 # NCERT Solution Class 12 Chapter 12- Linear Programming Exercise 12.2

## NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.2 – CBSE Term I Free PDF Download

The Exercise 12.2 of NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming is based on the following topics:

1. Different Types of Linear Programming Problems
1. Manufacturing problems
2. Diet problems
3. Transportation problems

Solve all the problems of this exercise to get thorough with the concepts and topics covered in the entire chapter.

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Exercise 12.1 Solutions 10 Questions

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#### Access Answers to NCERT Class 12 Maths Chapter 12 Exercise 12.2

1.Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units /kg of vitamin A and 5 units /kg of vitamin B while food Q contains 4 units /kg of vitamin A and 2 units /kg of vitamin B. Determine the minimum cost of the mixture?

Solution:

Let the mixture contain x kg of food P and y kg of food Q.

Hence, x ≥ 0 and y ≥ 0

The given information can be compiled in a table as given

 Vitamin A (units / kg) Vitamin B (units / kg Cost (Rs / kg) Food P 3 5 60 Food Q 4 2 80 Requirement (units / kg) 8 11

The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Hence, the constraints are

3x + 4y ≥ 8

5x + 2y ≥ 11

Total cost of purchasing food is, Z = 60x + 80y

So, the mathematical formulation of the given problem can be written as

Minimise Z = 60x + 80y (i)

Now, subject to the constraints,

3x + 4y ≥ 8 … (2)

5+ 2y ≥ 11 … (3)

xy ≥ 0 … (4)

The feasible region determined by the system of constraints is given below Clearly, we can see that the feasible region is unbounded

A (8 / 3, 0), B (2, 1 / 2) and C (0, 11 / 2)

The values of Z at these corner points are given below

 Corner point Z = 60x + 80 y A (8 / 3, 0) 160 Minimum B (2, 1 / 2) 160 Minimum C (0, 11 / 2) 440

Here the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.

For this purpose, we graph the inequality, 60x + 80y < 160 or 3x + 4y < 8, and check whether the resulting half plane has points in common with the feasible region or not

Here, it can be seen that the feasible region has no common point with 3x + 4y < 8

Hence, at the line segment joining the points (8 / 3, 0) and (2, 1 / 2), the minimum cost of the mixture will be Rs 160

2. One kind of cake requires 200g flour and 25g of fat, and another kind of cake requires 100g of flour and 50g 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?

Solution:

Let the first kind of cakes be x and second kind of cakes be y. Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as shown below

 Flour (g) Fat (g) Cakes of first kind, x 200 25 Cakes of second kind, y 100 50 Availability 5000 1000

So, 200x + 100y ≤ 5000

2x + y ≤ 50

25x + 50y ≤ 1000

x + 2y ≤ 40

Total number of cakes Z that can be made are

Z = x + y

The mathematical formulation of the given problem can be written as

Maximize Z = x + y (i)

Here, subject to the constraints,

2x + y ≤ 50 (ii)

x + 2y ≤ 40 (iii)

x, y ≥ 0 (iv)

The feasible region determined by the system of constraints is given as below A (25, 0), B (20, 10), O (0, 0) and C (0, 20) are the corner points

The values of Z at these corner points are as given below

 Corner point Z = x + y A (25, 0) 25 B (20, 10) 30 Maximum C (0, 20) 20 O (0,0) 0

Hence, the maximum numbers of cakes that can be made are 30 (20 cakes of one kind and 10 cakes of other kind)

3. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.

(ii) What number of rackets and bats must be made if the factory is to work at full capacity?

(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.

Solution:

Let x and y be the number of rackets and the number of bats to be made.

Given that the machine time is not available for more than 42 hours

Hence, 1.5x + 3y ≤ 42 ……………. (i)

Also, given that the craftsman’s time is not available for more than 24 hours

Hence, 3x + y ≤ 24 ………… (ii)

The factory is to work at full capacity. Hence,

1.5x + 3y = 42

3x + y = 24

On solving these equations, we get

x = 4 and y = 12

Therefore, 4 rackets and 12 bats must be made.

(i) The given information can be compiled in a table as give below

 Tennis Racket Cricket Bat Availability Machine Time (h) 1.5 3 42 Craftsman’s Time (h) 3 1 24

1.5x + 3y ≤ 42

3x + y ≤ 24

x, y ≥ 0

Since, the profit on a racket is Rs 20 and Rs 10

Hence, Z = 20x + 10y

The mathematical formulation of the given problem can be written as

Maximize Z = 20x + 10y ………….. (i)

Subject to the constraints,

1.5x + 3y ≤ 42 …………. (ii)

3x + y ≤ 24 …………….. (iii)

x, y ≥ 0 ………………… (iv)

The feasible region determined by the system of constraints is given below A (8, 0), B (4, 12), C (0, 14) and O (0, 0) are the corner points respectively.

The values of Z at these corner points are given below

 Corner point Z = 20x + 10y A (8, 0) 160 B (4, 12) 200 Maximum C (0, 14) 140

Therefore, the maximum profit of the factory when it works to its full capacity is Rs 200

4. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs 17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximize his profit, if he operates his machines for at the most 12 hours a day?

