NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 12 Linear Programming

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

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

1. Linear Programming Problem and its Mathematical Formulation
2. Different Types of Linear Programming 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 Miscellaneous Exercise

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Exercise 12.1 Solutions 10 Questions

Exercise 12.2 Solutions 11 Questions

Access Answers to NCERT Class 12 Maths Chapter 12 Miscellaneous Exercise

1. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?

Solution:

Let the diet contain x and y packets of foods P and Q respectively. Hence,

x ≥ 0 and y ≥ 0

The mathematical formulation of the given problem is given below

Maximize z = 6x + 3y ………….. (i)

Subject to the constraints,

4x + y ≥ 80 …………. (ii)

x + 5y ≥ 115 ………. (iii)

3x + 2y ≤ 150 ………… (iv)

x, y ≥ 0 …………… (v)

The feasible region determined by the system of constraints is given below

A (15, 20), B (40, 15) and C (2, 72) are the corner points of the feasible region

The values of z at these corner points are given below

So, the maximum value of z is 285 at (40, 15)

Hence, to maximize the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used.

The maximum amount of vitamin A in the diet is 285 units.

2. A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?

Solution:

Let the farmer mix x bags of brand P and y bags of brand Q respectively

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

The given problem can be formulated as given below

Minimize z = 250x + 200y ………… (i)

3x + 1.5y ≥ 18 ………….. (ii)

2.5x + 11.25y ≥ 45 ……….. (iii)

2x + 3y ≥ 24 ………….. (iv)

x, y ≥ 0 ………. (v)

The feasible region determined by the system of constraints is given below

A (18, 0), B (9, 2), C (3, 6) and D (0, 12) are the corner points of the feasible region

The values of z at these corner points is given below

Here, the feasible region is unbounded, hence, 1950 may or may not be the minimum value of z

For this purpose, we draw a graph of the inequality, 250x + 200y < 1950 or 5x + 4y < 39, 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 5x + 4y < 39

Hence, at point (3, 6) the minimum value of z is 1950

Therefore, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimize the cost to Rs 1950

3. A dietician 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 one kg food is given below:

One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?

Solution:

Let the mixture contain x kg of food X and y kg of food Y respectively

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

Minimize z = 16x + 20y …….. (i)

Subject to the constraints,

x + 2y ≥ 10 …………. (ii)

x + y ≥ 6 ………… (iii)

3x + y ≥ 8 …………. (iv)

x, y ≥ 0 ………… (v)

The feasible region determined by the system of constraints is given below

A (10, 0), B (2, 4), C (1, 5) and D (0, 8) are the corner points of the feasible region

The values of z at these corner points are given below

Since the feasible region is unbounded, hence, 112 may or may not be the minimum value of z

For this purpose, we draw a graph of the inequality, 16x + 20y < 112 or 4x + 5y < 28, 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 4x + 5y < 28

Hence, the minimum value of z is 112 at (2, 4)

Therefore, the mixture should contain 2 kg of food X and 4 kg of food Y.

The minimum cost of the mixture is Rs 112.

4. A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is give below:

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

Solution:

Let x and y toys of type A and type B be manufactured in a day respectively.

The given problem can be formulated as given below

Maximize z = 7.5x + 5y ………….. (i)

Subject to the constraints,

2x + y ≤ 60 …………. (ii)

x ≤ 20 ……….. (iii)

2x + 3y ≤ 120 ……….. (iv)

x, y ≥ 0 ……………. (v)

The feasible region determined by the constraints is given below

A (20, 0), B (20, 20), C (15, 30) and D (0, 40) are the corner points of the feasible region.

The values of z at these corner points are given below

262.5 at (15, 30) is the maximum value of z

Hence, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximize the profit.

5. An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?

Solution:

Let the airline sell x tickets of executive class and y tickets of economy class respectively.

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

Maximize z = 1000x + 600y ………… (i)

Subject to the constraints,

x + y ≤ 200 …………….. (ii)

x ≥ 20 ………… (iii)

y – 4x ≥ 0 …………… (iv)

x, y ≥ 0 ……………(v)

The feasible region determined by the constraints is given below

A (20, 80), B (40, 160) and C (20, 180) are the corner points of the feasible region

The values of z at these corner points are given below

136000 at (40, 160) is the maximum value of z

Therefore, 40 tickets of executive class and 160 tickets of economy class should be sold to maximize the profit and the maximum profit is Rs 136000.

6. Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost 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? What is the minimum cost?

Solution:

Let godown A supply x and y quintals of grain to the shops D and E

So, (100 – x – y) will be supplied to shop F.

Since, x quintals are transported from godown A, so the requirement at shop D is 60 quintals. Hence, the remaining (60 – x) quintals will be transported from godown B.

