3 (1)+ 4(3.5)=17 z atD ( 0,4 ) =3(0)+4 (4)=16 Maximum value of z is 17 at C(1,3.5)

181

(d) The LPP standard form is maximize Subject to

z=21 x1 +15 x 2+ 0 s 1+0 s2

x 1+2 x 2+ s 1=6, 4 x 1 +3 x2 +2 ≤=12, x1 , x 2 ≥ 0, s1 , s 2 ≥ 0

x 1∧x2 aredicisionvariables , s 1 , s2 arestackvariables The number of variables are n=4 and no of equations are m=2 Total possible basic solutions are nCm=4C2=6 184

185

(c) We know that for a given solution, a slack variable is equal to zero if the entire amount of resource with the constraints in which the slack variable appears has been consumed. (b) For maximization LP model , the simplex method terminated when

186

C j−z j ≤ 0

(b)

z atA ( 0,4 )=4 (0)+5( 4)=20 P a g e |113

z atB ( 3,2)=4(3)+5(2)=22 z atC ( 4.0)=4 (4)+5(0)=16 Maximum value of z is 22 at B(3,2)

187

(b)

4 x +3 y ≤1800, 5 x+7 y ≤1500, 0 ≤ x ≤ 300,0≤ y ≤ 400

188

Maximize z=50x+100y Where x= no of bats B1 Y=no of bats B2 (c)

z atA ( 0,0 )=4 (0)+5(0)=0 z

atB

( 83 , 43 )

=4 (8 /3)+5(4 /3)=53/3

z atC (2.0 )=4(2)+5(0)=8 Maximum value of z is

1713

190

at B(8/3,4/3)

(c)

z z

=3(8/3)+6(16 /3)=40

atA

( 83 , 163 )

atB

( 163 , 183 )

=3(16 / 3)+6 (18/ 3)=32

z atO ( 0.0)=3 (0)+6(0)=0 Maximum value of z is 40 at A(8/3,16/3) P a g e |114

191

(b) Let x is no of kg of lemans y is no of kg of orange

x+ y ≥ 6, x ≤ 2, y ≥ 4 Minimize z=50x+25y

z atA ( 0.6 )=50(0)+25(6)=150 z atB ( 2.4) =50(2)+ 25(4)=200 Minimum value of z is 150 at A(0,6)

192

(c)

x+ y ≤ 4, 3 x+ 8 y ≤ 24,5 x +2 y ≤ 18, x ≥0, y ≥0 Minimize z=50x+70y

z atA ( 0.3 )=50(0)+70(3)=210 z z

=50 (8/5)+ 70(12/5)=248

atB

( 85 . 25 )

atC

(103 . 23 )

=50(10 /3)+70(2 /3)=640 /3

z atD ( 3.6 .0 )=50(3.6)+70(0)=180 maximum value of z is 248 at B(8/5,2/5)

P a g e |115

193

(a) To maximize z=30x+20y subject to following constraints

10 x+6 y ≤ 60,5 x +4 y ≥ 35, x ≥0, y ≥ 0 194

(d) Let x is no of tin of tea y is no of tin of coffee

2 x +3 y ≤75, x+ y ≥ 30,2 y ≥ x Minimize z=40x+60y

195

(a) To Minimize z=x+y subject to

2 x + y ≥12, 5 x+ 8 y ≥ 26, x +6 y ≥ 24, x , y ≥ 0 196

(a) To Minimize z=20x+40y subject to

36 x+ 6 y ≥ 108,3 x +12 y ≥ 36,20 x +10 y ≥ 100, x , y ≥ 0 197

(c)

2 x + y ≥15, x + y ≥ 20, x+3 y ≥ 0, x , y ≥ 0 198

(a) To Maximize z=20x+15y subject to

350 x+250 y ≤ 4000, x + y ≤25, x , y ≥ 0 199

(b)

x 1 = no of product P. P a g e |116

x 2 = no of product Q. To Maximize z=5

x 1 +4 x 2 subject to

1 2 x 1 + x 2 ≤ 45, 4 x 1 +2 x 2 ≤150, 3 x1 + x 2 ≤ 250 x 1 , x 2 ≥ 0 2 200

(c) x= no of units of F1. y= no of units of F2. To form constraints, we make following table. Vitamin Food stuffs Requirements F1 F2 (in units)

≥6

V1

1

1

V2

2

1

≥7

V3

1

4

≥8

Thus the formulation of LPP is minimize z=2x+3y subject to constraints

x+ y ≥ 6, 2 x + y ≥7, x+ 4 y ≥ 8, x , y ≥ 0 201

(c)

2 x +3 y=18

The equalities are

x+ 4 y =16

x=3 ∴

…(i)

…(ii)

…(iii) By putting

x=0

and

Draw the line AB, CD and

y=0 ,we get A(9,0), B(0,6), C(16,0), D(0,4)

x=6

The feasible region shown in fig. Solving L1 and L2 we get P(4.8,2.8) and solving L1 and L3 ,we get Q(6,2)

∴ Zat (0,0)=5 ( 0 ) +10 ( 0 )=0

Zat ( 6,0 )=5 ( 6 ) +10 ( 0 ) =30 Zat ( 6,2 )=5 ( 6 ) +10 ( 2 )=50 Zat ( 4.8,2.8 )=5 ( 4.8 ) +10(2.8) ¿ 24+28=52

Zat ( 0,4 )=5 ( 0 )+ 10 ( 4 ) =40 ∴ Zismaximizeatx=4.8, y =2.8 .

∴ ( c ) iscorrectanswer .

P a g e |117

202

(a) Constraints are

203

2 x +3 y ≤36 ; 5 x +2 y ≤ 50; 2 x +6 y ≤ 60 ; x ≥ 0 ; y ≥0

(c) Given inequalities are

2 x + y =20 x +2 y=50 40 x +40 y =1400 x=0, y=20 x=0, y=25, x+ y=35

y=0, x=10 y=0 x=50, x=0, y=35 y=0, x=34 (0,20), (10,0) (0,25),(50,0) (0,35) (35,0) Point of intersection B

x+ 2 y =50 x+ y=35 --------------Y=15

∴ x +15=35→ x=20 ∴ B=( 20,15 ) z=4 x+ 8 y At A(0,35)

z=0+35 × 8=280

At B(20,15),

z=4 ×20+8 × 15=80+120=200

At C(50,0)

z=4 ×50+8 × 0=200

Minimum value of z is 200 throughout on seg BC .

P a g e |118

208

(b) Value at Z at A(0,4) Z= 0+4= 4 Value at Z at B(2,3) Z= 2+3= 5 Value at Z at C(4,0) Z= 4+0= 4 Maximum value Z at B(2,3) is 5 Answer is B.