Solution:

Let the manufacturer produce x package of nuts and y package of bolts. Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as given below

 Nuts Bolts Availability Machine A (h) 1 3 12 Machine B (h) 3 1 12

The profit on a package of nuts is Rs 17.50 and on a package of bolts is Rs 7

Hence, the constraints are

x + 3y ≤ 12

3x + y ≤ 12

Then, total profit, Z = 17.5x + 7y

The mathematical formulation of the given problem can be written as follows

Maximize Z = 17.5x + 7y …………. (1)

Subject to the constraints,

x + 3y ≤ 12 …………. (2)

3x + y ≤ 12 ………… (3)

x, y ≥ 0 …………….. (4)

The feasible region determined by the system of constraints is given below A (4, 0), B (3, 3) and C (0, 4) are the corner points

The values of Z at these corner points are given below

 Corner point Z = 17.5x + 7y O (0, 0) 0 A (4, 0) 70 B (3, 3) 73.5 Maximum C (0, 4) 28

Therefore, Rs 73.50 at (3, 3) is the maximum value of Z

Hence, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit of Rs 73.50

5. A factory manufacturers two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

Solution:

On each day, let the factory manufacture x screws of type A and y screws of type B.

Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as given below

 Screw A Screw B Availability Automatic Machine (min) 4 6 4 × 60 = 240 Hand Operated Machine (min) 6 3 4 × 60 = 240

The profit on a package of screws A is Rs 7 and on the package screws B is Rs 10

Hence, the constraints are

4x + 6y ≤ 240

6x + 3y ≤ 240

Total profit, Z = 7x + 10y

The mathematical formulation of the given problem can be written as

Maximize Z = 7x + 10y …………. (i)

Subject to the constraints,

4x + 6y ≤ 240 …………. (ii)

6x + 3y ≤ 240 …………. (iii)

x, y ≥ 0 ……………… (iv)

The feasible region determined by the system of constraints is given below A (40, 0), B (30, 20) and C (0, 40) are the corner points

The value of Z at these corner points are given below

 Corner point Z = 7x + 10y A (40, 0) 280 B (30, 20) 410 Maximum C (0, 40) 400

The maximum value of Z is 410 at (30, 20)

Hence, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410

6. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding / cutting machine and a sprayer. It takes 2 hours on grinding / cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding / cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding / cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shade that he produces, how should he schedule his daily production in order to maximize his profit?

Solution:

Let the cottage industry manufacture x pedestal lamps and y wooden shades respectively

Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table is given below

 Lamps Shades Availability Grinding / Cutting Machine (h) 2 1 12 Sprayer (h) 3 2 20

The profit on a lamp is Rs 5 and on the shades is Rs 3. Hence, the constraints are

2x + y ≤ 12

3x + 2y ≤ 20

Total profit, Z = 5x + 3y …………….. (i)

Subject to the constraints,

2x + y ≤ 12 …………. (ii)

3x + 2y ≤ 20 ………… (iii)

x, y ≥ 0 …………. (iv)

The feasible region determined by the system of constraints is given below A (6, 0), B (4, 4) and C (0, 10) are the corner points

The value of Z at these corner points are given below

 Corner point Z = 5x + 3y A (6, 0) 30 B (4, 4) 32 Maximum C (0, 10) 30

The maximum value of Z is 32 at point (4, 4)

Therefore, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximize his profits.

7. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

Solution:

Let the company manufacture x souvenirs of type A and y souvenirs of type B respectively

Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table is given below

 Type A Type B Availability Cutting (min) 5 8 3 × 60 + 20 = 200 Assembling (min) 10 8 4 × 60 = 240

The profit on type A souvenirs is Rs 5 and on type B souvenirs is Rs 6. Hence, the constraints are

5x + 8y ≤ 200

10x + 8y ≤ 240 i.e.,

5x + 4y ≤ 120

Total profit, Z = 5x + 6y

The mathematical formulation of the given problem can be written as

Maximize Z = 5x + 6y …………… (i)

Subject to the constraints,

5x + 8y ≤ 200 ……………. (ii)

5x + 4y ≤ 120 ………….. (iii)

x, y ≥ 0 ………….. (iv)

The feasible region determined by the system of constraints is given below A (24, 0), B (8, 20) and C (0, 25) are the corner points

The values of Z at these corner points are given below

 Corner point Z = 5x + 6y A (24, 0) 120 B (8, 20) 160 Maximum C (0, 25) 150

The maximum value of Z is 200 at (8, 20)

Hence, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of Rs 160

8. A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.

Solution:

Let the merchant stock x desktop models and y portable models respectively.

Hence,

x ≥ 0 and y ≥ 0

Given that the cost of desktop model is Rs 25000 and of a portable model is Rs 40000.

However, the merchant can invest a maximum of Rs 70 lakhs

Hence, 25000x + 40000y ≤ 7000000

5x + 8y ≤ 1400

The monthly demand of computers will not exceed 250 units.

