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Linear Programming

Class 12th Mathematics RS Aggarwal Solution
Exercise 33a
  1. x + y ≥ 4 Graph the solution sets of the following inequations:
  2. x - y≤3 Graph the solution sets of the following inequations:
  3. x + 2y1 Graph the solution sets of the following inequations:
  4. 2x - 3y4 Graph the solution sets of the following inequations:
  5. x ≥ y - 2 Graph the solution sets of the following inequations:
  6. y - 2≤3x Graph the solution sets of the following inequations:
  7. 2x + y1 and 2x - y ≥ - 3 Solve each of the following systems of
  8. x - 2y ≥ 0, 2x - y≤ - 2 Solve each of the following systems of simultaneous…
  9. 3x + 4y ≥ 12, x ≥ 0, y ≥ 1 and 4x + 7y≤28 Solve each of the following systems…
  10. Show that the solution set of the following linear constraints is empty:x - 2y…
  11. Find the linear constraints for which the shaded area in the figure given is…
Exercise 33b
  1. Find the maximum value of Z = 7X + 7Y, subject to the constraints.x ≥ 0, y ≥ 0,…
  2. Find the maximum value of Z = 7X + 7Y, subject to the constraints.x ≥ 0, y ≥ 0,…
  3. Maximize Z = 4x + 9y, subject to the constraintsx geater than or equal to 0, y…
  4. Maximize Z = 4x + 9y, subject to the constraintsx geater than or equal to 0, y…
  5. Find the minimum value of Z = 3x + 5y, subject to the constraints- 2x + y ≤ 4,…
  6. Find the minimum value of Z = 3x + 5y, subject to the constraints- 2x + y ≤ 4,…
  7. Minimize Z = 2x + 3y, subject to the constraintsx ≥ 0, y ≥ 0, x + 2y ≥ 1 and x…
  8. Minimize Z = 2x + 3y, subject to the constraintsx ≥ 0, y ≥ 0, x + 2y ≥ 1 and x…
  9. Maximize Z = 3x + 5y, subject to the constraintsX + 2y ≤ 2000, x + y ≤ 1500, y…
  10. Maximize Z = 3x + 5y, subject to the constraintsX + 2y ≤ 2000, x + y ≤ 1500, y…
  11. Find the maximum and minimum values of Z = 2x + y, subject to the constraintsX…
  12. Find the maximum and minimum values of Z = 2x + y, subject to the constraintsX…
  13. Mr.Dass wants to invest ₹12000 in public provident fund (PPF) and in national…
  14. Mr.Dass wants to invest ₹12000 in public provident fund (PPF) and in national…
  15. A small firm manufactures necklace and bracelets. The total number of necklace…
  16. A small firm manufactures necklace and bracelets. The total number of necklace…
  17. A man has ₹1500 to purchase rice and wheat. A bag of rice and a bag of wheat…
  18. A man has ₹1500 to purchase rice and wheat. A bag of rice and a bag of wheat…
  19. A manufacture produces nuts and bolts for industrial machinery. It takes 1…
  20. A manufacture produces nuts and bolts for industrial machinery. It takes 1…
  21. Two tailors, A and B, earn ₹300 and ₹400 per day respectively. A can stitch 6…
  22. Two tailors, A and B, earn ₹300 and ₹400 per day respectively. A can stitch 6…
  23. A dealer wishes to purchase a number of fans and sewing machines. He has only…
  24. A dealer wishes to purchase a number of fans and sewing machines. He has only…
  25. A firm manufactures two types of products, A and B, and sells them at a profit…
  26. A firm manufactures two types of products, A and B, and sells them at a profit…
  27. A manufactures produces two types of soap bars using two machines, A and B. A…
  28. A manufactures produces two types of soap bars using two machines, A and B. A…
  29. A manufacture of a line of patent medicines is preparing a production plan on…
  30. A manufacture of a line of patent medicines is preparing a production plan on…
  31. A toy company manufactures two types of dolls, A and B. Each doll of type B…
  32. A toy company manufactures two types of dolls, A and B. Each doll of type B…
  33. A small manufacture has employed 5 skilled men and 10 semiskilled men and…
  34. A small manufacture has employed 5 skilled men and 10 semiskilled men and…
  35. A company producing soft drinks has a contrast which requires a minimum of 80…
  36. A company producing soft drinks has a contrast which requires a minimum of 80…
  37. A small firm manufactures gold rings and chains. The combined number of rings…
  38. A small firm manufactures gold rings and chains. The combined number of rings…
  39. A manufacture makes two types, A and B, of teapots. Three machines are needed…
  40. A manufacture makes two types, A and B, of teapots. Three machines are needed…
  41. A manufacture makes two product, A and B. product A sells at ₹200 each and…
  42. A manufacture makes two product, A and B. product A sells at ₹200 each and…
  43. A man owns a field area 1000 m2. He wants to plant fruit trees in it. He has a…
  44. A man owns a field area 1000 m2. He wants to plant fruit trees in it. He has a…
  45. A publisher sells a hardcover edition of a book for ₹72 and a paperback…
  46. A publisher sells a hardcover edition of a book for ₹72 and a paperback…
  47. A gardener has a supply of fertilizers of the type 1 which consist of 10%…
  48. A gardener has a supply of fertilizers of the type 1 which consist of 10%…
  49. Two godowns, A and B, have a grain storage capacity of 100 quintals and 50…
  50. Two godowns, A and B, have a grain storage capacity of 100 quintals and 50…
  51. A brick manufacture has two depots, P and Q, with stocks of 30000 and 20000…
  52. A brick manufacture has two depots, P and Q, with stocks of 30000 and 20000…
  53. A medicine company has factories at two places, X and Y. From these places,…
  54. A medicine company has factories at two places, X and Y. From these places,…
  55. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L…
  56. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L…
  57. A firm is engaged in breeding pigs. The pigs are fed on various products grown…
  58. A firm is engaged in breeding pigs. The pigs are fed on various products grown…
  59. A dietician wishes to mix two types of food, X and Y, in such a way that the…
  60. A dietician wishes to mix two types of food, X and Y, in such a way that the…
  61. A diet for a sick person must contain at least 4000 units of vitamins, 50…
  62. A diet for a sick person must contain at least 4000 units of vitamins, 50…
  63. A housewife wishes to mix together two kinds of food, X and Y, in such a way…
  64. A housewife wishes to mix together two kinds of food, X and Y, in such a way…
  65. A firm manufactures two types of product, A and B, and sells them at a profit…
  66. A firm manufactures two types of product, A and B, and sells them at a profit…
  67. A small firm manufactures items A and B. The total number of items that it can…
  68. A small firm manufactures items A and B. The total number of items that it can…
  69. A manufacture produces two types of steel trunks. He has two machines, A and…
  70. A manufacture produces two types of steel trunks. He has two machines, A and…
  71. A company manufactures two types of toys, A and B. Type A requires 5 minutes…
  72. A company manufactures two types of toys, A and B. Type A requires 5 minutes…
  73. Kellogg is a new cereal formed of a mixture of bran and rice, that contains at…
  74. Kellogg is a new cereal formed of a mixture of bran and rice, that contains at…
  75. A dealer wishes to purchase a number of fans and sewing machines. He has only…
  76. A dealer wishes to purchase a number of fans and sewing machines. He has only…
  77. Anil wants to invest at the most ₹12000 in bonds A and B. According to rules,…
  78. Anil wants to invest at the most ₹12000 in bonds A and B. According to rules,…
  79. Maximize = 60x + 15y, subject to the constraintsx + y ≤ 50, 3x + ≤ y90, x, y ≥…
  80. Maximize = 60x + 15y, subject to the constraintsx + y ≤ 50, 3x + ≤ y90, x, y ≥…
  81. A company manufacture two types of toys A and B. type A requires 5 minutes…
  82. A company manufacture two types of toys A and B. type A requires 5 minutes…
  83. One kind of cake requires 200 g of flour and 25 g of fat, another kind of cake…
  84. One kind of cake requires 200 g of flour and 25 g of fat, another kind of cake…
  85. A manufacturing company makes two types of teaching aids A and B of…
  86. A manufacturing company makes two types of teaching aids A and B of…
Exercise 33a
Question 1.

Graph the solution sets of the following inequations:

x + y ≥ 4
Answer:

Given x + y ≥ 4


⇒ y ≥ 4 - x


Consider the equation y = 4 - x.


Finding points on the coordinate axes:


If x = 0, the y value is 4 i.e, y = 4


⇒ the point on the Y axis is A(0,4)


If y = 0, 0 = 4 - x


⇒x = 4


The point on the X axis is B(4,0)


Plotting the points on the graph: fig. 1a


Now consider the inequality y ≥ 4 - x


Here we need the y value greater than or equal to 4 - x


⇒ the required region is above point A.


Therefore the graph of the inequation x + y ≥ 4 is fig. 1b



Fig 1a



Fig 1b
Question 2.

Graph the solution sets of the following inequations:

x - y≤3
Answer:

Given x - y≤3


⇒ - y≤ 3 - x


Multiplying by minus on both the sides, we’ll get


y ≥ - 3 + x


y ≥ x - 3


Consider the equation y = x - 3.


Finding points on the coordinate axes:


If x = 0, the y value is - 3 i.e, y = - 3


⇒ the point on the Y axis is A(0, - 3)


If y = 0, 0 = x - 3


⇒x = 3


The point on the X axis is B(3,0)


Plotting the points on the graph: fig. 2a


Now consider the inequality y ≥ x - 3


Here we need the y value greater than or equal to x - 3


⇒ the required region is above point A.


Therefore the graph of the inequation x + y ≥ 4 is fig. 2b



Fig 2a



Fig 2b
Question 3.

Graph the solution sets of the following inequations:

x + 2y>1
Answer:

Given x + 2y>1


⇒ 2y>1 - x


⇒


Consider the equation


Finding points on the coordinate axes:


If x = 0, the y value is i.e., y = 4


⇒ the point on the Y axis is A(0,)


If y = 0, x = 1


The point on the X axis is B(1,0)


Plotting the points on the graph: fig. 3a


Now consider the inequality


Here we need the y value greater than


⇒ the required region is above point A.


Also , the line AB is represented in dotted line. This is s done because


Therefore the graph of the inequation is fig. 3b



Fig 3a



Fig 3b
Question 4.

Graph the solution sets of the following inequations:

2x - 3y<4
Answer:

Given 2x - 3y<4


⇒ 2x - 4<3y


⇒3y>2x - 4


⇒


Consider the equation


Finding points on the coordinate axes:


If x = 0, the y value is i.e., y =


⇒ the point on the Y axis is A(0,)


If y = 0, x = 2


The point on the X axis is B(2,0)


Plotting the points on the graph: fig. 4a


Now consider the inequality


Here we need the y value greater than


⇒ the required region is above point A.


Also , the line AB is represented in dotted line. This is s done because


Therefore the graph of the inequation is fig. 4b



Fig 4a



Fig 4b
Question 5.

Graph the solution sets of the following inequations:

x ≥ y - 2
Answer:

Given x ≥ y - 2


⇒ y≤x + 2


Consider the equation y = x + 2


Finding points on the coordinate axes:


If x = 0, the y value is 2 i.e, y = 2


⇒ the point on the Y axis is A(0,2)


If y = 0, 0 = x + 2


⇒x = - 2


The point on the X axis is B( - 2,0)


Plotting the points on the graph: fig. 5a


Now consider the inequality y≤x + 2


Here we need the y value less than or equal to x + 2


⇒ the required region is below point A.


