NCERT Solutions class 12 Maths Exercise 12.2 (Ex 12.2) Chapter 12 Linear Programming

NCERT Solutions for Class 12 Maths Exercise 12.2 Chapter 12 Linear Programming – FREE PDF Download

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NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming (Ex 12.2) 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.

 

Ans. Let food P consists of  kg and food Q consists of  kg. 

Minimum Z = 

According to question   and 

 and    0

ABC
04
25

Consider 

Let  

But (0, 0) does not satisfy this inequation, therefore the required half-plane does not contain (0, 0).

Again consider 

Let   

DEF
013
5.53-2

Again (0, 0) does not satisfy this inequation, therefore the required half-plane does not contain (0, 0).

The double shaded region is our solution set. The corners of this region are D (0, 5.5), Q (2, 1.5) and P

Now Z = 

At D(0,5.5)

Z = 60 x 0 + 80 x 5.5 = 440

At Q (2, 1.5)

Z = 60 x 2 + 80 x 1.5 = 160

AtP

Z = 60 x   80 x 0 = 160

Hence, minimum Z = 160 at  means minimum cost Z = Rs. 160 when Reshma mixes food P = 2 kg and food Q = 1.5 kg.

2. One kind of cake requires 200g of flour and 25 g of fat and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cake 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.

 

 

Ans. Let number of cakes made of first kind are  and that of second kind is  

Let to maximize Z = 

According to question   and 

Consider 

Let 

 

Here, (0, 0) satisfies this inequation, therefore the required half plane contains (0, 0).

Again consider 

Let 

 

 

Here, again (0, 0) satisfies this inequation, therefore the required half plane contains (0, 0).

The double shaded region is the feasible region which is solution set.

The corner points of this region are O (0, 0), A (25, 0), B (20, 10) and C (0, 20).

Z = 

At O (0, 0) Z = 0 + 0 = 0

At A (25, 0) Z = 25 + 0 = 25

At B (20, 10) Z = 20 + 10 = 30

AtC (0, 20) Z = 0 + 20 = 20

Hence, maximum number of cakes Z = 30 when 

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

 

(i) 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 maximum profit of the factory when it works at full capacity.

 

Ans. Let number of rackets =  and number of bats =  

ABC
28200
0414

(i) To maximize Z = 

According to question    and 

Consider 

Let 

 

Here (0, 0) satisfies this inequaiton, therefore the required half-plane contains (0, 0).

Again consider

Let 

 

Here (0, 0) satisfies this inequaiton, therefore the required half-plane contains (0, 0).

Now the feasible region is the double-shaded region i.e., the solution set. The corner points of this region are O (0, 0), P (8, 0), Q (4, 12) and R (0, 14).

Now Z = 

At O (0, 0) Z = 0 + 0 = 0

At P (8, 0) Z = 8 + 0 = 0

At Q (4, 12) Z = 4 + 12 = 16

At R (0, 14) Z = 0 + 14 = 14

Now maximum Z = 16 at 

(ii) The profit on a racket = Rs. 20 and the profit on a bat = Rs. 10

Let maximum profit P = 

Hence maximum profit P = 20 x 4 + 10 x 12 = 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?

 

 

Ans. Let number of package of nuts =  and number of packages of bolts =  

To maximum profit Z = 

According to question  and  

Consider 

Let  

Therefore, points are A (12, 0), B (0, 4). Also, (0, 0) satisfies this inequation, therefore the required half-plane contains (0, 0).

Again consider 

Let 

 

Therefore, points are C (4, 0), D (0, 12).

Again (0, 0) satisfies this inequation, therefore the required half-plane contains (0, 0).

The double shaded region is OBPCO and its corners are O (0, 0), B (0, 4), D (3, 3), C (4, 0).

Now Z = 

At O (0, 0) Z = 17.5 x 0 + 7 x 0 = 0

At B (0, 4) Z = 17.5 x 0 + 7 x 4 = 28

At P (3, 3) Z = 17.5 x 3 + 7 x 3 = 73.50

At C (4, 0) Z = 17.5 x 4 + 7 x 0 = 70

 Maximum profit = Rs. 73.50 at 

Hence maximum profit Z = Rs. 73.50 when he produces number of packets of nuts = 3 and number of packets of bolts = 3.