209

(c) The vertices of the common feasible region are A(0,5), B(18/11,10/11), C(3,0) Value of Z at A Z=0+5=5 Value of Z at B Z=18/11+10/11=28/11 Value of Z at C Z=3+0=3 Minimum value of Z at B (18/11, 10/11) is =28/11. Answer is C.

P a g e |119

211

(c) To find the feasible region Consider

x+ 2 y ≤ 8

∴ A =(8,0), B=(0,4) and ∴

3 x+2 y=12

C(4,0), D(0,6) satisfies the equation

P is the point of intersection of the line AB and CD solving equation

x+2 y =8,∧3 x+ 2 y =12 we get

x=2, y=3, p=( 2,3) The shaded region OCPB is a feasible region. The optimal solution is obtained at one of the corner point O(0,0), C(4,0), P(2,3) and B(0,4) Corner Points

∴

value of

O(0,0)

z=0

C(4,0)

z=4

P(2,3)

z=5

B(0,4)

z=4

z=x + y

The maximum value of z is 5 at the point P(2,3)

P a g e |120

212

(a) Consider the equation

2 x +3 y=13

∴ The points A(13/2,0), B(0,13/3) Consider the equation

x+ y=5

∴ C=(5,0) and D=(0,5) Draw the lines Q is the point of intersection of the lines AB and CD Solving the equation

x+ y=5∧2 x +3 y=13

We get Q(2,3) The shaded region OAQD is a feasible region Optimal solution is obtained at one of the points O(0,0), A(5,0), Q(2,3), D(0,13/3) Corner Points O(0,0) A(5,0) Q(2,3) D(0,13/3)

value of

z=6 x +3 y z=0

z=30+0=0 z=12+9=21

z=0+13=13

The maximum value of z is 30 at the point A(5,0)

P a g e |121

213

(d) Consider the equation

( i ) x + y=8 ∴ points A(8,0), B(0,8) (ii) 6 x+ 4 y =12

∴ C=(2,0) and D=(0,3) (iii) 5 x+8 y =20 E(4,0), F(0,5/2) Draw the lines Q is the point of intersection of the lines CD and EF Solving the equation

6 x+ 4 y =12∧5 x+ 8 y=20

We get Q(4/7,15/7) The shaded region ABDQE is a feasible region Optimal solution is obtained at one of the points A(8,0), D(0,3), B(0,8), Q(4/7,15/7), E(4,0) Corner Points

value of p=7 x +4 y

A(8,0)p=56+0=56 B(0,8) p=0+32=32 D(0,3) p=0+12=12 Q(4/7,15/7) p=28/7+60/7=12.57 E(4,0) p=28+0=28 The minimum value of P is 12 at the point D(0,3)

214

(d) P a g e |122

Consider the equation

( i ) 4 x+ y=4 ∴ points A(1,0), B(0,4) (ii) x+3 y =6

∴ C=(6,0) and D=(0,2) (iii) x+ y=3 E(3,0), F(0,3) Draw the lines G is the point of intersection of the lines CD and EF Solving the equation

x+ 3 y =6 andx + y=3

We get G(3/2,3/2) H is the point of intersection of the lines AB and EF Solving the equation

1 x= , y=8 /3 3 We get H(1/3,8/3) The shaded region CGHB is a feasible region Optimal solution is obtained at one of the points C(6,0), G(3/2,3/2), H(1/3,8/3), B(0,4) Corner Points

value of p=12 x+ 8 y

C(6,0) p=72+0=72 G(3/2,3/2) p=18+12=30 H(1/3,8/3) p=4+64/3=76/3=25.33 B(0,4) p=0+32=32 The minimum value of P is 25.33 at the point H(1/3,8/3)

215

(c) We draw the lines

x+ y=5

x=3

by joining two points A(5,0) and B(0,5)

passing through the points C(3,0) and parallel to y axis

P a g e |123

y=3

passing through the points D(0,3) and parallel to x axis

The point P point of intersection of the lines

x+ y=5 andx=3

solving these equation

x+ y=5 andy=3

solving these equation

x=3, y=2 P=(3,2) The point Q point of intersection of the lines

y=3, x=2 Q=(2,3) The shaded region OCPQD is a feasible region Optimal solution is obtained at one of the corner points O(0,0), P(3,2), Q(2,3), D(0,3) Corner Points

value of z=20 x+ 50 y

O(0,0) z=0+0=0 P(3,2) z=60+100=160 Q(2,3) z=40+150=190 D(0,3) z=0+150=150 The minimum value of z is 190 at the point Q(2,3)

216

(d) Common region is quadrilateral.

217

(c) We draw the lines

5 x+ y =10

by joining two points A(2,0) and B(0,10) P a g e |124

2 x +2 y=12

by joining two points C(6,0) and D(0,6)

x+ 4 y =12 by joining two points E(12,0) and F(0,3) The point G point of intersection of the lines

x+ 4 y =12∧2 x+2 y=12

solving these

equation

x=4, y=2 G=(4,2) The point H is the point of intersection of the lines

2 x +2 y=12∧5 x + y=10

solving these equation we get

x=1, y=5 Q=(1,5) The shaded region EGHB is a feasible region Optimal solution is obtained at one of the points E(12,0), G(4,2), H(1,5), B(0,10) Corner Points

value of z=4 x+5 y

E(12,0) p=48+0=48 G(4,2) p=16+10=26 H(1,5) z=4+25=29 B(0,10) z=0+50=50 The minimum value of p is 48 at the point G(4,2)

218

(d) We draw the lines

2 x + y =9

by joining two points A(9/2,0) and B(0,9/2)

x+ 2 y =9

by joining two points C(9,0) and D(0,9/2)

x+ y=7

by joining two points E(7,0) and F(0,7)

P is the point of intersection of the lines

x+ 2 y =9 andx+ y=7

solving these equation

x=5, y=2 G=(5,2) The point Q is the point of intersection of the lines

x+ y=7∧2 x + y=9 P a g e |125

solving these equation we get

x=2, y=5 Q=(2,5) The shaded region CPQB is a common region

∴ The corner points are C(9,0), P(5,2), Q(2,5), B(0,9)

219

(c) Option :(a)

Shaded region represents given inequality and is a convex set Option :(b)

Shaded region represents given inequality and is a convex set Option :(c)

x=5 → x=± 5 ∴

points on the lines

x=5, x =−5

This is not a convex set. Option :(d) P a g e |126

2

2

3 x +2 y ❑≤ 6 223

i.e.