Therefore the graph of the inequation x ≥ y - 2 is fig. 5b



fig 5a



Fig 5b
Question 6.

Graph the solution sets of the following inequations:

y - 2≤3x
Answer:

Given y - 2≤3x


⇒ y≤3x + 2


Consider the equation y = 3x + 2


Finding points on the coordinate axes:


If x = 0, the y value is 2 i.e, y = 2


⇒ the point on Y axis is A(0,2)


If y = 0, 0 = 3x + 2


⇒x =


The point on the X axis is B( - ,0)


Plotting the points on the graph: fig. 6a


Now consider the inequality y≤3x + 2


Here we need the y value less than or equal to 3x + 2


⇒ the required region is below point A.


Therefore the graph of the inequation y≤3x + 2 is fig. 5b



Fig 6a



Fig 6b
Question 7.

Solve each of the following systems of simultaneous inequations:

2x + y>1 and 2x - y ≥ - 3
Answer:

Consider the inequation 2x + y>1:


⇒ y>1 - 2x


Consider the equation y = 1 - 2x


Finding points on the coordinate axes:


If x = 0, the y value is 1 i.e, y = 1


⇒ the point on Y axis is A(0,1)


If y = 0, 0 = x + 2


⇒x =


The point on the X axis is B(,0)


Plotting the points on the graph: fig. 7a


Now consider the inequality y>1 - 2x


Here we need the y value greater than x + 2


⇒ the required region is below point A.


Therefore the graph of the inequation y>1 - 2x is fig. 7b



Fig 7a



Fig 7b


Consider the inequation 2x - y ≥ 3


⇒ y≤2x - 3


Consider the equation y = 2x - 3


Finding points on the coordinate axes:


If x = 0, the y value is - 3 i.e, y = - 3


⇒ the point on the Y axis is C(0, - 3)


If y = 0, 0 = 2x + 3


⇒x =


The point on the X axis is D(,0)


Plotting the points on the graph: fig. 7c


Now consider the inequality y≤2x - 3


Here we need the y value less than or equal to 2x - 3


⇒ the required region is below point C.


Therefore the graph of the inequation y≤2x - 3 is fig. 7d



Fig 7c



Fig 7d


Combining the graphs 7c and 7d, we’ll get,



The solution of the system of simultaneous inequations is the intersection region of the solutions of the two given inequations.
Question 8.

Solve each of the following systems of simultaneous inequations:

x - 2y ≥ 0, 2x - y≤ - 2
Answer:

Consider the inequation x - 2y ≥ 0:


⇒x ≥ 2y


⇒y≤


consider the equation y = . This equation’s graph is a straight line passing through origin.



Now consider the inequality y≤


Here we need the y value less than or equal to


⇒ the required region is below the origin.


Therefore the graph of the inequation y≤is fig.8a



Fig 8a


Consider the inequation 2x - y≤ - 2 :


⇒y ≥ 2x + 2


Consider the equation y = 2x + 2


Finding points on the coordinate axes:


If x = 0, the y value is 2 i.e, y = 2


⇒ the point on the Y axis is A(0,2)


If y = 0, 0 = 2x + 2


⇒x = - 1


The point on the X axis is B( - 1,0)


Plotting the points on the graph: fig. 8b.


Now consider the inequality y ≥ 2x + 2


Here we need the y value greater than or equal to 2x + 2


⇒ the required region is above point A.


Therefore the graph of the inequation y ≥ 2x + 2 is fig. 8c



Fig 8b



Fig 8c


Combining the graphs of 8a and 8c, we’ll get



The solution of the system of simultaneous inequations is the intersection region of the solutions of the two given inequations.
Question 9.

Solve each of the following systems of simultaneous inequations:

3x + 4y ≥ 12, x ≥ 0, y ≥ 1 and 4x + 7y≤28
Answer:

Consider the inequation 3x + 4y ≥ 12 :


⇒4y ≥ 12 - 3x


⇒y ≥ 3 -


Consider the equation y = 3 -


Finding points on the coordinate axes:


If x = 0, the y value is 3 i.e, y = 3


⇒ the point on the Y axis is A(0,3)


If y = 0, 0 = 3 -


⇒x = 4


The point on the X axis is B(4,0)


Now consider the inequality y ≥ 3 -


Here we need the y value greater than or equal to y ≥ 3 -


⇒ the required region is above point A.


Therefore the graph of the inequation y ≥ 3 - is fig. 9a



Fig 9a


Consider the inequation 4x + 7y≤28


⇒ 7y≤28 - 4x


⇒y≤4 -


Consider the equation y = 4 -


Finding points on the coordinate axes:


If x = 0, the y value is 4 i.e, y = 4


⇒ the point on the Y axis is C(0,4)


If y = 0, 0 = 4 -


⇒x = 7


The point on the X axis is D(7,0)


Now consider the inequality y≤4 -


Here we need the y value less than or equal to 4 -


⇒ the required region is below point C.


Therefore the graph of the inequation y≤4 - is fig. 9b



Fig 9b


x ≥ 0 is the region right side of Y - axis.


y ≥ 1 is the region above the line y = 1


Combining all the above results in a single graph , we’ll get



The solution of the system of simultaneous inequations is the intersection region of the solutions of the two given inequations.
Question 10.

Show that the solution set of the following linear constraints is empty:

x - 2y ≥ 0, 2x - y≤ - 2, x ≥ 0 and y ≥ 0
Answer:

Consider the inequation x - 2y ≥ 0:


⇒x ≥ 2y


⇒y≤


consider the equation y = . This equation’s graph is a straight line passing through origin.


Now consider the inequality y≤


Here we need the y value less than or equal to


⇒ the required region is below origin.


Therefore the graph of the inequation y≤ is fig.10a



Fig 10a


Consider the inequation 2x - y≤ - 2 :


⇒y ≥ 2x + 2


Consider the equation y = 2x + 2


Finding points on the coordinate axes:


If x = 0, the y value is 2 i.e, y = 2


⇒ the point on Y axis is A(0,2)


If y = 0, 0 = 2x + 2


⇒x = - 1


The point on X axis is B( - 1,0)


Now consider the inequality y ≥ 2x + 2


Here we need the y value greater than or equal to 2x + 2


⇒ the required region is above point A.


Therefore the graph of the inequation y ≥ 2x + 2 is fig. 10b



Fig 10b


y ≥ 0 is the region above X - axis


x ≥ 0 is the region right side of Y - axis


Combining the above results, we’ll get



As they is no common area of intersection , there is no solution for the given set of simultaneous inequations.
Question 11.

Find the linear constraints for which the shaded area in the figure given is the solution set.
Answer:


Consider A:


Given line x - y = 1


⇒y = x - 1


As the region given in the figure is above the y - intercept’s coordinates (0, - 1),


⇒y ≥ x - 1


⇒x - y≤1


Consider B:


Given line 2x + y = 2


⇒y = 2 - 2x


As the region given in the figure is above the y - intercept’s coordinates (0,2),


⇒y ≥ 2 - 2x


⇒2x + y ≥ 2


Consider C:


Given line x + 2y = 8


⇒2y = 8 - x


⇒y = 4 -


As the region given in the figure is below the y - intercept’s coordinates (0,4),


⇒y≤4 -


⇒2y≤8 - x


⇒x + 2y≤8


Consider D:


It is the region right side of the Y - axis.


It is x ≥ 0.


All the results derived:


x - y≤1


2x + y ≥ 2


x + 2y≤8


x ≥ 0
Exercise 33b
Question 1.

Find the maximum value of Z = 7X + 7Y, subject to the constraints.

x ≥ 0, y ≥ 0, x + y ≥ 2 and 2x + 3y ≤ 6.
Answer:

The feasible region determined by the constraints x , ,


x + y , 2x + 3y ≤ 6 is given by



The corner points of the feasible region is A(0,2),B(2,0),C(3,0).


The values of Z at the following points is



The maximum value of Z is 21 at point C(3,0) .
Question 2.

Find the maximum value of Z = 7X + 7Y, subject to the constraints.

x ≥ 0, y ≥ 0, x + y ≥ 2 and 2x + 3y ≤ 6.
Answer:

The feasible region determined by the constraints x , ,


x + y , 2x + 3y ≤ 6 is given by



The corner points of the feasible region is A(0,2),B(2,0),C(3,0).


The values of Z at the following points is



The maximum value of Z is 21 at point C(3,0) .
Question 3.

Maximize Z = 4x + 9y, subject to the constraints

x0, y0, x + 5y ≤ 200, 2x + 3y ≤ 134.
Answer:

The feasible region determined by the constraints x0, y0, x + 5y ≤ 200, 2x + 3y ≤ 134 is given by



The corner points of feasible region are A(10,38) ,B(0,40) ,C(0,0), D(67,0) . The values of Z at the following points is



The maximum value of Z is 382 at point A(10,38) .
Question 4.

Maximize Z = 4x + 9y, subject to the constraints

x0, y0, x + 5y ≤ 200, 2x + 3y ≤ 134.
Answer:

The feasible region determined by the constraints x0, y0, x + 5y ≤ 200, 2x + 3y ≤ 134 is given by



The corner points of feasible region are A(10,38) ,B(0,40) ,C(0,0), D(67,0) . The values of Z at the following points is



The maximum value of Z is 382 at point A(10,38) .
Question 5.

Find the minimum value of Z = 3x + 5y, subject to the constraints

- 2x + y ≤ 4, x + y ≥ 3, x - 2y ≤ 2, x ≥ 0 and y ≥ 0
Answer:

The feasible region determined by the - 2x + y ≤ 4, x + y ≥ 3, x - 2y ≤ 2, x ≥ 0 and y ≥ 0 is given by



Here the feasible region is unbounded. The vertices of the region are A(0,4) ,B(0,3) ,C. The values of Z at the following points is



The minimum value of Z is at point C.
Question 6.

Find the minimum value of Z = 3x + 5y, subject to the constraints

- 2x + y ≤ 4, x + y ≥ 3, x - 2y ≤ 2, x ≥ 0 and y ≥ 0
Answer:

The feasible region determined by the - 2x + y ≤ 4, x + y ≥ 3, x - 2y ≤ 2, x ≥ 0 and y ≥ 0 is given by



Here the feasible region is unbounded. The vertices of the region are A(0,4) ,B(0,3) ,C. The values of Z at the following points is



The minimum value of Z is at point C.
Question 7.

Minimize Z = 2x + 3y, subject to the constraints

x ≥ 0, y ≥ 0, x + 2y ≥ 1 and x + 2y ≤ 10.
Answer:

The feasible region determined by the x ≥ 0, y ≥ 0, x + 2y ≥ 1 and x + 2y ≤ 10 is given by



The corner points of the feasible region is A(0,), B(0,5), C(10,0), D(1,0).The value of Z at corner points are



The minimum value of Z is at point A(0,).
Question 8.

Minimize Z = 2x + 3y, subject to the constraints

x ≥ 0, y ≥ 0, x + 2y ≥ 1 and x + 2y ≤ 10.
Answer:

The feasible region determined by the x ≥ 0, y ≥ 0, x + 2y ≥ 1 and x + 2y ≤ 10 is given by



The corner points of the feasible region is A(0,), B(0,5), C(10,0), D(1,0).The value of Z at corner points are



The minimum value of Z is at point A(0,).
Question 9.