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 screw A, while it takes 6 minutes on automatic and 3 minutes on 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 a day in order to maximize his profit? Determine the maximum profit.

 

 

Ans. Let the manufacturers produces  packages of screws A and  packages of screws B, then time taken by  packages of screws A and  packages of screws B on automatic machine =  minutes. And time taken by  packages of screws A and  packages of screws B on hand operated machine =  minutes. 

Since, each machine is available for at the most 4 hours, i.e., 4 x 60 = 240 minutes.

Therefore, we have    and   

Profit on selling  packages of screws A and  packages of screws B = Z = 

 To find:  and  such that Z =  is maximum subject to ,   and 

Consider 

Let 

   A (60, 0) and B (0, 40)

Here, (0, 0) satisfies this inequation, therefore the required hal-plane contains (0, 0).

Again 

Let 

 

   C (40, 0) and D (0, 80)

Here also (0, 0) satisfies this inequation, therefore the required hal-plane contains (0, 0).

The feasible portion of the graph satisfying the inequalities  and  is OABC which is shown shaded in the figure.

Co-ordinates of O, A, B and C are (0, 0), (0, 40), (30, 20) and (40, 0) respectively.

Now Z = 

At O (0, 0) Z = 7 x 0 + 10 x 0 = 0

At A (0, 40) Z = 7 x 0 + 10 x 40 = 400

At B (30, 20) Z = 7 x 30 + 10 x 20 = 410

At C (40, 0) Z = 7 x 40 + 10 x 0 = 280

 Maximum profit Z = Rs. 410 at 

Hence, if the manufacturer produces 30 screws of type A and 20 screws of type B, he earn a 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 shades that he produces, how should he schedule his daily production in order to maximize his profit?

 

 

Ans. Let the manufacturer produces  pedestal lamps and  wooden shades, then the time taken by  pedestal lamps and  wooden shades on grinding/cutting machines =  hours and the time taken by  pedestal lamps and  wooden shades on the sprayer =  hours. 

Since grinding/cutting machine is available for at the most 12 hours, i.e.,  and sprayer is available for at the most 20 hours, i.e., 

Profit from the sale of  lamps and  shades Z = 

 To find:  and  such that Z =  is maximum subject to constraints , , 

Consider 

Let 

ABC
024
1074

Now the area represented by  is the half-plane containing (0, 0) as (0, 0) satisfies the inequaiton.

Consider 

Let 

 

DEC
024
1284

The inequation consists of the half-plane containing (0, ) as (0, 0) satisfies this inequation.

The double shaded region OPCAO is our solution where O (0, 0), P (6, 0), C (4, 4), A (0, 10).

Now Z = 

At O (0, 0) Z = 5 x 0 + 3 x 0 = 0

At P (6, 0) Z = 5 x 6 + 3 x 0 = 30

At C (4, 4) Z = 5 x 4 + 3 x 4 = 32

At A (0, 10) Z = 5 x 0 + 3 x 10 = 30

Now maximum Z = 32 at 

Hence, maximum profit Z = Rs. 32 when he manufactures 4 pedestal lamps and 4 wooden shades.

7. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A requires 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?

 

 

Ans. Let the company manufactures  souvenirs of type A and  souvenirs of type B, then time taken for cutting  souvenirs of type A and  souvenirs of type B =  minutes and time taken for assembling  souvenirs of type A and  souvenirs of type B =  minutes. 

Since 3 hours 29 minutes i.e., 200 minutes are available for cutting, therefore we have 

Also since 4 hours i.e., 240 minutes are available for assembling, therefore we have   

Thus, our L.P.P. is to maximize profit Z =  subject to constraints , , 

Now 

Let  

400
025

Again 

Let 

 

24200
0530

In both the equations origin (0, 0) satisfies them and therefore the required half planes are there which contains (0, 0).