Similarly, (50 – y) quintals and 40 – (100 – x – y) = (x + y – 60) quintals will be transported from godown B to shop E and F

The given problem can be represented diagrammatically as given below

x ≥ 0, y ≥ 0, and 100 – x – y ≥ 0

Then, x ≥ 0, y ≥ 0, and x + y ≤ 100

60 – x ≥ 0, 50 – y ≥ 0, and x + y – 60 ≥ 0

Then, x ≤ 60, y ≤ 50, and x + y ≥ 60

Total transportation cost z is given by,

z = 6x + 3y + 2.5 (100 – x – y) + 4 (60 – x) + 2 (50 – y) + 3 (x + y – 60)

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

= 2.5x + 1.5y + 410

The given problem can be formulated as given below

Minimize z = 2.5x + 1.5y + 410 …………. (i)

Subject to the constraints,

x + y ≤ 100 ……….. (ii)

x ≤ 60 ……….. (iii)

y ≤ 50 ………. (iv)

x + y ≥ 60 ……… (v)

x, y ≥ 0 …………..(vi)

The feasible region determined by the system of constraints is given below

A (60, 0), B (60, 40), C (50, 50) and D (10, 50) are the corner points

The values of z at these corner points are given below

The minimum value of z is 510 at (10, 50)

Hence, the amount of grain transported from A to D, E and F is 10 quintals, 50 quintals and 40 quintals respectively and from B to D, E and F is 50 quintals, 0 quintals, 0 quintals respectively

Thus, the minimum cost is Rs 510

7. 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 petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:

Assuming that the transportation cost of 10 litres of oil is Rs 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?

Solution:

Let x and y litres of oil be supplied from A to the petrol pumps, D and E. So, (7000 – x – y) will be supplied from A to petrol pump F.

The requirement at petrol pump D is 4500L. Since x L are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.

Similarly, (3000 – y)L and 3500 – (7000 – x – y) = (x + y – 3500) L will be transported from depot B to petrol pump E and F respectively.

The given problem can be represented diagrammatically as given below

x ≥ 0, y ≥ 0, and (7000 – x – y) ≥ 0

Then, x ≥ 0, y ≥ 0, and x + y ≤ 7000

4500 – x ≥ 0, 3000 – y ≥ 0, and x + y – 3500 ≥ 0

Then, x ≤ 4500, y ≤ 3000, and x + y ≥ 3500

Cost of transporting 10L of petrol = Rs 1

Cost of transporting 1L of petrol = Rs 1 / 10

Hence, total transportation cost is given by,

z = (7 / 10) x + (6 / 10) y + 3 / 10 (7000 – x – y) + 3 / 10 (4500 – x) + 4 / 10 (3000 – y) + 2 / 10 (x + y – 3500)

= 0.3x + 0.1y + 3950

The problem can be formulated as given below

Minimize z = 0.3x + 0.1y + 3950 ………. (i)

Subject to constraints,

x + y ≤ 7000 ………. (ii)

x ≤ 4500 ……….. (iii)

y ≤ 3000 ……. (iv)

x + y ≥ 3500 ……… (v)

x, y ≥ 0 ………… (vi)

The feasible region determined by the constraints is given below

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

The values of z at these corner points are given below

The minimum value of z is 4400 at (500, 3000)

Hence, the oil supplied from depot A is 500 L, 3000 L and 3500 L and from depot B is 4000 L, 0 L and 0 L to petrol pumps D, E and F respectively.

Therefore, the minimum transportation cost is Rs 4400.

8. A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table, Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.

If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

Solution:

Let the fruit grower use x bags of brand P and y bags of brand Q respectively

The problem can be formulated as given below

Minimize z = 3x + 3.5y …………. (i)

Subject to the constraints,

x + 2y ≥ 240 ……….. (ii)

x + 0.5y ≥ 90 …….. (iii)

1.5x + 2y ≤ 310 ……….. (iv)

x, y ≥ 0 …………. (v)

The feasible region determined by the system of constraints is given below

A (240, 0), B (140, 50) and C (20, 140) are the corner points of the feasible region

The value of z at these corner points are given below

The maximum value of z is 470 at (40, 100)

Therefore, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimize the amount of nitrogen

Hence, the minimum amount of nitrogen added to the garden is 470 kg.

9. Refer to Question 8. If the grower wants to maximize the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?

Solution:

Let the fruit grower use x bags of brand P and y bags of brand Q respectively

The problem can be formulated as given below

Maximize z = 3x + 3.5y ……….. (i)

Subject to the constraints,

x + 2y ≥ 240 ……….. (ii)

x + 0.5y ≥ 90 ……….. (iii)

1.5x + 2y ≤ 310 ……….. (iv)

x, y ≥ 0 …………. (v)

The feasible region determined by the system of constraints is given below

A (140, 50), B (20, 140) and C (40, 100) are the corner points of the feasible region

The values of z at these corner points are given below

The maximum value of z is 595 at (140, 50)

Hence, 140 bags of brand P and 50 bags of brand Q should be used to maximize the amount of nitrogen.

Thus, the maximum amount of nitrogen added to the garden is 595 kg.

10. A toy company manufactures two types of dolls, A and B. Market research and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximize the profit?

Solution:

Let x and y be the number of dolls of type A and B respectively that are produced in a week.

The given problem can be formulated as given below

Maximize z = 12x + 16y ……….. (i)

Subject to the constraints,

x + y ≤ 1200 ………… (ii)

y ≤ x / 2 or x ≥ 2y ………. (iii)

x – 3y ≤ 600 …………. (iv)

x, y ≥ 0 …………… (v)

The feasible region determined by the system of constraints is given below

A (600, 0), B (1050, 150) and C (800, 400) are the corner points of the feasible region

The values of z at these corner points are given below

The maximum value of z is 16000 at (800, 400)

Hence, 800 and 400 dolls of type A and type B should be produced respectively to get the maximum profit of Rs 16000.