Maximize Z = 3x + 5y, subject to the constraints

X + 2y ≤ 2000, x + y ≤ 1500, y ≤ 00, y ≤ 600, x ≥ 0 and y ≥ 0.
Answer:

The feasible region determined by the X + 2y ≤ 2000, x + y ≤ 1500, y ≤ 600, x ≥ 0 and y ≥ 0 is given by



The corner points of the feasible region are A(0,0), B(0,600), C(800,600), D(1000,500), E(1500,0).The value of Z at the corner points are



The maximum value of Z is 5500 at point D(1000,500).
Question 10.

Maximize Z = 3x + 5y, subject to the constraints

X + 2y ≤ 2000, x + y ≤ 1500, y ≤ 00, y ≤ 600, x ≥ 0 and y ≥ 0.
Answer:

The feasible region determined by the X + 2y ≤ 2000, x + y ≤ 1500, y ≤ 600, x ≥ 0 and y ≥ 0 is given by



The corner points of the feasible region are A(0,0), B(0,600), C(800,600), D(1000,500), E(1500,0).The value of Z at the corner points are



The maximum value of Z is 5500 at point D(1000,500).
Question 11.

Find the maximum and minimum values of Z = 2x + y, subject to the constraints

X + 3y ≥ 6, x - 3y ≤ 3, 3x + 4y ≤ 24,

- 3x + 2y ≤ 6, 5x + y ≥ 5, x ≥ 0 and y ≥ 0.
Answer:

The feasible region determined by X + 3y ≥ 6, x - 3y ≤ 3, 3x + 4y ≤ 24,


- 3x + 2y ≤ 6, 5x + y ≥ 5, x ≥ 0 and y ≥ 0 is given by



The corner points of the feasible region are A(4/3,5) , B(4/13,45/13), C(9/14,25/14) , D(9/2,1/2) , E(84/13,15/13).The value of Z at corner points are



The maximum and minimum value of Z is 183/13 and 43/14 at points E(84/13,15/13) and C(9/14,25/14).
Question 12.

Find the maximum and minimum values of Z = 2x + y, subject to the constraints

X + 3y ≥ 6, x - 3y ≤ 3, 3x + 4y ≤ 24,

- 3x + 2y ≤ 6, 5x + y ≥ 5, x ≥ 0 and y ≥ 0.
Answer:

The feasible region determined by X + 3y ≥ 6, x - 3y ≤ 3, 3x + 4y ≤ 24,


- 3x + 2y ≤ 6, 5x + y ≥ 5, x ≥ 0 and y ≥ 0 is given by



The corner points of the feasible region are A(4/3,5) , B(4/13,45/13), C(9/14,25/14) , D(9/2,1/2) , E(84/13,15/13).The value of Z at corner points are



The maximum and minimum value of Z is 183/13 and 43/14 at points E(84/13,15/13) and C(9/14,25/14).
Question 13.

Mr.Dass wants to invest ₹12000 in public provident fund (PPF) and in national bonds. He has to invest at least ₹1000 in PPF and at least ₹2000 in bonds. If the rate of interest on PPF is 12% per annum and that on bonds is 15% per annum, how should he invest the money to earn maximum annual income? Also find the maximum annual income.
Answer:

Let the invested money in PPF be x and in national bonds be y.


∴According to the question,


X + y


x , y


Maximize Z = 0.12x + 0.15y


The feasible region determined by X + y x ,


y is given by



The corner points of the feasible region are A(1000,11000) , B(1000,2000) and C(10000,2000) . The value of Z at the corner point are



The maximum value of Z is 1770 at point A(1000,11000).


So, he must invest Rs.1000 in PPF and Rs.11000 in national bonds.


The maximum annual income is Rs.1770 .
Question 14.

Mr.Dass wants to invest ₹12000 in public provident fund (PPF) and in national bonds. He has to invest at least ₹1000 in PPF and at least ₹2000 in bonds. If the rate of interest on PPF is 12% per annum and that on bonds is 15% per annum, how should he invest the money to earn maximum annual income? Also find the maximum annual income.
Answer:

Let the invested money in PPF be x and in national bonds be y.


∴According to the question,


X + y


x , y


Maximize Z = 0.12x + 0.15y


The feasible region determined by X + y x ,


y is given by



The corner points of the feasible region are A(1000,11000) , B(1000,2000) and C(10000,2000) . The value of Z at the corner point are



The maximum value of Z is 1770 at point A(1000,11000).


So, he must invest Rs.1000 in PPF and Rs.11000 in national bonds.


The maximum annual income is Rs.1770 .
Question 15.

A small firm manufactures necklace and bracelets. The total number of necklace and bracelet that it can handle per day is at most 24. It takes 1 hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ₹100 and that on a bracelet is ₹300, how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced.
Answer:

Let the firm manufacture x number of necklaces and y number of bracelets a day.


∴According to the question,


X + y , 0.5x + y x


Maximize Z = 100x + 300y


The feasible region determined by X + y , 0.5x + y xis given by



The corner points of the feasible region are A(1,1) , B(1,15.5) , C(16,8) , D(23,1).The number of bracelets should be whole number. Therefore, considering point (2,15). The value of Z at corner point is



The maximum value of Z is 4700 at point B(2,15).


∴ The firm should make 2 necklaces and 15 bracelets.
Question 16.

A small firm manufactures necklace and bracelets. The total number of necklace and bracelet that it can handle per day is at most 24. It takes 1 hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ₹100 and that on a bracelet is ₹300, how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced.
Answer:

Let the firm manufacture x number of necklaces and y number of bracelets a day.


∴According to the question,


X + y , 0.5x + y x


Maximize Z = 100x + 300y


The feasible region determined by X + y , 0.5x + y xis given by



The corner points of the feasible region are A(1,1) , B(1,15.5) , C(16,8) , D(23,1).The number of bracelets should be whole number. Therefore, considering point (2,15). The value of Z at corner point is



The maximum value of Z is 4700 at point B(2,15).


∴ The firm should make 2 necklaces and 15 bracelets.
Question 17.

A man has ₹1500 to purchase rice and wheat. A bag of rice and a bag of wheat cost ₹180 and 120 respectively. He has storage capacity of 10 bags only. He earns a profit of ₹11 and ₹8 per bag of rice and wheat respectively. How many bags of each must he buy to make maximum profit?
Answer:

Let the number of wheat and rice bags be x and y.


∴According to the question,


120x + 180y, x + y,x


Maximize Z = 8x + 11y


The feasible region determined by 120x + 180y, x + y,xis given by



The corner points of feasible region are A(0,8), B(0,0), C(10,0), D(5,5) .


The value of Z at corner point is



The maximum value of Z is 95 at point (5,5).


Hence, the man should 5 bags each of wheat and rice to earn maximum profit.
Question 18.

A man has ₹1500 to purchase rice and wheat. A bag of rice and a bag of wheat cost ₹180 and 120 respectively. He has storage capacity of 10 bags only. He earns a profit of ₹11 and ₹8 per bag of rice and wheat respectively. How many bags of each must he buy to make maximum profit?
Answer:

Let the number of wheat and rice bags be x and y.


∴According to the question,


120x + 180y, x + y,x


Maximize Z = 8x + 11y


The feasible region determined by 120x + 180y, x + y,xis given by



The corner points of feasible region are A(0,8), B(0,0), C(10,0), D(5,5) .


The value of Z at corner point is



The maximum value of Z is 95 at point (5,5).


Hence, the man should 5 bags each of wheat and rice to earn maximum profit.
Question 19.

A manufacture produces nuts and bolts for industrial machinery. It takes 1 hour of work on machine A and 3 hours on machine B to produces a packet of nuts while it takes 3 hours on machine A and 1 hours on machine B to produce a packet of bolts. He earns a profit ₹17.50 per packet on nuts and ₹7 per packet on bolts. How many packets of each should be produced each day so as to maximize his profit if he operates each machine for at the most 12 hours a day? Also find the maximum profit.
Answer:

Let the number of packets of nuts and bolts be x and y respectively.


∴According to the question,


X + 3y


Maximize Z = 17.50x + 7y


The feasible region determined by X + 3y is given by



The corner points of the feasible region are A(0,0), B(0,4), C(3,3)), D(4,0).The value of Z at the corner point is



The maximum value of Z is 73.50 at (3,3).


The manufacturer should make 3 packets each of nuts and bolts to make maximum profit of Rs.73.50.
Question 20.

A manufacture produces nuts and bolts for industrial machinery. It takes 1 hour of work on machine A and 3 hours on machine B to produces a packet of nuts while it takes 3 hours on machine A and 1 hours on machine B to produce a packet of bolts. He earns a profit ₹17.50 per packet on nuts and ₹7 per packet on bolts. How many packets of each should be produced each day so as to maximize his profit if he operates each machine for at the most 12 hours a day? Also find the maximum profit.
Answer:

Let the number of packets of nuts and bolts be x and y respectively.


∴According to the question,


X + 3y


Maximize Z = 17.50x + 7y


The feasible region determined by X + 3y is given by



The corner points of the feasible region are A(0,0), B(0,4), C(3,3)), D(4,0).The value of Z at the corner point is



The maximum value of Z is 73.50 at (3,3).


The manufacturer should make 3 packets each of nuts and bolts to make maximum profit of Rs.73.50.
Question 21.

Two tailors, A and B, earn ₹300 and ₹400 per day respectively. A can stitch 6 shirts and 4 pair of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. How many days should each of them work if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labor cost?
Answer:

Let the total number of days tailor A work be x and tailor B be y.


∴According to the question,


6x + 10 y , 4x + 4y


Minimize Z = 300x + 400y


The feasible region determined by 6x + 10 y , 4x + 4y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,8),B(5,3),C(10,0) .The value of Z at corner point is



The minimum value of Z is 2700 at point (5,3).


∴Tailor A must work for 5 days and tailor B must work for 3 days for minimum expenses.
Question 22.

Two tailors, A and B, earn ₹300 and ₹400 per day respectively. A can stitch 6 shirts and 4 pair of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. How many days should each of them work if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labor cost?
Answer:

Let the total number of days tailor A work be x and tailor B be y.


∴According to the question,


6x + 10 y , 4x + 4y


Minimize Z = 300x + 400y


The feasible region determined by 6x + 10 y , 4x + 4y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,8),B(5,3),C(10,0) .The value of Z at corner point is



The minimum value of Z is 2700 at point (5,3).


∴Tailor A must work for 5 days and tailor B must work for 3 days for minimum expenses.
Question 23.

A dealer wishes to purchase a number of fans and sewing machines. He has only ₹5760 to invest and space for at most 20 items. A fan costs him ₹360 and a sewing machine, ₹240. He expects to gain ₹22 on a fan and ₹18 on a sewing machine. Assuming that he can sell all the items he can buy, how should he invest the money in order to maximize the profit?
Answer:

Let the number of fans bought be x and sewing machines bought be y.


∴According to the question,


360x + 240y,x + y


Maximize Z = 22x + 18y


The feasible region determined by 360x + 240y,x + y is given by



The corner points of the feasible region are A(0,0) , B(0,20),C(8,12) , D(16,0).The value of Z at corner points is



The maximum value of Z is 392 at point (8,12).


The dealer must buy 8 fans and 12 sewing machines to make the maximum profit.
Question 24.

A dealer wishes to purchase a number of fans and sewing machines. He has only ₹5760 to invest and space for at most 20 items. A fan costs him ₹360 and a sewing machine, ₹240. He expects to gain ₹22 on a fan and ₹18 on a sewing machine. Assuming that he can sell all the items he can buy, how should he invest the money in order to maximize the profit?
Answer:

Let the number of fans bought be x and sewing machines bought be y.