2

2

x /2+ y ❑ /3 ≤ 1

Shaded region represents given inequality and is a convex set (d) Feasible area ABCDO Vertices A(0,3), B(2,3), C(3,2), D(3,0), O(0,0)

p=20 x +50 y At At At At At

A(0,3),=P=150 B(2,3),=P=40+150=190 C(3,2),=P=60+100=160 D(3,0),=P=60 A(0,0),=P=0

∴

Maximum is 190 at (2,3)

Minimum at O at (0,0)

264

(d)

2 x + y =2

x y → + =1 1 2 This line meets the axes at (1,0) and (0.2). Join these points to obtain the line Now,

2 x + y =2

x− y=3

P a g e |127

x y + =1 3 −3 This line meets the axes at (3, 0) and (0.-3). Join these points to obtain the line

x− y=3

To find common vertex solve

2 x + y =2… (i)

x− y=3 … ( ii ) On solving equations (i) and (ii), we get X=5/3, y=-4/3 Thus common vertex of linear inequalities is (5/3,-4/3)

288

(c) The four vertices of the feasible region are (0, 3), (1.6, 2.4). (3.5, 0); (7/3, 5/3). The corresponding value of z at these points are 21,24.8, 17.5 and 70/3 respectively. Thus the maximum value of z is 24.8.

290

(b) The positive x=2, and the line

x+ y=4

gives common region with end points

O(0,0),A(2,0),B(2,2), c(1,3) and D(0,3) which enclose the feasible region shown shaded, applying

p=3 x +5 y

to these point we get P(0)=0, P(A)=3(2)+5(0)=6,

P(B)=3(2)+5(2)=16, P(C)=3(1)+5(3)=18 and P(D)=3(0)+5(3)=15.thus maximum value of P is 18.

P a g e |128

291

(a)

Z ( 0,20 )=25× 0+30 ×20=600, Z (12,4 )=25 ×12+30 × 4=420 Z ( 18,0 )=25 ×18+ 30× 0=450 ∴ Minimum value of Z is 420.

292

(a)

Z

( 32 ,1)=3 × 32 +1× 1= 112 , Z ( 3,0) =3× 3+1 ×0=9

Z ( 0,4 )=3 ×0+1 × 4=4 ∴ Minimum value of Z is 4.

P a g e |129

293

(d)

Z ( Max ) at

294

( 125 , 25 )

(d) Draw the lines

4 x + y =4, x+3 y =6, x + y=3

The shaded portion gives the feasible region the vertices are A(0,4), for B solving

4 x + y =4 andx+ y =3, we get x=3/2, y=3/2

∴ C=(3/2,3/2) and D is (6,0) ∴

values of objective function

z=12 x +8 y

at vertices are

1 8 76 Z ( A )=12 ×0+ 8 ×4=32, Z ( B )=12 × + 8× = 3 3 3 3 3 Z ( C ) =12× +8 × =30, Z ( D )=12 ×6+ 8× 0=72 2 2 ∴ minimum value of Z is 76/3 at (1/3,8/3)

P a g e |130

295

(c)

Giventhatz=2 x1 +3 x 2 The shaded region is the required region. The maximum value may be on the point A,B,C and

Z A =Z(11,0)=2 (11 ) +3 ( 0 )=22 Z B=Z(4,2) =2 ( 4 ) +3 ( 2 )=14 Z C =Z (1,5 )=2 ( 1 )+ 3 ( 5 )=17 Z D=Z (0,10) =2 ( 0 ) +3 (10 )=30 The minimum value is 14.

296

(d) Draw the lines

x+ y=5, x +2 y=4, 4 x + y=12

The feasible region is shown by shaded portion. The vertices are B(7/3,8/3), obtained by solving

∴

x+ y=5∧4 x + y=12,

C=(0,5) and D is (0,2)

∴ values of objective function

z=6 x +3 y

at vertices are

7 8 Z ( A )=6× 20/ 7+3 × 4 /7=132/7, Z ( B ) =6 × + 3× =22 3 3 P a g e |131

Z ( C ) =6 ×0+3 × 5=15, Z ( D )=6 × 0+3 ×2=6 ∴ 297

minimum value of Z is 22 at (7/3,8/3)

(b)

Z ( 0,1 )=1 ×0+ 2× 1=2, Z

( 125 , 25 )=1 × 125 +2 × 25 = 165

Z ( 0,2 )=1 ×0+ 2× 2=4 ∴

299

Maximum value of Z is 4.

(c) To find the feasible region Consider

∴

x+ 2 y ≤ 8

A =(8,0), B=(0,4) and

3 x+2 y=12

∴ C(4,0), D(0,6) satisfies the equation

P is the point of intersection of the line AB and CD solving equation

x+ 2 y =8,∧3 x+ 2 y =12 we get

x=2, y=3, p=( 2,3) The shaded region OCPB is a feasible region. The optimal solution is obtained at one of the corner point O(0,0), C(4,0), P(2,3) and B(0,4) Corner Points

value of

z=x + y

P a g e |132

z=0

O(0,0)

∴ 300

C(4,0)

z=4

P(2,3)

z=5

B(0,4)

z=4

The maximum value of z is 5 at the point P(2,3)

(a) Consider the equation

2 x +3 y=13

∴ The points A(13/2,0), B(0,13/3) Consider the equation

x+ y=5

∴ C=(5,0) and D=(0,5) Draw the lines

Q is the point of intersection of the lines AB and CD Solving the equation

x+ y=5∧2 x +3 y=13

We get Q(2,3) The shaded region OAQD is a feasible region Optimal solution is obtained at one of the points O(0,0), A(5,0), Q(2,3), D(0,13/3) Corner Points

value of

O(0,0) A(5,0) Q(2,3) D(0,13/3) 301 302

z=6 x +3 y

z=0 z=30+0=0

z=12+9=21 z=0+13=13

The maximum value of z is 30 at the point A(5,0) (b) The objective function of a L.P.P. is a function to be optimized. (d) Consider the equation

( i ) x + y=8 ∴ points A(8,0), B(0,8) P a g e |133

(ii) 6 x+ 4 y =12

∴ C=(2,0) and D=(0,3) (iii) 5 x+8 y =20 E(4,0), F(0,5/2) Draw the lines

Q is the point of intersection of the lines CD and EF Solving the equation

6 x+ 4 y =12∧5 x+ 8 y=20

We get Q(4/7,15/7) The shaded region ABDQE is a feasible region Optimal solution is obtained at one of the points A(8,0), D(0,3), B(0,8), Q(4/7,15/7), E(4,0) Corner Points

303

value of p=7 x +4 y

A(8,0)p=56+0=56 B(0,8) p=0+32=32 D(0,3) p=0+12=12 Q(4/7,15/7) p=28/7+60/7=12.57 E(4,0) p=28+0=28 The minimum value of P is 12 at the point D(0,3) (d) Consider the equation

( i ) 4 x+ y=4 ∴ points A(1,0), B(0,4) (ii) x+3 y =6

∴ C=(6,0) and D=(0,2) (iii) x+ y=3 E(3,0), F(0,3) Draw the lines G is the point of intersection of the lines CD and EF Solving the equation

x+ 3 y =6 andx + y=3

We get G(3/2,3/2) H is the point of intersection of the lines AB and EF Solving the equation P a g e |134

