The Exercise 12.2 of NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming is based on the following topics:
 Different Types of Linear Programming Problems
 Manufacturing problems
 Diet problems
 Transportation problems
Solve all the problems of this exercise to get thorough with the concepts and topics covered in the entire chapter.
Download PDF of NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.2
Access other exercises of Class 12 Maths Chapter 12
Exercise 12.1 Solutions 10 Questions
Miscellaneous Exercise On Chapter 12 Solutions 10 Questions
Access Answers of Maths NCERT Class 12 Chapter 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)
5x + 2y ≥ 11 … (3)
x, y ≥ 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 F_{1} and F_{2} are available. Food F_{1} costs Rs 4 per unit food and F_{2} costs Rs 6 per unit. One unit of food F_{1} contains 3 units of vitamin A and 4 units of minerals. One unit of food F_{2} 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 F_{1} and y units of food F_{2}. 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 F_{1} (x) 
3 
4 
4 
Food F_{2} (y) 
6 
3 
6 
Requirement 
80 
100 
The cost of food F_{1} is Rs 4 per unit and of food F_{2} 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 F_{1} and F_{2}. F_{1} consists of 10% nitrogen and 6% phosphoric acid and F_{2} 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 F_{1} costs Rs 6 / kg and F_{2} 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 F_{1} and y kg of fertilizer F_{2}. Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table is given below
Nitrogen (%) 
Phosphoric Acid (%) 
Cost (Rs / kg) 

F_{1} (x) 
10 
6 
6 
F_{2} (y) 
5 
10 
5 
Requirement (kg) 
14 
14 
F_{1} consists of 10% nitrogen and F_{2} 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
F_{1} consists of 6% phosphoric acid and F_{2} 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 F_{1} and 80 kg of fertilizer F_{2} 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)