∴According to the question,


360x + 240y,x + y


Maximize Z = 22x + 18y


The feasible region determined by 360x + 240y,x + y is given by



The corner points of the feasible region are A(0,0) , B(0,20),C(8,12) , D(16,0).The value of Z at corner points is



The maximum value of Z is 392 at point (8,12).


The dealer must buy 8 fans and 12 sewing machines to make the maximum profit.
Question 25.

A firm manufactures two types of products, A and B, and sells them at a profit of ₹2 on type A and ₹2 on type B. Each product is processed on two machines, M1 and M2. Type A requires one minute of processing time on M1 and two minutes on M2. Type B requires one minute on M1 and one minute on M2 is available for not more than 6 hours 40 minutes while M2 is available for at most 10 hours a day.

Find how many products of each type the firm should produce each day in order to get maximum profit.
Answer:

Let the firm manufacture x number of Aand y number of B products.


∴According to the question,


X + y


Maximize Z = 2x + 2y


The feasible region determined by X + y is given by



The corner points of feasible region are A(0,0) , B(0,400) , C(200,200) , D(300,0).The value of Z at corner point is



The maximum value of Z is 800 and occurs at two points. Hence the line BC is a feasible solution.


The firm should produce 200 number of Aproducts and 200 number of B products.
Question 26.

A firm manufactures two types of products, A and B, and sells them at a profit of ₹2 on type A and ₹2 on type B. Each product is processed on two machines, M1 and M2. Type A requires one minute of processing time on M1 and two minutes on M2. Type B requires one minute on M1 and one minute on M2 is available for not more than 6 hours 40 minutes while M2 is available for at most 10 hours a day.

Find how many products of each type the firm should produce each day in order to get maximum profit.
Answer:

Let the firm manufacture x number of Aand y number of B products.


∴According to the question,


X + y


Maximize Z = 2x + 2y


The feasible region determined by X + y is given by



The corner points of feasible region are A(0,0) , B(0,400) , C(200,200) , D(300,0).The value of Z at corner point is



The maximum value of Z is 800 and occurs at two points. Hence the line BC is a feasible solution.


The firm should produce 200 number of Aproducts and 200 number of B products.
Question 27.

A manufactures produces two types of soap bars using two machines, A and B. A is operated for 2 minutes and B for 3 minutes to manufacture the first type, while it takes 3 minutes on machine A and 5 minutes on machine B to manufacture the second type. Each machine can be operated at the most for 8 hours per day. The two types of soap bars are sold at a profit of ₹0.25 and ₹0.50 each. Assuming that the manufacture can sell all the soap bars he can manufacture, how many bars of soap of each type should be manufactured per day so as to maximize his profit?
Answer:

Let x and y be number of soaps be manufactured of 1st and 2nd type.


∴According to the question,


2x + 3y 480 , 3x + 5y 480, x , y


Maximize Z = 0.25x + 0.50y


The feasible region determined by 2x + 3y 480 , 3x + 5y 480, x , y is given by



The corner points of feasible region are A(0,96) , B(0,0) , C(160,0).


The value of Z at corner points are



The maximum value of Z is 48 at point (0,96).


Hence, the manufacturer should make 96 soaps of the 2nd type to make maximum profit.
Question 28.

A manufactures produces two types of soap bars using two machines, A and B. A is operated for 2 minutes and B for 3 minutes to manufacture the first type, while it takes 3 minutes on machine A and 5 minutes on machine B to manufacture the second type. Each machine can be operated at the most for 8 hours per day. The two types of soap bars are sold at a profit of ₹0.25 and ₹0.50 each. Assuming that the manufacture can sell all the soap bars he can manufacture, how many bars of soap of each type should be manufactured per day so as to maximize his profit?
Answer:

Let x and y be number of soaps be manufactured of 1st and 2nd type.


∴According to the question,


2x + 3y 480 , 3x + 5y 480, x , y


Maximize Z = 0.25x + 0.50y


The feasible region determined by 2x + 3y 480 , 3x + 5y 480, x , y is given by



The corner points of feasible region are A(0,96) , B(0,0) , C(160,0).


The value of Z at corner points are



The maximum value of Z is 48 at point (0,96).


Hence, the manufacturer should make 96 soaps of the 2nd type to make maximum profit.
Question 29.

A manufacture of a line of patent medicines is preparing a production plan on medicines A and B. There are sufficient ingredients available to make 20000 bottles of A and 40000 bottles of B but there are only 45000 bottles into which either of the medicines can be put. Furthermore, it takes 3 hours to prepare enough material to fill 1000 bottles of A and it takes 1 hour to prepare enough material to fill 1000 bottles of B, and there are 66 hours available for this operation. The profit is ₹8 per bottle for A and ₹7 per bottle for B.

How should the manufacture schedule the production in order to maximize his profit? Also, find the maximum profit.
Answer:

Let x and y be number of bottles of medicines A and B be prepared.


∴According to the question,


x + y 45000 , 3x + y 66000, x , y,


Maximize Z = 8x + 7y


The feasible region determined by x + y 45000 , 3x + y 66000, x , y, is given by



The corner points of feasible region are A(0,0) , B(0,40000) , C(5000,40000),D(10500,34500),E(20000,6000),F(20000,0).


The value of Z at corner points are



The maximum value of Z is 325500 at point (10500,34500).


Hence, the manufacturer should produce 10500 bottles of medicine A and 34500 bottles of medicine B.
Question 30.

A manufacture of a line of patent medicines is preparing a production plan on medicines A and B. There are sufficient ingredients available to make 20000 bottles of A and 40000 bottles of B but there are only 45000 bottles into which either of the medicines can be put. Furthermore, it takes 3 hours to prepare enough material to fill 1000 bottles of A and it takes 1 hour to prepare enough material to fill 1000 bottles of B, and there are 66 hours available for this operation. The profit is ₹8 per bottle for A and ₹7 per bottle for B.

How should the manufacture schedule the production in order to maximize his profit? Also, find the maximum profit.
Answer:

Let x and y be number of bottles of medicines A and B be prepared.


∴According to the question,


x + y 45000 , 3x + y 66000, x , y,


Maximize Z = 8x + 7y


The feasible region determined by x + y 45000 , 3x + y 66000, x , y, is given by



The corner points of feasible region are A(0,0) , B(0,40000) , C(5000,40000),D(10500,34500),E(20000,6000),F(20000,0).


The value of Z at corner points are



The maximum value of Z is 325500 at point (10500,34500).


Hence, the manufacturer should produce 10500 bottles of medicine A and 34500 bottles of medicine B.
Question 31.

A toy company manufactures two types of dolls, A and B. Each doll of type B take twice as long to produce as one of type A, and the company would have time to make a maximum of 2000 per day, if it produces only type A. the supply of plastic is sufficient to produce 1500 dolls per day (both A and B combined ). Type B requires a fancy dress of which there are only 600 per day available. If the company makes profit of ₹3 and ₹5 per dolls respectively on dolls A and B, how many of each should be produced per day in order to maximize the profit? Also, find the maximum profit.
Answer:

Let x and y be number of doll A manufactured and doll B manufactured.


∴According to the question,


x + y 1500, x + 2y y


Maximize Z = 3x + 5y


The feasible region determined by x + y 1500, x + 2y yis given by



The corner points of feasible region are A(0,0) , B(0,600) , C(800,600),D(1000,500),E(1500,0).


The value of Z at corner points are



The maximum value of Z is 5500 at point (1000,500).


Hence, the manufacturer should produce 1000 types of doll A and 500 types of doll B to make maximum profit of Rs.5500.
Question 32.

A toy company manufactures two types of dolls, A and B. Each doll of type B take twice as long to produce as one of type A, and the company would have time to make a maximum of 2000 per day, if it produces only type A. the supply of plastic is sufficient to produce 1500 dolls per day (both A and B combined ). Type B requires a fancy dress of which there are only 600 per day available. If the company makes profit of ₹3 and ₹5 per dolls respectively on dolls A and B, how many of each should be produced per day in order to maximize the profit? Also, find the maximum profit.
Answer:

Let x and y be number of doll A manufactured and doll B manufactured.


∴According to the question,


x + y 1500, x + 2y y


Maximize Z = 3x + 5y


The feasible region determined by x + y 1500, x + 2y yis given by



The corner points of feasible region are A(0,0) , B(0,600) , C(800,600),D(1000,500),E(1500,0).


The value of Z at corner points are



The maximum value of Z is 5500 at point (1000,500).


Hence, the manufacturer should produce 1000 types of doll A and 500 types of doll B to make maximum profit of Rs.5500.
Question 33.

A small manufacture has employed 5 skilled men and 10 semiskilled men and makes an article in two qualities, a deluxe model and an ordinary model. The making of a deluxe model requires 2hours work by a skilled man and 2hours work by a semiskilled man. The ordinary model requires 1 hour by a skilled man and 3 hours by a semiskilled man. By union rules, no man can work more than 8 hours per day. The manufacture gains ₹15 on the deluxe model and ₹10 on the ordinary model. How many of each type should be made in order to maximize his total daily profit? Also, find the maximum daily profit.
Answer:

Let x and y be number of deluxe article manufactured and ordinary article manufactured.


∴According to the question,


2x + y , 2x + 3y


Maximize Z = 15x + 10y


The feasible region determined by 2x + y , 2x + 3y is given by



The corner points of feasible region are A(0,0) , B(0,80/3) , C(10,20),D(20,0).


The value of Z at corner points are



The maximum value of Z is 350 at point (10,20).


Hence, the manufacturer should produce 10 types of deluxe article and 20 types of ordinary article to make maximum profit of Rs.350.
Question 34.

A small manufacture has employed 5 skilled men and 10 semiskilled men and makes an article in two qualities, a deluxe model and an ordinary model. The making of a deluxe model requires 2hours work by a skilled man and 2hours work by a semiskilled man. The ordinary model requires 1 hour by a skilled man and 3 hours by a semiskilled man. By union rules, no man can work more than 8 hours per day. The manufacture gains ₹15 on the deluxe model and ₹10 on the ordinary model. How many of each type should be made in order to maximize his total daily profit? Also, find the maximum daily profit.
Answer:

Let x and y be number of deluxe article manufactured and ordinary article manufactured.


∴According to the question,


2x + y , 2x + 3y


Maximize Z = 15x + 10y


The feasible region determined by 2x + y , 2x + 3y is given by



The corner points of feasible region are A(0,0) , B(0,80/3) , C(10,20),D(20,0).


The value of Z at corner points are



The maximum value of Z is 350 at point (10,20).


Hence, the manufacturer should produce 10 types of deluxe article and 20 types of ordinary article to make maximum profit of Rs.350.
Question 35.

A company producing soft drinks has a contrast which requires a minimum of 80 units of chemical A and 60 units of chemical B to go in each bottle of the drink. The chemical are available in a prepared mix from two different suppliers. Supplier X has a mix of 4 units of A and 2 units of B that costs ₹10, and the supplier Y has a mix of 1 unit of A and 1 unit of B that costs ₹4. How many mixes from X and Y should the company purchase to honor the contract requirement and yet minimize the cost?
Answer:

Let x and y be number of mixes from suppliers X and Y.


∴According to the question,


4x + y , 2x + y


Minimize Z = 10x + 4y


The feasible region determined by 4x + y , 2x + y is given by



The feasible region is unbounded .The corner points of feasible region are A(0,80) , B(10,40) , C(30,0).