1 x= , y=8 /3 3 We get H(1/3,8/3)

The shaded region CGHB is a feasible region Optimal solution is obtained at one of the points C(6,0), G(3/2,3/2), H(1/3,8/3), B(0,4) Corner Points

304

value of p=12 x+ 8 y

C(6,0) p=72+0=72 G(3/2,3/2) p=18+12=30 H(1/3,8/3) p=4+64/3=76/3=25.33 B(0,4) p=0+32=32 The minimum value of P is 25.33 at the point H(1/3,8/3) (c) We draw the lines

x+ y=5

by joining two points A(5,0) and B(0,5)

x=3

passing through the points C(3,0) and parallel to y axis

y=3

passing through the points D(0,3) and parallel to x axis

The point P point of intersection of the lines

x+ y=5 andx=3

solving these equation

x+ y=5 andy=3

solving these equation

x=3, y=2 P=(3,2) The point Q point of intersection of the lines

y=3, x=2 Q=(2,3)

P a g e |135

The shaded region OCPQD is a feasible region Optimal solution is obtained at one of the corner points O(0,0), P(3,2), Q(2,3), D(0,3) Corner Points

305

O(0,0) z=0+0=0 P(3,2) z=60+100=160 Q(2,3) z=40+150=190 D(0,3) z=0+150=150 The minimum value of z is 190 at the point Q(2,3) (c) Solving

x=

306

value of z=20 x+ 50 y

2 x +3 y=6 andx+ 4 y=4, weget

12 2 , y= 5 5

Hence vertex is (12/5,2/5) (d) Common region is quadrilateral.

P a g e |136

307

(d) We draw the lines

x+ 2 y =8

by joining two points A(8,0) and B(0,4)

2 x +2 y=9 2 x + y =7

by joining two points C(9/2,0) and D(0,9/2) by joining two points E(7/2,0) and F(0,7/2)

The point P point of intersection of the lines

2 x +2 y=9∧2 x+ y =7

solving these

equation

x=5/2, y=2 P=(5/2,2)

The point Q is the point of intersection of the lines

x+ 2 y =8∧2 x +2 y =9

solving these equation we get

x=1, y=7/2

308

Q=(1,7/2) The point Q(1,7/2) =(1,3.5) provide the solution to the L.P.P. (c) We draw the lines

5 x+ y =10

2 x +2 y=12

by joining two points A(2,0) and B(0,10) by joining two points C(6,0) and D(0,6)

x+ 4 y =12 by joining two points E(12,0) and F(0,3) The point G point of intersection of the lines

x+ 4 y =12∧2 x+2 y=12

solving these

equation

x=4, y=2 P a g e |137

G=(4,2) The point H is the point of intersection of the lines

2 x +2 y=12∧5 x + y=10

solving these equation we get

x=1, y=5 H=(1,5)

The shaded region EGHB is a feasible region Optimal solution is obtained at one of the points E(12,0), G(4,2), H(1,5), B(0,10) Corner Points

309

value of z=4 x+5 y

E(12,0) p=48+0=48 G(4,2) p=16+10=26 H(1,5) z=4+25=29 B(0,10) z=0+50=50 The minimum value of p is 48 at the point G(4,2) (d) We draw the lines

2 x + y =9

by joining two points A(9/2,0) and B(0,9/2)

x+ 2 y =9

by joining two points C(9,0) and D(0,9/2)

x+ y=7

by joining two points E(7,0) and F(0,7)

P is the point of intersection of the lines

x+ 2 y =9 andx+ y=7

solving these equation

x=5, y=2 G=(5,2)

P a g e |138

The point Q is the point of intersection of the lines

x+ y=7∧2 x + y=9

solving these equation we get

x=2, y=5 Q=(2,5) The shaded region CPQB is a common region

∴ The corner points are C(9,0), P(5,2), Q(2,5), B(0,9) 310

(c) Option :(a)

Shaded region represents given inequality and is a convex set Option :(b)

Shaded region represents given inequality and is a convex set Option :(c)

x=5 → x=± 5

∴ points on the lines

x=5, x =−5

This is not a convex set. Option :(d)

3 x2 +2 y 2❑≤ 6 i.e.

x 2 /2+ y 2❑ /3 ≤ 1

Shaded region represents given inequality and is a convex set P a g e |139

329

(c)

Z =3 x +2 y Zat ( 0,2 )=3 ( 0 ) +2 ( 2 )=4 Zat ( 0,0 )=3 ( 0 ) +2 ( 0 ) =0 0=¿ Zat (2,0 )=3 ( 2 ) +2 ¿

∴ Zismaximizeat (2,0) ∴ ( c ) iscorrectanswer .

330

(b)

Z =3 x +2 y Zat ( 0,3 )=3 ( 0 ) +2 ( 3 )=6 Zat ( 0,1 )=3 ( 0 ) +2 ( 1 )=2 Zat

( 23 , 73 )=3 ( 32 )+2( 73 )= 203

2 7 ∴ Zismaximizeat ( , ) 3 3 ∴ ( B ) iscorrectanswer .

331

(d)

Z =9 x+13 y

P a g e |140

Zat ( 0,6 )=9 ( 0 )+ 13 ( 6 )=78 Zat (5,0 )=9 ( 5 ) +13 ( 0 )=45 Zat (3,4 )=9 ( 3 )+13 ( 4 ) =79 ∴ Zismaximizeat (3,4 )

∴ ( D ) iscorrectanswer .

332

(b)

P=30 x +20 y Pat ( 0,3 )=30 ( 0 ) +20 ( 3 ) =60 Pat ( 1,3/2 )=30 ( 1 )+ 20 (3 /2 )=60 Pat ( 4,0 ) =30 ( 4 ) +20 ( 0 )=120

Pat ( 0,8 )=30 ( 0 )+20 ( 8 )=160

Pat ( 8,0 )=30 ( 8 )+20 ( 0 )=240 ∴ Zisminimiumat (1,3/2)

∴ ( B ) iscorrectanswer .

339

(a) 2

Function x +2 x +3 340

satisfies all the points.