The portion of graph satisfying the inequalities  and  is OABC and is shown in shaded in the figure. Coordinates of the points O, A, B and C are (0, 0), (24, 0), (8, 20) and (0, 25) respectively.

Now Z = 

At O (0, 0) Z = 5 x 0 + 6 x 0 = 0

At A (24, 0) Z = 5 x 24 + 6 x 0 = 120

At B (8, 20) Z = 5 x 8 + 6 x 20 = 160

At C (0, 25) Z = 5 x 0 + 6 x 25 = 150

Now maximum Z = 160 at 

Hence, maximum profit = Rs. 160 when he manufactures 8 souvenirs of type A and 20 souvenirs of type B.

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.

 

 

Ans. Let number of desktop model computer =  and number of a portable model computer =  

To maximum profit Z =  subject to ,  

Consider  

Let 

 

 Points are A (280, 0), B (0, 175)

Now (0, 0) satisfies the inequation, therefore the required half-plane contains (0, 0).

Again consider 

Let 

 

 Points are C (250, 0), D (0, 250).

Again (0, 0) satisfies the inequation, therefore the required half-plane contains (0, 0).

The double shaded region OCPBO is the solution set. Its corners are O (0, 0), C (280, 0), P (200, 50), B (0, 200).

Now Z = 

AtO (0, 0) Z = 4500 x 0 + 5000 x 0 = 0

At C (280, 0) Z = 4500 x 280 + 5000 x 0 = 12,60,000

(Rejected as )

At P (200, 50) Z = 4500 x 200 + 5000 x 50 = 11,50,000

At B (0, 200) Z = 4500 x 0 + 5000 x 200 = 10,00,000

Now maximum Z = Rs. 11,50,000 at 

Hence, maximum profit = Rs. 11,50,000 when he sells 200 desktop models and 50 portable models.

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.

 

 

Ans. Let the number of units of food F1 =  and number of units of food F2 =  

To minimize Z =  subject to ,  , 

Consider 

Let   

ABC
0510

Clearly, (0, 0) is not included in the half-plane.

Now again consider 

Let    

DEF
252210
0420

Here, also (0, 0) is not included in the required half-plane.

The double shaded region is our feasible region and its corners are

A P and Q

Now Z = 

At A Z = 4 x  + 6 x 0 = 106.67

At P Z = 4 x 24 + 6 x  = 104

At Q Z = 4 x 0 + 6 x  = 200

Hence Z = Rs. 104 is minimum at .

Therefore, minimum cost is Rs.104 when 24 units of food F1 and mixed with 4/3 units of food F2.

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 at least 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?

 

 

Ans. Let the quantity of F1 =  kg and the quantity of F2 =  kg 

We have to minimize Z =   subject to , 

Consider 

Let  

 

 

ABCD
050100140
280180800

Here, (0, 0) is not included in the required half plane and (0, 0) does not satisfy this inequation.

Again consider 

Let    

     

ECG
0100200
1408020

Again Here, (0, 0) is not included in the required half plane and (0, 0) does not satisfy this inequation.

The shaded region XPCAY is our feasible region. Its corners are P, C (100, 80) and A (0, 280).

Now Z = 

At P  Z = 6 x  + 5 x 0 = 1400

At C (100, 80) Z = 6 x 100 + 5 x 80 = 1000

At A (0, 280) Z = 6 x 0 + 5 x 280 = 1400

Now minimum cost Z = Rs. 1000 at 

Therefore, Minimum cost is Rs.1000 when the farmer used 100 kg of fertilizer F1 and 80 kg of fertilizer F2.

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

 

 are (0, 0), (5, 0), (3, 4) and (0, 5). Let  where  Condition on  and  so that the maximum of Z occurs at both (3, 4) and (0, 5) is:

(A)  

(B)  

(C)  

(D)  

 

Ans. To maximize Z =  subject to ,  

The corner points are (0, 0), (5, 0), (3, 4) and (0, 5).

Maximum of Z occurs at both (3, 4) and (0, 5)

Z = 

At (3, 4) Z = 

At (0, 5) Z = 

Now maxima occurs at both points,

 

 

Hence option (D) is correct.