The value of Z at corner points are



The minimum value of Z is 260 at point (10,40).


Hence, the company should buy 10 mixes from supplier X and 40 mixes from supplier Y to minimize the cost.
Question 36.

A company producing soft drinks has a contrast which requires a minimum of 80 units of chemical A and 60 units of chemical B to go in each bottle of the drink. The chemical are available in a prepared mix from two different suppliers. Supplier X has a mix of 4 units of A and 2 units of B that costs ₹10, and the supplier Y has a mix of 1 unit of A and 1 unit of B that costs ₹4. How many mixes from X and Y should the company purchase to honor the contract requirement and yet minimize the cost?
Answer:

Let x and y be number of mixes from suppliers X and Y.


∴According to the question,


4x + y , 2x + y


Minimize Z = 10x + 4y


The feasible region determined by 4x + y , 2x + y is given by



The feasible region is unbounded .The corner points of feasible region are A(0,80) , B(10,40) , C(30,0).


The value of Z at corner points are



The minimum value of Z is 260 at point (10,40).


Hence, the company should buy 10 mixes from supplier X and 40 mixes from supplier Y to minimize the cost.
Question 37.

A small firm manufactures gold rings and chains. The combined number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and half an hour for a chain. The maximum number of hour to available per day is 16. If the profit on a ring is ₹300 and that on a chain is ₹190, how many of each should be manufactured daily so as to maximize the profit?
Answer:

Let x and y be number of gold rings and chains.


∴According to the question,


x + y , x + 0.5y


Maximize Z = 300x + 190y


The feasible region determined by x + y , x + 0.5y is given by



The corner points of feasible region are A(0,0) , B(0,24) , C(8,16), D(16,0).The value of Z at corner points are



The maximum value of Z is 5440 at point (8,16).


Hence, the firm should manufacture 8 gold rings and 16 gold chains to maximize their profit.
Question 38.

A small firm manufactures gold rings and chains. The combined number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and half an hour for a chain. The maximum number of hour to available per day is 16. If the profit on a ring is ₹300 and that on a chain is ₹190, how many of each should be manufactured daily so as to maximize the profit?
Answer:

Let x and y be number of gold rings and chains.


∴According to the question,


x + y , x + 0.5y


Maximize Z = 300x + 190y


The feasible region determined by x + y , x + 0.5y is given by



The corner points of feasible region are A(0,0) , B(0,24) , C(8,16), D(16,0).The value of Z at corner points are



The maximum value of Z is 5440 at point (8,16).


Hence, the firm should manufacture 8 gold rings and 16 gold chains to maximize their profit.
Question 39.

A manufacture makes two types, A and B, of teapots. Three machines are needed for the manufacture and the time required for each teapot on the machines is given below.

Each machine is available for a maximum of 6 hours per day. If the profit on each teapot of type A is 75 paise and that on each teapot of type B is 50 paise, show that 15 teapots of type A and 30 of type B should be manufactured in a day to get the maximum profit.
Answer:

Let x teapots of type A and y teapots of type B manufactured.


Then,


x ≥ 0, y ≥ 0


Also,


12x + 6y ≤ 6 × 60


12x + 6y ≤ 360


2x + y ≤ 60…..(1)


And,


18x + 0y ≤ 6 × 60


X ≤ 20……(2)


Also,


6x + 9y ≤ 6 × 60


2x + 3y ≤ 120…..(3)


The profit will be given by: Z


On plotting the constraints, we get,



Profit will be maximum when x = 30 and y = 15


Hence, Proved.
Question 40.

A manufacture makes two types, A and B, of teapots. Three machines are needed for the manufacture and the time required for each teapot on the machines is given below.

Each machine is available for a maximum of 6 hours per day. If the profit on each teapot of type A is 75 paise and that on each teapot of type B is 50 paise, show that 15 teapots of type A and 30 of type B should be manufactured in a day to get the maximum profit.
Answer:

Let x teapots of type A and y teapots of type B manufactured.


Then,


x ≥ 0, y ≥ 0


Also,


12x + 6y ≤ 6 × 60


12x + 6y ≤ 360


2x + y ≤ 60…..(1)


And,


18x + 0y ≤ 6 × 60


X ≤ 20……(2)


Also,


6x + 9y ≤ 6 × 60


2x + 3y ≤ 120…..(3)


The profit will be given by: Z


On plotting the constraints, we get,



Profit will be maximum when x = 30 and y = 15


Hence, Proved.
Question 41.

A manufacture makes two product, A and B. product A sells at ₹200 each and takes hour to make. Product B sells at ₹300 each and takes 1 hour to make. There is a permanent order for 14 of product A and 16 of product B. A working week consist of 40 hours of production and the weekly turnover must not be less than ₹10000. If the profit on each of the product A is ₹20 and on product B, it is ₹30 then how many of each should be produced so that the profit is maximum? Also, find the maximum profit.
Answer:

Let x and y be number of A and B products.


∴According to the question,


0.5x + y , 200x + 300y


Maximize Z = 20x + 30y


The feasible region determined by 0.5x + y , 200x + 300y is given by



The corner points of feasible region are A(14,33) , B(14,24) , C(26,16), D(48,16).The value of Z at corner points are



The maximum value of Z is 1440 at point (48,16).


Hence, the manufacturer should manufacture 48 A products and 16 B products to maximize their profit of Rs.1440.
Question 42.

A manufacture makes two product, A and B. product A sells at ₹200 each and takes hour to make. Product B sells at ₹300 each and takes 1 hour to make. There is a permanent order for 14 of product A and 16 of product B. A working week consist of 40 hours of production and the weekly turnover must not be less than ₹10000. If the profit on each of the product A is ₹20 and on product B, it is ₹30 then how many of each should be produced so that the profit is maximum? Also, find the maximum profit.
Answer:

Let x and y be number of A and B products.


∴According to the question,


0.5x + y , 200x + 300y


Maximize Z = 20x + 30y


The feasible region determined by 0.5x + y , 200x + 300y is given by



The corner points of feasible region are A(14,33) , B(14,24) , C(26,16), D(48,16).The value of Z at corner points are



The maximum value of Z is 1440 at point (48,16).


Hence, the manufacturer should manufacture 48 A products and 16 B products to maximize their profit of Rs.1440.
Question 43.

A man owns a field area 1000 m2. He wants to plant fruit trees in it. He has a sum of ₹1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 m2 of ground per trees and costs ₹20 per tree, and type B requires 20 m2 of ground per tree and costs ₹25 per tree. When full grown, a type - A tree produces an average of 20 kg of fruit which can be sold at a profit ₹2 per kg and type - B tree produces an average of 40 kg of fruit which can be sold at a profit of ₹1.50 per kg. How many of each type should be planted to achieve maximum profit when tree are full grown? What is the maximum profit?
Answer:

Let x and y be number of A and B trees.


∴According to the question,


20x + 25y , 10x + 20y


Maximize Z = 40x + 60y


The feasible region determined by 20x + 25y , 10x + 20y is given by



The corner points of feasible region are A(0,0) , B(0,50) , C(20,40), D(70,0).The value of Z at corner points are



The maximum value of Z is 3200 at point (20,40).


Hence, the man should plant 20 A trees and 40 B trees to make maximum profit of Rs.3200.
Question 44.

A man owns a field area 1000 m2. He wants to plant fruit trees in it. He has a sum of ₹1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 m2 of ground per trees and costs ₹20 per tree, and type B requires 20 m2 of ground per tree and costs ₹25 per tree. When full grown, a type - A tree produces an average of 20 kg of fruit which can be sold at a profit ₹2 per kg and type - B tree produces an average of 40 kg of fruit which can be sold at a profit of ₹1.50 per kg. How many of each type should be planted to achieve maximum profit when tree are full grown? What is the maximum profit?
Answer:

Let x and y be number of A and B trees.


∴According to the question,


20x + 25y , 10x + 20y


Maximize Z = 40x + 60y


The feasible region determined by 20x + 25y , 10x + 20y is given by



The corner points of feasible region are A(0,0) , B(0,50) , C(20,40), D(70,0).The value of Z at corner points are



The maximum value of Z is 3200 at point (20,40).


Hence, the man should plant 20 A trees and 40 B trees to make maximum profit of Rs.3200.
Question 45.

A publisher sells a hardcover edition of a book for ₹72 and a paperback edition of the same for ₹40. Costs to the publisher are ₹56 and ₹28 respectively in addition to weekly costs of ₹9600. Both types require 5 minutes of printing time although the hardcover edition requires 10 minutes of binding time and the paperback edition requires only 2 minutes. Both the printing and binding operations have 4800 minutes available each week. How many of each type of books should be produced in order to maximize the profit? Also, find the maximum profit per week.
Answer:

Let x and y be number of hardcover and paperback edition of the book.


∴According to the question,


5x + 5y , 10x + 2y


Maximize Z = (72x + 40y) – (56x + 28y + 9600)


= 16x + 12y - 9600


The feasible region determined by 5x + 5y , 10x + 2y is given by



The corner points of feasible region are A(0,0) , B(0,960) , C(360,600), D(480,0).The value of Z at corner points are



The maximum value of Z is 3360 at point (360,600).


Hence, the publisher should publish 360 hardcover edition and 600 and paperback edition of the book to earn maximum profit of Rs.3360.
Question 46.

A publisher sells a hardcover edition of a book for ₹72 and a paperback edition of the same for ₹40. Costs to the publisher are ₹56 and ₹28 respectively in addition to weekly costs of ₹9600. Both types require 5 minutes of printing time although the hardcover edition requires 10 minutes of binding time and the paperback edition requires only 2 minutes. Both the printing and binding operations have 4800 minutes available each week. How many of each type of books should be produced in order to maximize the profit? Also, find the maximum profit per week.
Answer:

Let x and y be number of hardcover and paperback edition of the book.


∴According to the question,


5x + 5y , 10x + 2y


Maximize Z = (72x + 40y) – (56x + 28y + 9600)


= 16x + 12y - 9600


The feasible region determined by 5x + 5y , 10x + 2y is given by



The corner points of feasible region are A(0,0) , B(0,960) , C(360,600), D(480,0).The value of Z at corner points are



The maximum value of Z is 3360 at point (360,600).


Hence, the publisher should publish 360 hardcover edition and 600 and paperback edition of the book to earn maximum profit of Rs.3360.
Question 47.

A gardener has a supply of fertilizers of the type 1 which consist of 10% nitrogen and 6% phosphoric acid, and of the type II which consist of 5% nitrogen and 10% phosphoric acid. After testing the soil condition, he finds that he needs at least 14kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type - I fertilizer costs 60 paise per kg and the type - II fertilizer costs 40 paise per kg, determine how many kilograms of each type of fertilizer should be used so that the nutrient requirement are met at a minimum cost. What is the minimum cost?
Answer:

Let x and y be number of kilograms of fertilizer I and II


∴According to the question,


0.10x + 0.05y , 0.06x + 0.10y


Minimize Z = 0.60x + 0.40y


The feasible region determined by 0.10x + 0.05y , 0.06x + 0.10y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,280) , B(100,80) , C(700/3,0).The value of Z at corner points are



The minimum value of Z is 92 at point (100,80).


Hence, the gardener should by 100 kilograms o fertilizer I and 80 kg of fertilizer II to minimize the cost which is Rs.92.
Question 48.