(a)

P a g e |141

200 x+100 y=2000

i. e .2 x+ y=20 …(1) Represent a line through the points A(10,0) and B(0,20)

x+ 2 y =50 …(2) Represent a line through the points C(50,0) and D(0,25)

40 x +40 y=1400 i. e . x + y=35 … ( 3 ) Represent a line through the points E(35,0) and F(0,35) Lines (2) and (3) intersect at points G(20,15) Feasible region is the shaded portion FGC with vertices F(0,35),G(20,15),C(50,0)

z=4 x+ 8 y At At At At At

F(0,35), G(20,15), C(50,0), D(3,0),=P=60 A(0,0),=P=0

∴ 341

Z=4(0)+8(35)=280 Z=4(20)+8(15)=200 Z=4(50)+8(0)=200

minimum value of Z is 200 at seg GC

(b)

Consider the equation,

2 x − y=1

P a g e |142

342

We may write it as

x y + =1 1 −1 2

This shows the line

2 x − y=1

makes intercepts of ½ and -1

On the axes. Thus, the line meets X-axis at (1/2,0) and Y-axis at (0,-1). We plot these points and join them by thick line. Consider (0, 0).Clearly (0, 0) does not satisfy the given in equation. Therefore, out of portions divided by this line, the one not containing (0, 0) together with the points on the line forms a solution set. (b)

Consider the equation,

x+ y=1

It meets the axes at A (1, 0) and B (0, 1). Join these points. Clearly, the portion not containing (0,0) is the solution set of Consider

x+ y ≥ 1

x y 7 x+ 9 y=63 i. e . + =1 9 7

Thus, the line meets axes at C (9,0) and D(0,7). Join these points. Clearly the portion containing (0,0) is the solution set of

7 x+ 9 y ≤ 63

Y=5, is a line parallel to X-axis at the distance 5 from X-axis and the portion containing (0,0) is the solution set of the in equation

y ≤ 5.

x=6, is a line parallel to Y-axis at the distance 6 from Y-axis and the portion containing (0,0) is the solution set of the in equation Clearly, x ≥ 0 Also, 343

y≥0

x ≤ 6.

has a solution represented by Y-axis and the portion on its right. a solution represented by X-axis and the portion above it.

The shaded region represents the solution of the given system of inequations. (b)

P a g e |143

X:distance travelled with speed 25km/hr y:distance travelled with speed 40km/hr

∴ maximizez=x + ysubjecttoconstraints

2 x +5 y ≤100,

x y + ≤1, x ≥ 0, y ≥ 0 25 40

2 x +5 y=100 … ( 1 ) Represents a line through the points A(50,0) and B(0,20)

x y + =1 … ( 2 ) 25 40 Represents a line through the point C(25,0) and D(0,40) Lines (1) and (2) interacts the point E(50/3,40/3) The feasible region is OCEB with corner point O(0,0),C(25,0),E(50/3,40/3),B(0,20) At O(0,0) Z=0+0=0 At C(25,0) Z=25+0=25 At E(50/3, 40/3) Z=50/3+40/3=30 At B(0,20) Z=0+20=20

∴ 344

Maximum distance covered in 1 hour is 30 km with 50/3 km at 25km/hr and

40/3 km at 40 km/hr. (b)

Let x kg of wheat and y kg of rice be mixed. Then we have to minimize the cost function

C=2 x +8 y

subject to the constraints

given below. P a g e |144

X kg of wheat and y kg of rice must give at least 50 gm i.e.0.05 kg of proteins.

∴ 0.1 x +0.05 y ≥ 0.05 ∴10 x +5 y ≥ 5∨2 x + y ≥ 1 Similarly, x kg of wheat and y kg of rice must give at least 200gm i.e.0.2kg of carbohydrates.

∴ 0.25 x +0.5 y ≥0.2 ∴25 x +50 y ≥ 20∨5 x +10 y ≥ 4 Thus, we have to minimize

C=2 x +8 y

subject to the constraints

x ≥ 0, y ≥ 0, 2 x + y ≥1, 5 x+10 y ≥ 4 2 x + y =1… ( 1 ) x y + =1 i.e. 1 1 meets the axis at A(1/2,0) and B(0,1) 2 5 x+10 y=4 … ( 2 ) x y + =1 4 2/5 i.e. meets the axis at C(4/5,0) and D(0,2/5) 5 x=0 is the y-axis and y=0 is the x-axis. Lines(1) and (2) meets at E (6/15,3/15) The feasible region is DEB with corner point D(0,2/5),E(6/15,3/15),B(0,1) The vales of

c=2 x+ 8 y

The minimum value of 345

at these points are 3.2, 2.4 and 8 respectively.

c=2 x+ 8 y

is 2.4 obtained when x=2/5 and y=1/5 i.e. when

x=400 gm and y=200 gm .then the diet cost is minimum. (c)

X=2 represents a line through the points A (2, 0) and parallel to Y-axis. y=3 represents a line through the points B (0, 3) and parallel to X-axis. Line

x+ y=4

meets the axes at C(4,0) and D(0,4)

x=2 andx+ y=4

Intersect at the point E (2,2) and the lines

y=3 andx + y=4

intersect at the point F(1,3). P a g e |145

∴ The shaded portion OAEFB is the feasible region with vertices O(0,0),A(2,0),E(2,2),F(1.3),B(0,3).

P=3 x +5 y

346

P(A)=3(2)+5(0)=6 P(E)=3(2)+5(2)=16 P(A)=3(1)+5(3)=18 P(A)=3(0)+5(3)=15 Hence the maximum value of p is 18. (d)

2 x + y =4

meets the axes at A(2,0) and B(0,4)

x+ 2 y =4

meets the axes at C(4,0) and D(0,2)

Lines

2 x + y =4 andx+2 y=4 intersect at the points

E(4/3,4/3)

∴ The feasible region is shaded portion OAED with vertices O(0,0),A(2,0),E(4/3,4/3),D(0,2) P= x+y P(A)=2+0=2 P(E)=(4/3+4/3)=(8/3) P(D)=0+2=2

∴ 347

Maximum value of P is (8/3).

(a)

P a g e |146

2 x + y =6 meets the axes at A(3,0) and B(0,6) x+ y=4

348

meets the axes at C(4,0) and D(0,4)

X=y represents a line through origin. Hence only those points which lie in the region between 2x+y=6 and x=y are feasible. Lines 2x+6=6 and x=y intersect at the points E (2, 2) which is obvious that the point at which E (2, 2) =2+3(2) =8 is least. (b)

x+ y=4 meets the axes at A(4,0) and B(0,4) 3 x+ 8 y =24

10 x+7 y =35

meets the axes at C(8,0) and D(0,3) meets the axes at E(35/10,0) and F(0,5)

Lines 10x+7y=35 and x+y=4 intersect at point G(7/3,5/3) Lines 3x+8y=24 and x+y=4 intersect at point H(8/5,12/5)

∴

The shaded portion OEGHD is the feasible region with vertices O(0,0),

E(3.5,0),G(2.4,1.6), H(1.6,2.4), D(0,3) The value of z=5x+5y at these points are Z(E)=17.5 Z(G)=20 Z(H)=20 Z(D)=15

∴ The maximum value is 20. 349

(b)

P a g e |147

x+ 2 y =2000 meets the axis at A(2000,0) and B(0,1000) x+ y=1500

meets the axis at C(1500,0) and D(0,1500)

X=600 represents a line parallel to Y-axis passing through E(600,0) Lines x=600 and x+2y=2000 intersect at point F(600,700)

∴ The shaded portion OEFB is the feasible region with vertices O(0,0), E(600,0),F(600,700), B(0,1000), D(0,3) The corresponding value of z=3x+5y at these points are Z(E)=1800 Z(G)=5300 Z(H)=5000

∴ 350

The maximum value is 5300.