Hence, x + y ≤ 250

The profit on a desktop model is 4500 and the profit on a portable model is Rs 5000

Total profit, Z = 4500x + 5000y

Therefore, the mathematical formulation of the given problem is

Maximum Z = 4500x + 5000y ………… (i)

Subject to the constraints,

5x + 8y ≤ 1400 ………… (ii)

x + y ≤ 250 ………….. (iii)

x, y ≥ 0 …………. (iv)

The feasible region determined by the system of constraints is given below A (250, 0), B (200, 50) and C (0, 175) are the corner points.

The values of Z at these corner points are given below

 Corner point Z = 4500x + 5000y A (250, 0) 1125000 B (200, 50) 1150000 Maximum C (0, 175) 875000

The maximum value of Z is 1150000 at (200, 50)

Therefore, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs 1150000.

9. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs 4 per unit food and F2 costs Rs 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.

Solution:

Let the diet contain x units of food F1 and y units of food F2. Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table is given below

 Vitamin A (units) Mineral (units) Cost per unit (Rs) Food F1 (x) 3 4 4 Food F2 (y) 6 3 6 Requirement 80 100

The cost of food F1 is Rs 4 per unit and of food F2 is Rs 6 per unit

Hence, the constraints are

3x + 6y ≥ 80

4x + 3y ≥ 100

x, y ≥ 0

Total cost of the diet, Z = 4x + 6y

The mathematical formulation of the given problem can be written as

Minimise Z = 4x + 6y …………… (i)

Subject to the constraints,

3x + 6y ≥ 80 ………… (ii)

4x + 3y ≥ 100 ………. (iii)

x, y ≥ 0 …………. (iv)

The feasible region determined by the constraints is given below We can see that the feasible region is unbounded.

A (80 / 3, 0), B (24, 4 / 3), and C (0, 100 / 3) are the corner points

The values of Z at these corner points are given below

 Corner point Z = 4x + 6y A (80 / 3, 0) 320 / 3 = 106.67 B (24, 4 / 3) 104 Minimum C (0, 100 / 3) 200

Here, the feasible region is unbounded, so 104 may or not be the minimum value of Z.

For this purpose, we draw a graph of the inequality, 4x + 6y < 104 or 2x + 3y < 52, and check whether the resulting half plane has points in common with the feasible region or not

Here, it can be seen that the feasible region has no common point with 2x + 3y < 52

Hence, the minimum cost of the mixture will be Rs 104

10. There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6 / kg and F2 costs Rs 5 / kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Solution:

Let the farmer buy x kg of fertilizer F1 and y kg of fertilizer F2. Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table is given below

 Nitrogen (%) Phosphoric Acid (%) Cost (Rs / kg) F1 (x) 10 6 6 F2 (y) 5 10 5 Requirement (kg) 14 14

F1 consists of 10% nitrogen and F2 consists of 5% nitrogen.

However, the farmer requires at least 14 kg of nitrogen

So, 10% of x + 5% of y ≥ 14

x / 10 + y / 20 ≥ 14

By L.C.M we get

2x + y ≥ 280

F1 consists of 6% phosphoric acid and F2 consists of 10% phosphoric acid.

However, the farmer requires at least 14 kg of phosphoric acid

So, 6% of x + 10 % of y ≥ 14

6x / 100 + 10y / 100 ≥ 14

3x + 5y ≥ 700

Total cost of fertilizers, Z = 6x + 5y

The mathematical formulation of the given problem can be written as

Minimize Z = 6x + 5y ………….. (i)

Subject to the constraints,

2x + y ≥ 280 ……… (ii)

3x + 5y ≥ 700 ………. (iii)

x, y ≥ 0 …………. (iv)

The feasible region determined by the system of constraints is given below Here, we can see that the feasible region is unbounded.

A (700 / 3, 0), B (100, 80) and C (0, 280) are the corner points

The values of Z at these points are given below

 Corner point Z = 6x + 5y A (700 / 3, 0) 1400 B (100, 80) 1000 Minimum C (0, 280) 1400

Here, the feasible region is unbounded, hence, 1000 may or may not be the minimum value of Z.

For this purpose, we draw a graph of the inequality, 6x + 5y < 1000, and check whether the resulting half plane has points in common with the feasible region or not.

Here, it can be seen that the feasible region has no common point with 6x + 5y < 1000

Hence, 100 kg of fertilizer F1 and 80 kg of fertilizer F2 should be used to minimize the cost. The minimum cost is Rs 1000

11. The corner points of the feasible region determined by the following system of linear inequalities:

2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is

(A) p = q

(B) p = 2q

(C) p = 3q

(D) q = 3p

Solution:

The maximum value of Z is unique

Here, it is given that the maximum value of Z occurs at two points, (3, 4) and (0, 5)

Value of Z at (3, 4) = Value of Z at (0, 5)

p (3) + q (4) = p (0) + q (5)

3p + 4q = 5q

3p = 5q – 4q

3p = q or q = 3p

Therefore, the correct answer is option (D)