A gardener has a supply of fertilizers of the type 1 which consist of 10% nitrogen and 6% phosphoric acid, and of the type II which consist of 5% nitrogen and 10% phosphoric acid. After testing the soil condition, he finds that he needs at least 14kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type - I fertilizer costs 60 paise per kg and the type - II fertilizer costs 40 paise per kg, determine how many kilograms of each type of fertilizer should be used so that the nutrient requirement are met at a minimum cost. What is the minimum cost?
Answer:

Let x and y be number of kilograms of fertilizer I and II


∴According to the question,


0.10x + 0.05y , 0.06x + 0.10y


Minimize Z = 0.60x + 0.40y


The feasible region determined by 0.10x + 0.05y , 0.06x + 0.10y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,280) , B(100,80) , C(700/3,0).The value of Z at corner points are



The minimum value of Z is 92 at point (100,80).


Hence, the gardener should by 100 kilograms o fertilizer I and 80 kg of fertilizer II to minimize the cost which is Rs.92.
Question 49.

Two godowns, A and B, have a grain storage capacity of 100 quintals and 50 quintals respectively. Their supply goes to three ration shops, D, E and F, whose requirements are 60, 50 and 40 quintals respectively. The costs of transportation per quintal from the godowns to the shops are given in the following table.



How should the supplies be transported in order that the transportation cost is minimum?
Answer:

Let x quintals of supplies be transported from A to D and y quintals be transported from A to E.


Therefore, 100 - (x + y) will be transported to F.


Also, (60 - x) quintals, (50 - y) quintals and (40 – (100 – (x + y))) quintals will be transported to D, E, F by godown B.


∴According to the question,


x


Minimize Z = 6x + 4(60 - x) + 3y + 2(50 – y) + 2.50(100 - (x + y)) + 3((x + y ) – 60 )


Z = 6x + 240 – 4x + 3y + 100 – 2y + 250 - 2.5x - 2.5y + 3x + 3y – 180


Z = 2.5x + 1.5y + 210


The feasible region represented by x is given by



The corner points of feasible region are A(10,50) , B(50,50) , C(60,40) , D(60,0)



The minimum value of Z is 310 at point (10,50).


Hence, 10, 50, 40 quintals of supplies should be transported from A to D, E, F and 50, 0, 0 quintals of supplies should be transported from B to D, E, F.
Question 50.

Two godowns, A and B, have a grain storage capacity of 100 quintals and 50 quintals respectively. Their supply goes to three ration shops, D, E and F, whose requirements are 60, 50 and 40 quintals respectively. The costs of transportation per quintal from the godowns to the shops are given in the following table.



How should the supplies be transported in order that the transportation cost is minimum?
Answer:

Let x quintals of supplies be transported from A to D and y quintals be transported from A to E.


Therefore, 100 - (x + y) will be transported to F.


Also, (60 - x) quintals, (50 - y) quintals and (40 – (100 – (x + y))) quintals will be transported to D, E, F by godown B.


∴According to the question,


x


Minimize Z = 6x + 4(60 - x) + 3y + 2(50 – y) + 2.50(100 - (x + y)) + 3((x + y ) – 60 )


Z = 6x + 240 – 4x + 3y + 100 – 2y + 250 - 2.5x - 2.5y + 3x + 3y – 180


Z = 2.5x + 1.5y + 210


The feasible region represented by x is given by



The corner points of feasible region are A(10,50) , B(50,50) , C(60,40) , D(60,0)



The minimum value of Z is 310 at point (10,50).


Hence, 10, 50, 40 quintals of supplies should be transported from A to D, E, F and 50, 0, 0 quintals of supplies should be transported from B to D, E, F.
Question 51.

A brick manufacture has two depots, P and Q, with stocks of 30000 and 20000 bricks respectively. He receives order from three building A, B, C, for 15000, 20000 and 15000 bricks respectively. The costs of transporting 1000 bricks to the building from the depots are given below.



How should the manufacture fulfill the orders so as to keep the cost of transportation minimum?
Answer:

Let x bricks be transported from P to A and y bricks be transported from P to B.


Therefore, 30000 - (x + y) will be transported to C.


Also, (15000 - x) bricks, (20000 - y) bricks and (15000 - (30000 - (x + y))) bricks will be transported to A, B, C from Q.


∴According to the question,


x


Minimize Z = 0.04x + 0.02(15000 - x) + 0.02y + 0.06(20000 - y) + 0.03(30000 - (x + y)) + 0.04((x + y) – 15000)


Z = 0.03x - 0.03y + 1800


The feasible region represented by x is given by



The corner points of feasible region are A(0,15000) , B(0,20000) , C(10000,20000) , D(15000,15000), E(15000,0).



The minimum value of Z is 1200 at point (0,20000).


Hence, 0, 20000, 10000 bricks should be transported from P to A, B, C and 15000, 0, 5000 bricks should be transported from Q to A, B, C.
Question 52.

A brick manufacture has two depots, P and Q, with stocks of 30000 and 20000 bricks respectively. He receives order from three building A, B, C, for 15000, 20000 and 15000 bricks respectively. The costs of transporting 1000 bricks to the building from the depots are given below.



How should the manufacture fulfill the orders so as to keep the cost of transportation minimum?
Answer:

Let x bricks be transported from P to A and y bricks be transported from P to B.


Therefore, 30000 - (x + y) will be transported to C.


Also, (15000 - x) bricks, (20000 - y) bricks and (15000 - (30000 - (x + y))) bricks will be transported to A, B, C from Q.


∴According to the question,


x


Minimize Z = 0.04x + 0.02(15000 - x) + 0.02y + 0.06(20000 - y) + 0.03(30000 - (x + y)) + 0.04((x + y) – 15000)


Z = 0.03x - 0.03y + 1800


The feasible region represented by x is given by



The corner points of feasible region are A(0,15000) , B(0,20000) , C(10000,20000) , D(15000,15000), E(15000,0).



The minimum value of Z is 1200 at point (0,20000).


Hence, 0, 20000, 10000 bricks should be transported from P to A, B, C and 15000, 0, 5000 bricks should be transported from Q to A, B, C.
Question 53.

A medicine company has factories at two places, X and Y. From these places, supply is made to each of its three agencies situated at P, Q and R. the monthly requirement of the agencies are respectively 40 packets, 40 packets and 50 packets of medicine, while the production capacity of the factories at X and Y are 60 packets and 70 packets respectively. The transportation costs per packet from the factories to the agencies are given as follows.



How many packets from each factory should be transported to each agency so that the cost of transportation is minimum? Also, find the minimum cost.
Answer:

Let x packets of medicines be transported from X to P and y packets of medicines be transported from X to Q.


Therefore, 60 - (x + y) will be transported to R.


Also, (40 - x) packets, (40 - y) packets and (50 - (60 - (x + y))) packets will be transported to P, Q, R from Y.


∴According to the question,


x


Minimize Z = 5x + 4(40 - x) + 4y + 2(40 - y) + 3(60 - (x + y)) + 5((x + y) – 10)


Z = 3x + 4y + 370


The feasible region represented by x is given by



The corner points of feasible region are A(0,10) , B(0,40) , C(20,40) , D(40,20), E(10,0).



The minimum value of Z is 40 at point (10,0).


Hence, 10, 0, 50 packets of medicines should be transported from X to P, Q, R and 30, 40, 0 packets of medicines should be transported from Y to P, Q, R.
Question 54.

A medicine company has factories at two places, X and Y. From these places, supply is made to each of its three agencies situated at P, Q and R. the monthly requirement of the agencies are respectively 40 packets, 40 packets and 50 packets of medicine, while the production capacity of the factories at X and Y are 60 packets and 70 packets respectively. The transportation costs per packet from the factories to the agencies are given as follows.



How many packets from each factory should be transported to each agency so that the cost of transportation is minimum? Also, find the minimum cost.
Answer:

Let x packets of medicines be transported from X to P and y packets of medicines be transported from X to Q.


Therefore, 60 - (x + y) will be transported to R.


Also, (40 - x) packets, (40 - y) packets and (50 - (60 - (x + y))) packets will be transported to P, Q, R from Y.


∴According to the question,


x


Minimize Z = 5x + 4(40 - x) + 4y + 2(40 - y) + 3(60 - (x + y)) + 5((x + y) – 10)


Z = 3x + 4y + 370


The feasible region represented by x is given by



The corner points of feasible region are A(0,10) , B(0,40) , C(20,40) , D(40,20), E(10,0).



The minimum value of Z is 40 at point (10,0).


Hence, 10, 0, 50 packets of medicines should be transported from X to P, Q, R and 30, 40, 0 packets of medicines should be transported from Y to P, Q, R.
Question 55.

An oil company has two depots, A and B, with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three pumps, D, E, F, whose requirements are 4500 L, 3000 L, and 3500 L respectively. The distances (in km) between the depots and the petrol pumps are given in the following table:



Assuming that the transportation cost per km is re 1 per litre, how should the delivery be scheduled in order that the transportation cost is minimum?
Answer:

Let x liters of petrol be transported from A to D and y liters of petrol be transported from A to E.


Therefore, 7000 - (x + y) will be transported to F.


Also, (4500 - x) liters of petrol, (3000 - y) liters of petrol and (3500 – (7000 – (x + y))) liters of petrol will be transported to D, E, F by B.


∴According to the question,


x


Minimize Z = 7x + 3(4500 – x) + 6y + 4(3000 – y) + 3(7000 –(x + y)) + 2 ( (x + y) – 3500)


Z = 3x + y + 39500


The feasible region represented by x is given by



The corner points of feasible region are A(500,3000) , B(4000,3000) , C(4500,2500) , D(4500,0) , E(3500,0)



The minimum value of Z is 44000 at point (500,3000).


Hence, 500,3000,3500 liters of petrol should be transported from A to D, E, F and 4000, 0, 0 liters of petrol should be transported from B to D, E, F.
Question 56.

An oil company has two depots, A and B, with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three pumps, D, E, F, whose requirements are 4500 L, 3000 L, and 3500 L respectively. The distances (in km) between the depots and the petrol pumps are given in the following table:



Assuming that the transportation cost per km is re 1 per litre, how should the delivery be scheduled in order that the transportation cost is minimum?
Answer:

Let x liters of petrol be transported from A to D and y liters of petrol be transported from A to E.


Therefore, 7000 - (x + y) will be transported to F.


Also, (4500 - x) liters of petrol, (3000 - y) liters of petrol and (3500 – (7000 – (x + y))) liters of petrol will be transported to D, E, F by B.


∴According to the question,


x


Minimize Z = 7x + 3(4500 – x) + 6y + 4(3000 – y) + 3(7000 –(x + y)) + 2 ( (x + y) – 3500)


Z = 3x + y + 39500


The feasible region represented by x is given by



The corner points of feasible region are A(500,3000) , B(4000,3000) , C(4500,2500) , D(4500,0) , E(3500,0)



The minimum value of Z is 44000 at point (500,3000).


Hence, 500,3000,3500 liters of petrol should be transported from A to D, E, F and 4000, 0, 0 liters of petrol should be transported from B to D, E, F.
Question 57.

A firm is engaged in breeding pigs. The pigs are fed on various products grown on the farm. They need certain nutrients, named as X,Y,Z. the pigs are fed on two products, A and B. One unit of product A contain 36 unit of X, 3 units of Y and 20 units of Z, while one unit of product B contain 6 units of X, 12 units of Y and 10 units of Z. the minimum requirement of X, Y, Z are 108 units, 36 units and 100 units respectively. Product A costs ₹20 per unit and product B costs ₹40 per unit. How many units of each product must be taken to minimize the cost? Also, find the minimum cost.
Answer:

Let x and y be number of units of products of A and B.