(d)

4 x +2 y=46 meets the axis at the points A(23/2,0) and B(0,23) X+3y=24 meets the axis at the points C(24,0) and D(0,8) Lines 4x+2y=46 and x+3y=24 intersect at point E(9,5)

∴ The shaded portion OAED is the feasible region with vertices A(23/2,0),E(9,5), D(0,8) The corresponding values of P=4x+2y are P(A)=46 P(E)=46 P(D)=16

∴ 351

Hence The maximum value is 46.

(c)

P a g e |148

x+ y=8 meets the axis at A(8,0) and B(0,4) 6 x+ 8 y=12

meets the axis at C(2,0) and D(0,3)

5 x+8 y =20

meets the axis at E(4,0) and F(0,20/8)

Lines 6x+4y=12 and 5x+8y=20 intersect at point P(4/7,15/7)

∴

The shaded portion EABGDP is the feasible region with vertices E(4,0),

A(8,0),B(0,8), D(0,3), P(4/7,15/7) The value of function z=30x+20y at the above vertices are Z(E)=120 Z(A)=240 Z(B)=160 Z(D)=60 Z(P)=60

∴ The Z has minimum value 60 at P(4/7,15/7) and C(0,3). 352

(a)

x+y=5 …(1) Represent a line through the points A(5,0) and B(0,5) x+2y=4 …(2) Represent a line through the points C(4,0) and D(0,2) 4x+y=12 …(2) Represent a line through the points E(3,0) and F(0,12) Lines (2) and (3) intersect a point G(20/7,4/7) Lines (1) and (3) intersect a point H(7/3,8/3) The feasible region is the shaded portion DGHB with vertices D(0,2),G(20/7,4/7),H(7/3,8/3),B(0,5) Values of objective function z=6x+3y at these vertices are

z ( D ) =6 ×0+3 × 2=6 4 132 Z ( D )=6 × 0+3 × = 7 7 7 8 Z ( D )=6 × +3 × =22 3 3 z ( B )=6 × 0+3 ×5=15 ∴ maximum value of Z is 22 at (7/3,8/3) P a g e |149

353

(b)

x=3 Represents a line through the points A (3, 0) and parallel to Y-axis. y=3 Represents a line through the points B (0, 3) and parallel to X-axis. x+y=5 meets the axes at point C(5,0) and D(0,5) 2x+y=4 meets the axes at point E(2,0) and F(0,4) Lines x=3 and x+y=5 intersect at point G(3,2) Lines y=3 and x+y=5 intersect at point H(2,3) Lines y=3 and 2x+y=4 intersect at point I(1/2,3) The shade portion EAGHI is feasible region with vertices E(2,0),A(3,0),G(3,2),H(2,3) and I(1/2,3) Values of objective function z=3x+2y at these vertices are Z(E)=6 Z(A)=9 Z(G)=13 Z(H)=12 Z(I)=15/2

∴ Z has maximum value 13 at C(3,2) 354

(c)

2x+3y=6 meets the axes at points A(3,0) and B(2,0) 2x+y=4 meets the axes at point C(2,0) and D(0,4) x=4 Represents a line through the points E (4, 0) and parallel to Y-axis. y=6 Represents a line through the points F (0, 6) and parallel to X-axis. Lines y=6 and x=4 intersect at point G(4,6) Lines 2x+y=4 and 2x+3y=6 intersect at point H(3/2,1) The shade portion AEGFDH is feasible region with vertices A(3,0), E(4,0),G(4,6),F(0,6),D(0,4) and H(3/2,1) Values of objective function z=3x+2y at the above vertices are P a g e |150

Z(A)=9 Z(B)=12 Z(G)=24 Z(F)=12 Z(D)=8 Z(H)=13/2

∴ Z has minimum value 13/2 355

(c)

2 x 1 + x 2=104 x 1+2 x 2=76

meets the axes at points A(52,0) and B(0,104) meets the axes at points C(76,0) and B(0,38)

The origin (0, 0) satisfy both the given in equalities, so the regions below the two lines containing the origin are the regions satisfy these constants. Lines

2 x 1 + x 2=104∧x 1+2 x 2=76

intersect at point E(44,16)

The shaded portion OAED is feasible region with vertices O(0,0),A(52,0), E(44,16), D(0,38) Values of objective function

z=6 x 1+11 x 2

at the above vertices are

Z(O)=0 Z(A)=312 Z(E)=440 Z(D)=418

∴ 356

The maximum value of z is 440.

(c) In the given figure origin is present in shaded area, so equations are

2 x +5 y ≤80, x+ y ≤ 20, x ≥ 0, y ≥ 0 357

(a) The line passes through origin and it represents the half plane that contains the positive X-axis.

P a g e |151

358

(b)

Required region is rectangle ABCD Point A is point of intersection of -3x+2y=3 and 6x+4y=24

3 15 ∴ A ≡( , ) 2 14 Point B is point of intersection of -3x+2y=3 and 2x+3y=6

∴ B ≡(

3 24 , ) 13 13

Point C is point of intersection of x-2y=2 and 2x+3y=6

∴C ≡(

18 2 , ) 7 7

Point D is point of intersection of x-2y=2 and 6x+4y=24

7 3 ∴ D ≡( , ) 2 4 Z=5x+2y

atA ≡

( 32 , 1514 ) , z=15, AtB ≡ ( 133 , 2413 ) , z=4.84

atC ≡

( 187 , 27 ), z=13.42, AtD ≡( 72 , 34 ) , z =19

∴ for the maximum value of Z=5x+2y, x=7/2, y=3/4 359

(a) P a g e |152

Let x unit of food A and y unit of food B be used to give the sick person the least quantities of vitamins, mineral and calories. Total cost of food is C=4x+3y which is to be minimized. For x unit of food A and y units of food B a total of 200x+100y units of vitamins is obtained and its minimum requirement is 1400.