∴According to the question,


36x + 6y , 3x + 12y


Minimize Z = 20x + 40y


The feasible region determined 36x + 6y , 3x + 12y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,18) , B(2,6) , C(4,2) , D(12,0).The value of Z at corner points are



The minimum value of Z is 160 at point (4,2).


Hence, the firm should buy 4 units of fertilizer A and 2 units of fertilizer B to achieve minimum expense of Rs.160.
Question 58.

A firm is engaged in breeding pigs. The pigs are fed on various products grown on the farm. They need certain nutrients, named as X,Y,Z. the pigs are fed on two products, A and B. One unit of product A contain 36 unit of X, 3 units of Y and 20 units of Z, while one unit of product B contain 6 units of X, 12 units of Y and 10 units of Z. the minimum requirement of X, Y, Z are 108 units, 36 units and 100 units respectively. Product A costs ₹20 per unit and product B costs ₹40 per unit. How many units of each product must be taken to minimize the cost? Also, find the minimum cost.
Answer:

Let x and y be number of units of products of A and B.


∴According to the question,


36x + 6y , 3x + 12y


Minimize Z = 20x + 40y


The feasible region determined 36x + 6y , 3x + 12y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,18) , B(2,6) , C(4,2) , D(12,0).The value of Z at corner points are



The minimum value of Z is 160 at point (4,2).


Hence, the firm should buy 4 units of fertilizer A and 2 units of fertilizer B to achieve minimum expense of Rs.160.
Question 59.

A dietician wishes to mix two types of food, X and Y, in such a way that the vitamin contents of the mixture contains at least 8 units of vitamin A and 10 units of vitamin C. Food X contains 2 units/kg of vitamin A and 1 unit /kg of vitamin C, while food Y contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs ₹5 per kg to purchase the food X and ₹7 per kg to purchase the food Y. Determine the minimum cost of such a mixture.
Answer:

Let x and y be number of units of X and Y.


∴According to the question,


2x + y , x + 2y


Minimize Z = 5x + 7y


The feasible region determined 2x + y , x + 2y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,8) , B(2,4) , C(10,0).The value of Z at corner points are



The minimum value of Z is 160 at point (4,2).


Hence, the dietician should mix 2 units of X and 4 units of Y to meet the requirements at minimum cost of Rs.38.
Question 60.

A dietician wishes to mix two types of food, X and Y, in such a way that the vitamin contents of the mixture contains at least 8 units of vitamin A and 10 units of vitamin C. Food X contains 2 units/kg of vitamin A and 1 unit /kg of vitamin C, while food Y contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs ₹5 per kg to purchase the food X and ₹7 per kg to purchase the food Y. Determine the minimum cost of such a mixture.
Answer:

Let x and y be number of units of X and Y.


∴According to the question,


2x + y , x + 2y


Minimize Z = 5x + 7y


The feasible region determined 2x + y , x + 2y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,8) , B(2,4) , C(10,0).The value of Z at corner points are



The minimum value of Z is 160 at point (4,2).


Hence, the dietician should mix 2 units of X and 4 units of Y to meet the requirements at minimum cost of Rs.38.
Question 61.

A diet for a sick person must contain at least 4000 units of vitamins, 50 units of mineral and 1400 calories. Two food, A and B, are available at a cost of ₹4 and ₹3 per unit respectively. If one unit of A contains 200 units of vitamins, 1 unit of mineral and 40 calories, and 1 unit of B contains 100 units of vitamins, 2 units of mineral and 40 calories, find what combination of foods should be used to have the least cost.
Answer:

Let x and y be number of units of food A and B.


∴According to the question,


200x + 100y , x + 2y


Minimize Z = 4x + 3y


The feasible region determined 200x + 100y , x + 2y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,40) , B(5,30) , C(20,15) , D(50,0).The value of Z at corner points are



The minimum value of Z is 110 at point (5,30).


Hence, the diet should contain 5 units of food A and 30 units of food B for the least cost.
Question 62.

A diet for a sick person must contain at least 4000 units of vitamins, 50 units of mineral and 1400 calories. Two food, A and B, are available at a cost of ₹4 and ₹3 per unit respectively. If one unit of A contains 200 units of vitamins, 1 unit of mineral and 40 calories, and 1 unit of B contains 100 units of vitamins, 2 units of mineral and 40 calories, find what combination of foods should be used to have the least cost.
Answer:

Let x and y be number of units of food A and B.


∴According to the question,


200x + 100y , x + 2y


Minimize Z = 4x + 3y


The feasible region determined 200x + 100y , x + 2y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,40) , B(5,30) , C(20,15) , D(50,0).The value of Z at corner points are



The minimum value of Z is 110 at point (5,30).


Hence, the diet should contain 5 units of food A and 30 units of food B for the least cost.
Question 63.

A housewife wishes to mix together two kinds of food, X and Y, in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C.

The vitamin contents of 1 kg of each food are given below.



If 1 kg of food X cost ₹6 and 1 kg of food Y costs ₹10, find the minimum cost of the mixture which will produce the diet.
Answer:

Let x and y be number of kilograms of food X and Y.


∴According to the question,


x + 2y , 2x + 2y


Minimize Z = 6x + 10y


The feasible region determined x + 2y , 2x + 2y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,8) , B(1,5) , C(2,4) , D(10,0).The value of Z at corner points are



The minimum value of Z is 52 at point (2,4).


Hence, the diet should contain 2 kgs of food X and 4 kgs of food Y for the least cost of Rs. 52.
Question 64.

A housewife wishes to mix together two kinds of food, X and Y, in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C.

The vitamin contents of 1 kg of each food are given below.



If 1 kg of food X cost ₹6 and 1 kg of food Y costs ₹10, find the minimum cost of the mixture which will produce the diet.
Answer:

Let x and y be number of kilograms of food X and Y.


∴According to the question,


x + 2y , 2x + 2y


Minimize Z = 6x + 10y


The feasible region determined x + 2y , 2x + 2y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,8) , B(1,5) , C(2,4) , D(10,0).The value of Z at corner points are



The minimum value of Z is 52 at point (2,4).


Hence, the diet should contain 2 kgs of food X and 4 kgs of food Y for the least cost of Rs. 52.
Question 65.

A firm manufactures two types of product, A and B, and sells them at a profit of ₹5 per unit of type A and ₹3 per unit of type B. Each product is processed on two machines, M1 and M2 . one unit of type A requires one minute of processing time on M1 and two minutes of processing time on M2; whereas one unit of type B requires one minute of processing time on M1 and one minute on M2. Machines M1 and M2 are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product the firm should produce a day in order to maximize the profit. Solve the problem graphically.
Answer:

Let the firm manufacture x number of Aand y number of B products.


∴According to the question,


X + y


Maximize Z = 5x + 3y


The feasible region determined X + y is given by



The corner points of feasible region are A(0,0) , B(0,300) , C(60,240) , D(180,0).The value of Z at corner point is



The maximum value of Z is 1020 and occurs at point (60,240).


The firm should produce 60 Aproducts and 240 B products to earn maximum profit of Rs.1020.
Question 66.

A firm manufactures two types of product, A and B, and sells them at a profit of ₹5 per unit of type A and ₹3 per unit of type B. Each product is processed on two machines, M1 and M2 . one unit of type A requires one minute of processing time on M1 and two minutes of processing time on M2; whereas one unit of type B requires one minute of processing time on M1 and one minute on M2. Machines M1 and M2 are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product the firm should produce a day in order to maximize the profit. Solve the problem graphically.
Answer:

Let the firm manufacture x number of Aand y number of B products.


∴According to the question,


X + y


Maximize Z = 5x + 3y


The feasible region determined X + y is given by



The corner points of feasible region are A(0,0) , B(0,300) , C(60,240) , D(180,0).The value of Z at corner point is



The maximum value of Z is 1020 and occurs at point (60,240).


The firm should produce 60 Aproducts and 240 B products to earn maximum profit of Rs.1020.
Question 67.

A small firm manufactures items A and B. The total number of items that it can manufacture in a day is at most 24. Item A takes one hour to make while item B take only half an hour. The maximum time available per day is 16 hours. If the profit on one unit item A be ₹300 and that on one unit of item B be ₹160, how many of each type of item should be produced to maximize the profit? Solve the problem graphically.

Ans. Z is maximum at (8, 16) and its maximum value is ₹4960
Answer:

Let the firm manufacture x number of A and y number of B products.


∴According to the question,


X + y


Maximize Z = 300x + 160y


The feasible region determined X + y is given by



The corner points of feasible region are A(0,0) , B(0,24) , C(8,16) , D(16,0).The value of Z at corner point is



The maximum value of Z is 4960 and occurs at point (8,16).


The firm should produce 8 Aproducts and 16 B products to earn maximum profit of Rs.4960.
Question 68.

A small firm manufactures items A and B. The total number of items that it can manufacture in a day is at most 24. Item A takes one hour to make while item B take only half an hour. The maximum time available per day is 16 hours. If the profit on one unit item A be ₹300 and that on one unit of item B be ₹160, how many of each type of item should be produced to maximize the profit? Solve the problem graphically.

Ans. Z is maximum at (8, 16) and its maximum value is ₹4960
Answer:

Let the firm manufacture x number of A and y number of B products.


∴According to the question,


X + y


Maximize Z = 300x + 160y


The feasible region determined X + y is given by



The corner points of feasible region are A(0,0) , B(0,24) , C(8,16) , D(16,0).The value of Z at corner point is



The maximum value of Z is 4960 and occurs at point (8,16).


The firm should produce 8 Aproducts and 16 B products to earn maximum profit of Rs.4960.
Question 69.

A manufacture produces two types of steel trunks. He has two machines, A and B. The first type of trunk requires 3 hours on machine A and 3 hours on machine B. The second type required 3 hours on machine A and 2 hours on Machine A and 2 hours on machine B. Machine A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of ₹30 and ₹25 per trunk of the first type and second type respectively. How may trunks of each type must he make each day to make the maximum profit?
Answer:

Let the manufacturer manufacture x and y numbers of type 1 and type 2trunks.


∴According to the question,


3X + 3y


Maximize Z = 30x + 25y


The feasible region determined 3X + 3y is given by



The corner points of feasible region are A(0,0) , B(0,6) , C(3,3) , D(5,0).The value of Z at corner point is



The maximum value of Z is165 and occurs at point (3,3).


The manufacturer should manufacture 3 trunks of each type to earn maximum profit of Rs.165.
Question 70.

A manufacture produces two types of steel trunks. He has two machines, A and B. The first type of trunk requires 3 hours on machine A and 3 hours on machine B. The second type required 3 hours on machine A and 2 hours on Machine A and 2 hours on machine B. Machine A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of ₹30 and ₹25 per trunk of the first type and second type respectively. How may trunks of each type must he make each day to make the maximum profit?
Answer:

Let the manufacturer manufacture x and y numbers of type 1 and type 2trunks.


∴According to the question,


3X + 3y


Maximize Z = 30x + 25y


The feasible region determined 3X + 3y is given by



The corner points of feasible region are A(0,0) , B(0,6) , C(3,3) , D(5,0).The value of Z at corner point is



The maximum value of Z is165 and occurs at point (3,3).


The manufacturer should manufacture 3 trunks of each type to earn maximum profit of Rs.165.
Question 71.

A company manufactures two types of toys, A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B required 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is ₹50 each on type A and ₹60 each on type B. how many toys of each types should the company manufactures in a day to maximize the profit?
Answer:

Let the company manufacture x and y numbers of toys A and B.


∴According to the question,


5X + 8y


Maximize Z = 50x + 60y


The feasible region determined 5X + 8y is given by



The corner points of feasible region are A(0,0) , B(0,22.5) , C(12,15) , D(24,0).The value of Z at corner point is



The maximum value of Z is1500 and occurs at point (12,15).


The company should manufacture 12 A toys and 15 B toys to earn profit of rupees 1500.
Question 72.

A company manufactures two types of toys, A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B required 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is ₹50 each on type A and ₹60 each on type B. how many toys of each types should the company manufactures in a day to maximize the profit?
Answer:

Let the company manufacture x and y numbers of toys A and B.


∴According to the question,


5X + 8y


Maximize Z = 50x + 60y


The feasible region determined 5X + 8y is given by



The corner points of feasible region are A(0,0) , B(0,22.5) , C(12,15) , D(24,0).The value of Z at corner point is



The maximum value of Z is1500 and occurs at point (12,15).


The company should manufacture 12 A toys and 15 B toys to earn profit of rupees 1500.
Question 73.

Kellogg is a new cereal formed of a mixture of bran and rice, that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilograms, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost producing this new cereal if bran costs ₹5 per kilogram and rice costs ₹4 per kilogram.
Answer:

Let x and y be number of kilograms of bran and rice.


∴According to the question,


80x + 100y , 40x + 30y


Minimize Z = 5x + 4y


The feasible region determined 80x + 100y , 40x + 30y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,1.2) , B(0.6,0.4) , C(1.1,0).The value of Z at corner points are



The minimum value of Z is 4.6 at point (0.6,0.4).


Hence, the diet should contain 0.6 kgs of bran and 0.4 kgs of rice for achieving minimum cost of Rs.4.6.
Question 74.

Kellogg is a new cereal formed of a mixture of bran and rice, that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilograms, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost producing this new cereal if bran costs ₹5 per kilogram and rice costs ₹4 per kilogram.
Answer:

Let x and y be number of kilograms of bran and rice.


∴According to the question,


80x + 100y , 40x + 30y


Minimize Z = 5x + 4y


The feasible region determined 80x + 100y , 40x + 30y is given by



The feasible region is unbounded. The corner points of feasible region are A(0,1.2) , B(0.6,0.4) , C(1.1,0).The value of Z at corner points are



The minimum value of Z is 4.6 at point (0.6,0.4).


Hence, the diet should contain 0.6 kgs of bran and 0.4 kgs of rice for achieving minimum cost of Rs.4.6.
Question 75.

A dealer wishes to purchase a number of fans and sewing machines. He has only ₹5760 to invest and has space for at most 20 items. A fan costs him ₹360 and a sewing machine ₹240. He expects to sell a fan at a profit of ₹22 and a sewing machine at a profit of ₹18. Assuming that he can sell all the items that he buys, how should he invest his money to maximize the profit? Solve the graphically and find the maximum profit.
Answer:

Let the number of fans bought be x and sewing machines bought be y.


∴According to the question,


360x + 240y,x + y


Maximize Z = 22x + 18y


The feasible region determined by 360x + 240y,x + y is given by



The corner points of the feasible region are A(0,0) , B(0,20),C(8,12) , D(16,0).The value of Z at corner points is



The maximum value of Z is 392 at point (8,12).


The dealer must buy 8 fans and 12 sewing machines to make the maximum profit.
Question 76.

A dealer wishes to purchase a number of fans and sewing machines. He has only ₹5760 to invest and has space for at most 20 items. A fan costs him ₹360 and a sewing machine ₹240. He expects to sell a fan at a profit of ₹22 and a sewing machine at a profit of ₹18. Assuming that he can sell all the items that he buys, how should he invest his money to maximize the profit? Solve the graphically and find the maximum profit.
Answer:

Let the number of fans bought be x and sewing machines bought be y.


∴According to the question,


360x + 240y,x + y


Maximize Z = 22x + 18y


The feasible region determined by 360x + 240y,x + y is given by



The corner points of the feasible region are A(0,0) , B(0,20),C(8,12) , D(16,0).The value of Z at corner points is



The maximum value of Z is 392 at point (8,12).


The dealer must buy 8 fans and 12 sewing machines to make the maximum profit.
Question 77.

Anil wants to invest at the most ₹12000 in bonds A and B. According to rules, he has to invest at least ₹2000 in bond A and at least ₹4000 in bond B. if the rate of interest of bond A is 8% per annum and on bond B, it is 10% per annum, how should he invest his money for maximum interest? Solve the problem graphically.
Answer:

Let the invested money in bond A be x and in bond B be y.


∴According to the question,


X + y , x , y


Maximize Z = 0.08x + 0.10y


The feasible region determined by X + y , x , yis given by



The corner points of the feasible region are A(2000,4000) , B(2000,10000) and C(8000,4000) . The value of Z at the corner point are



The maximum value of Z is 116770 at point (2000,10000)


So, he must invest Rs.2000 in bond A and Rs.10000 in bond B.


The maximum annual income is Rs.1160 .
Question 78.

Anil wants to invest at the most ₹12000 in bonds A and B. According to rules, he has to invest at least ₹2000 in bond A and at least ₹4000 in bond B. if the rate of interest of bond A is 8% per annum and on bond B, it is 10% per annum, how should he invest his money for maximum interest? Solve the problem graphically.
Answer:

Let the invested money in bond A be x and in bond B be y.


∴According to the question,


X + y , x , y


Maximize Z = 0.08x + 0.10y


The feasible region determined by X + y , x , yis given by



The corner points of the feasible region are A(2000,4000) , B(2000,10000) and C(8000,4000) . The value of Z at the corner point are



The maximum value of Z is 116770 at point (2000,10000)


So, he must invest Rs.2000 in bond A and Rs.10000 in bond B.


The maximum annual income is Rs.1160 .
Question 79.

Maximize = 60x + 15y, subject to the constraints

x + y ≤ 50, 3x + ≤ y90, x, y ≥ 0.
Answer:

The feasible region determined by the constraints x + y ≤ 50, 3x + y ≤ 90, x, y ≥ 0. is given by



The corner points of feasible region are A(0,0) ,B(0,50) ,C(20,30), D(30,0) . The values of Z at the following points is



The maximum value of Z is 1800 at point A(30,0) .
Question 80.

Maximize = 60x + 15y, subject to the constraints

x + y ≤ 50, 3x + ≤ y90, x, y ≥ 0.
Answer:

The feasible region determined by the constraints x + y ≤ 50, 3x + y ≤ 90, x, y ≥ 0. is given by



The corner points of feasible region are A(0,0) ,B(0,50) ,C(20,30), D(30,0) . The values of Z at the following points is



The maximum value of Z is 1800 at point A(30,0) .
Question 81.

A company manufacture two types of toys A and B. type A requires 5 minutes each for cutting and 10 minutes for each assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. He earns a profit of ₹50 each on type A and ₹60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?
Answer:

Let the company manufacture x and y numbers of toys A and B.


∴According to the question,


5X + 8y


Maximize Z = 50x + 60y


The feasible region determined 5X + 8y is given by



The corner points of feasible region are A(0,0) , B(0,22.5) , C(12,15) , D(24,0).The value of Z at corner point is



The maximum value of Z is1500 and occurs at point (12,15).


The company should manufacture 12 A toys and 15 B toys to earn profit of rupees 1500.
Question 82.

A company manufacture two types of toys A and B. type A requires 5 minutes each for cutting and 10 minutes for each assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. He earns a profit of ₹50 each on type A and ₹60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?
Answer:

Let the company manufacture x and y numbers of toys A and B.


∴According to the question,


5X + 8y


Maximize Z = 50x + 60y


The feasible region determined 5X + 8y is given by



The corner points of feasible region are A(0,0) , B(0,22.5) , C(12,15) , D(24,0).The value of Z at corner point is



The maximum value of Z is1500 and occurs at point (12,15).


The company should manufacture 12 A toys and 15 B toys to earn profit of rupees 1500.
Question 83.

One kind of cake requires 200 g of flour and 25 g of fat, another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat, assuming that there is no shortage of the other ingredients used in making the cakes. Make it an LPP and solve it graphically.
Answer:

Let the company make x no of 1st kind and y no of 2nd cakes.


∴According to the question,


200x + 100y


Maximize Z = x + y


The feasible region determined by 200x + 100y is given by



The corner points of feasible region are A(0,0) , B(0,20) , C(20,10) , D(25,0).The value of Z at corner point is



The maximum value of Z is 30 and occurs at point (20,10).


The company should make 20 of 1st type and 10 of 2nd type.
Question 84.

One kind of cake requires 200 g of flour and 25 g of fat, another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat, assuming that there is no shortage of the other ingredients used in making the cakes. Make it an LPP and solve it graphically.
Answer:

Let the company make x no of 1st kind and y no of 2nd cakes.


∴According to the question,


200x + 100y


Maximize Z = x + y


The feasible region determined by 200x + 100y is given by



The corner points of feasible region are A(0,0) , B(0,20) , C(20,10) , D(25,0).The value of Z at corner point is



The maximum value of Z is 30 and occurs at point (20,10).


The company should make 20 of 1st type and 10 of 2nd type.
Question 85.

A manufacturing company makes two types of teaching aids A and B of mathematics for class XII. Each type of A requires 9 labor hours of fabricating and 1 labor hour for finishing. Each type of B requires 12 labors hour for fabricating and 3 labor hour for finishing. For fabricating and finishing, the maximum labor hours available per week are 180 and 30 respectively. The company makes a profit of ₹80 on each piece of type A and ₹120 on each piece of type B. how many pieces of type A and type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LLP and solve graphically. What is the maximum profit per week?
Answer:

Let the company make x no of 1st type of teaching aid and y no of 2nd type of teaching aid.


∴According to the question,


9x + 12y


Maximize Z = 80x + 120y


The feasible region determined by 9x + 12y is given by



The corner points of feasible region are A(0,0) , B(0,10) , C(12,6) , D(20,0).The value of Z at corner point is



The maximum value of Z is 1680 and occurs at point (12,6).


The company should make 12 of 1st type and 6 of 2nd type of teaching aid. Maximum profit is Rs.1680.
Question 86.

A manufacturing company makes two types of teaching aids A and B of mathematics for class XII. Each type of A requires 9 labor hours of fabricating and 1 labor hour for finishing. Each type of B requires 12 labors hour for fabricating and 3 labor hour for finishing. For fabricating and finishing, the maximum labor hours available per week are 180 and 30 respectively. The company makes a profit of ₹80 on each piece of type A and ₹120 on each piece of type B. how many pieces of type A and type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LLP and solve graphically. What is the maximum profit per week?
Answer:

Let the company make x no of 1st type of teaching aid and y no of 2nd type of teaching aid.


∴According to the question,


9x + 12y


Maximize Z = 80x + 120y


The feasible region determined by 9x + 12y is given by



The corner points of feasible region are A(0,0) , B(0,10) , C(12,6) , D(20,0).The value of Z at corner point is



The maximum value of Z is 1680 and occurs at point (12,6).


The company should make 12 of 1st type and 6 of 2nd type of teaching aid. Maximum profit is Rs.1680.