∴ we have 200 x+100 y ≥1400 Similarly for minerals, we have:

x+ 2 y ≥50 And for calories

40 x +40 y ≥1400

Since the number of units purchases cannot be negative.

x ≥ 0, y ≥ 0

360

After plotting the constraints we get the feasible region as FGC with vertices F(0,35), G(20,15) and C(50,0) Values of C=4x+3y at above vertices are C(F)=4(0)+3(35)=105 C(G)=4(20)+3(15)=125 C(H)=4(50)+3(0)=200 Thus the minimum cost is 105 at x=0 and y=35. (c)

Let A=number of product A and y=number of product B, which the carpenter produces. Total profit is p=30x+20y which is to be maximized subject to the constraints P a g e |153

2 x + y ≤20, x +3 y ≤ 15, x ≥ 0, y ≥ 0 Now 2x+y=20 …(1)

i. e .

x y + =1 is a line through the points A(10,0) and B(0,20) 10 2

Now x+3y=15 …(1)

i. e .

x y + =1 is a line through the points C(15,0) and D(0,5) 15 5

Moreover lines (1) and (2) intersect at R(9,2) Clearly shaded the portion OARD in the figure is the feasible and its corner points are O(0,0), A(10,0), R(9,2) and D(0,5) At O(0,0) P=30(0)+20(0)=0 At A(10,0) P=30(10)+20(0)=300 At R(9,2) P=30(9)+20(2)=310 At O(0,0) P=30(0)+20(5)=100

∴ 361

maximum profit is Rs.310 at R(9,2)

i.e. when 9 units of product A and 2 units of product B are produced. (a)

x+4y=4 …(1) Represents a line through the points A(4,0) and B(0,1) 2x+3y=6 …(1) Represents a line through the points C(3,0) and D(0,2) Lines (1) and (2) intersect at the points P(12/5,2/5) Also, x ≥ 0, y ≥ 0.

362

The feasible region of the given L.P.P. is the shaded region BPD of figure and its corner points are B(0,1), P(12/5,2/5), D(0,2) The objective function is f(x,y)=x+2y F(0,1)=0+2=2 f(0,2)=0+2(2)=4 Clearly maximum value of f(x,y) is 4 at (0,2) and its minimum value is 2 at (0,1). (d)

P a g e |154

Draw the lines x+y=4, 3x+4y=12 Shaded region is feasible region The vertices are A (4,0), D(0,3), O(0,0). The value of objective function Z=5x+7y at the vertices. Z(O)=0 Z(A)=5x0+7x3=21

∴ Z has maximum value 21 at (0,3) 363

(b)

364

x+2y=4 …(1) Represents a line through the points A(4,0) and B(0,2) x-y=-2 …(2) Represents a line through the points C(-2,0) and B(0,2) Lines (1) and (2) intersect at the points B(0,2) x=2 …(3) Is a line parallel to Y-axis through point D(2,0) and it meets lines (1) and (2) in the points P(2,1) and Q(2,4) respectively. y=-2 …(4) Is a line parallel to X-axis through point E(0,-2) and it meets lines (1) , (2) and (3) in the points L(8,-2) M(-4,-2) and N(2,-2) respectively. The feasible region of the given L.P.P. is the convex polygon MNPB(shaded portion) in the figure and its corner points are M(-4,-2), N(2,-2), P(2,1),B(0,2) The objective function is f(x,y)=x+2y f(-4,-2)=-4-4=-8 f(2,-2)=2-4=-2 f(2,1)=2+2=4 f(0,2)=0+4=4 Clearly maximum value of ‘f’ is 4 at (2, 1) and (0,2) and minimum value f is -8 at (-4,2). (a) P a g e |155

Let

x 1 , x 2 denote the number of questions to be attempted from section A, B

respectively. Then mathematically the problem can be written as maximize

M =10 x 1+15 x 2 [Total

marks] subject to constraints;

x 1 ≥ 2, x 2 ≥ 3, 15 x1 +25 x 2 ≤ 180, [3 hours = 180 minutes]

x 1+ x 2 ≤10∧x1 ≥ 0, x 2 ≥ 0 Consider a set of rectangular Cartesian axes OX1, OX2 in the plane. Each point has coordinate of the type

( x 1 , x 2)

and conversely. Clearly any point

which satisfies the conditions

x 1 ≥ 0, x 2 ≥0,

pair

(x 1 , x 2)

lies in the first quadrant only. Thus our search for the desired number

x (¿ ¿ 1 , x 2 ) ¿

Will be restricted to the first quadrant only. Also since

x 1 ≥ 2, x 2 ≥ 3, 15 x1 +25 x 2 ≤ 180, x 1+ x 2 ≤10 ∴

The desired point

(x 1 , x 2)

lies some where in the convex region GHI which is

shown (shaded) in the figure. This is feasible region i.e. solution space of the problem. The extreme points are G(2,3), H(7,3), I(2,6) At G(2,3), M=10(2)+15(3)=65 At H(7,3), M=10(7)+15(3)=115 At I(2,6), M=10(2)+15(6)=110 Since the maximum value occurs at H,

∴ x 1=7

is the optimal solution.

x 2=3∧115

is the maximum marks. P a g e |156

365

(a)

2 x 1 +7 x 2=22 x 1+ x 2=6

meets the axes at points A(11,0) and B(0,22/7)

meets the axes at points C(6,0) and D(0,6)

5 x1 + x 2=10

meets the axes at points E(2,0) and F(0,10)

Lines

2 x 1 +7 x 2=22∧x 1 + x 2=6

Lines

5 x1 + x 2=10∧x1 + x 2=6

intersect at point G(4,2) intersect at point H(1,5)

The shaded region FHGA is feasible region with vertices F(0,10),H(1,5), G(4,2), A(11,0) Values of objective function

z=2 x1 +3 x 2

at the above vertices are

Z(F)=30 Z(H)=17 Z(G)=14 Z(A)=11

∴ The minimum value is 14. 366

(c) Let process I and II be x and y times. Then

5 x+6 y ≤ 250,3 x +5 y ≤ 200

5 x+ 4 y ≤150, 8 x+ 4 y ≥ 80, x ≥0, y ≥0 Clearly 367

8 x+ 4 y ≥ 80 does not affect.

So, the number of constraints is 5. (b)

P a g e |157

x y 2 x + y =2i . e . + =1 1 2 This line meets the axes at (1,0) and (0,2).

x y x− y=3 i . e . + =1 3 −3 This line meets the axes at (3,0) and (0,-3) To find common vertex solve 2x+y=2 …(i) And x-y=3 …(iI) On solving equations (i) and (ii), we get

5 −4 x= , y= 3 3 Thus common vertex of linear in equalities is 368

5 4 ( ,− ) 3 3

(b) The time taken in two simple question

3 ¿ 2× =3 minute and in one difficult 2

questions =3 minute.

∴ → 369

For three successive questions total time taken= 6 minute. Total solved question in

3 hours=

180 ×3=30 ×3=90 . 6

(a)

i=1, x 11 + x 12 + x 13+...+ x 1 n i=2, x21 +x 22 +x 23+ ...+ x2 n i=3, x31 + x 32+ x 33+ ...+ x 3 n ……………………………………………….

i=m , x m 1 + x m 2+ xm 3 +...+ x mn Thus number of constraints is m (II) condition

j=1, x 11 +x 21+ x 31 +...+ xm 1 P a g e |158

j=2, x 12+ x 22 + x 32+...+ x m 2 ……………………………………………….

j=n , x 1 n + x 2n + x 3 n +...+ x mn Thus number of constraints is n

∴ Total number of constraints = m+n 370 371

372

(d) In the given figure the feasible region for given constraints is line segment GI. (d) A column in the simplex table that contains all the variables in the solution is called pivot or key column. (c)

(A)

(C)

380

(B)

(D)

From the above figure only C is not a convex polygon. (b)

P a g e |159

Consider the equation,

381

2 x − y=1

We may write it as

x y + =1 1 −1 2

This shows the line

2 x − y=1

makes intercepts of ½ and -1

On the axes. Thus, the line meets X-axis at (1/2,0) and Y-axis at (0,-1). We plot these points and join them by thick line. Consider (0, 0).Clearly (0, 0) does not satisfy the given in equation. Therefore, out of portions divided by this line, the one not containing (0, 0) together with the points on the line forms a solution set. (c)

x=2 represents a line through the points A (2, 0) and parallel to Y-axis. y=3 represents a line through the points B (0, 3) and parallel to X-axis. Line

x+ y=4

meets the axes at C(4,0) and D(0,4)

x=2 andx+ y=4

Intersect at the point E (2, 2) and the lines

y=3 andx + y=4

intersect at the point F (1, 3).

∴ The shaded portion OAEFB is the feasible region with vertices O(0,0),A(2,0),E(2,2),F(1,3),B(0,3).

P=3 x +5 y

382

P(A)=3(2)+5(0)=6 P(E)=3(2)+5(2)=16 P(A)=3(1)+5(3)=18 P(A)=3(0)+5(3)=15 Hence the maximum value of p is 18. (a) If we puts the given points in the given constraints, we find that (2,3), (3,1), (2,2) are not satisfying

x+ 2 y ≥11

because

¿ ( 2,3 ) , ( 2+2 ×3 ) ≥ 11 isfalse ¿ ( 3,1 ) , ( 3+3 ×1 ) ≥ 11 isfalse ¿ ( 2,2 ) , ( 2+2 ×2 ) ≥11 isfalse ¿ ( 3,4 ) , issatisfyingas 3+2× 4 ≥ 11 istrue 383

(c) P a g e |160

We draw the graph of

x 1 ≤ 4, x2 ≤ 6,

x1 x 2 + ≤1 6 9

Different point of convex polygon are (0,0), (4,0), (4,3), (2,6) and (0,6) Maximum value of

z=3 x 1+5 x 2

is at (2,6),z=3x2+5x6=36

∴ x 1=2, x 2=6, z=36 isthesolution. 384

(a) A set S is called convex if for all The set

386

x 1 , x 2 , ϵSand 0 ≤ λ ≤1, λ x 1+ (1−λ ) x 2 ϵS .

{x : x =5 } is not a convex set, can be seen by drawing the region.

(d)

On drawing graphs of given in equations, We have Now, Z=3x+4y

Z ( 0,0)=0 Z ( 40,0)=120++ 0=120 Z ( 0,30)=0++120=120 Z ( 20,20)=60++80=140 ∴ Hence the maximum value =140. 387

Which is obtained at x=20, and y=20 (c) P a g e |161

Graph of given constraints is shown below: Co-ordinates of B from 2x+2y=9 and 2x+y=7 are (2.5,2) Co-ordinates of C from 2x+2y=9 and x+2y=8 are (1, 3.5). value of objective function P=2x+3y at various points P(0,0)=0+0=0, P(2.5,2)=11, P(3.5,0)=7, P(1,3.5)=12.5

∴ 388

P (max.)=12.5 which is at (1,3.5)

(d) The L.P.P. in standard form is

z=21 x1 +15 x 2+ 0 s 1+0 s2

Maximize Subject to ,

x 1∧x2

x 1+2 x 2+ s 1=6, 4 x 1 +3 x2 + s2=12, x 1 , x 2 ≥ 0, s 1 , s2 ≥0

are decision variable.

s 1∧s2

are slack variables are n=4 and number of

equations are m=2.

∴ 389 391

The total possible basic solution are

(c) Obviously the ans is (C) (b) For maximization L.P. model, the simplex method is terminated when all values

c j −z j 392

nCm=4 C 2=6

are less than or equal to zero.

(a) (3,1), (2,0) are vertices of Min z for (2,0)

P a g e |162

Hence

x 1=2

393

(a)

394

c (c) The graph of linear programming problem is as given below.

Hence the required feasible region by the graph whose vertices are A(1.2,2.6), B(4.5,1.5) and C(8/3,10/3) Thus objective function is minimum at A(1.2,2.6) so

x 1=1.2, x 2=2.6 andz =2× 1.2+ 3× 2.6=10.2 395

(d) Clearly point D is outside.

396

(b)

(

A= 0,

134 , B ( ¿ 0,40 ) andC (10,38) 3

)

P a g e |163

397

399

Max z for C i.e. =40+342=382. (d) The shaded region represents the bounded region. (3, 3) satisfies, so x=3, y=3 and z=15.

(a) The equations, corresponding to in equalities

3 x+2 y ≤6∧6 x +4 y ≥ 20 are 3 x +2 y=6∧6 x +4 y =20.

So the line represented by these

equations are parallel.

400

(c) Solving 2x+3y=6 and x+4y=4, we get

x=

12 2 , y= 5 5

Hence vertex is

(

12 2 , ) 5 5

P a g e |164

402 403

404

(d) Min z= 2(0)+ 2 (40)80 (a) Min z= 2(2)+2(3)=C=10

(b) To test the origin for 2x+y=2, x-y=1, and x+2y=8 in reference to shaded area,

0+0< 2 is true for 2x+y=8 in reference to shaded area,

0+0< 2 is true for

2x+y=2 So for the region does not include origin (0,0), Again for x-y=1,

2x+ y ≥2

0−05

∴2 x + y−5> 0 Substituting origin in above equation, we get

2× 0+0−5>0 ∴−5>0 It does not satisfied the inequality Inequality does not contain origin Required solution set is open half plane not containing the origin.

425

(c) Constraints are

2 x +3 y ≤36, 5 x+2 y ≤50, 2 x +6 y ≤ 60, x ≥ 0, y ≥0

∴Thenumberofconstraintsare 5. 426

(a)

P a g e |168

The inequality is

3 x+ 4 y ≤12

x y ∴ + ≤1 4 3 ∴ 427

The half plane containing the origin and the points of the line

3 x+ 4 y=12

is

the required solution set. (a)

The feasible region is the shaded portion with vertices (1,0), (4,0), (4,4), (3/4,3/4)

P a g e |169