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NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2

NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-12 (Linear Programming)Exercise 12.2 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2

Exercise 12.2

Q1. 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 ot vitamin A and 11 units of vitamin 3. 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 3 while food Q contains 4 units /kg of vitamin A and 2 units Jkg of vitamin 3. Determine the minimum cost of the mixture?

Answer. Let the mixture contain x kg of food p and y kg of food Q. Therefore, x0 and y0 The given information can be compiled in a table as follows. NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2 The mixture must contain at least g units of vitamin A and 11 units of vitamin Therefore, the constraints are 3x+4y85x+2y11 Total cost, Z , of purchasing food is, z=60x+80y The mathematical formulation of the given problem is  Minimise Z=60x+80y(1) subject to the constraints, 3x+4y8(2)5x+2y11(3)x,y0(4) The feasible region determined by the system of constraints is as follows. NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2 It can be seen that the feasible region is unbounded. A(83,0),B(2,12), and C(0,112) The values of z at these corner points are as follows. NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2  As the feasible region is unbounded, therefore, 160 may or may not be the minimum  value of z .   For this, 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.  It can be seen that the feasible region has no common point with 3x+4y<8 Therefore, the minimum cost of the mixture will be Rs 160 at the line segment joining  

Q2. One kind of cake requires 200g of 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.

Answer.  Let there be x cakes of first kind and y cakes of second kind. Therefore, x0 and y0 The given information can be complied in a table as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2 200x+100y50002x+y5025x+50y1000x+2y40  Total numbers of cakes, z , that can be made are, z=x+y The mathematical formulation of the given problem is  Maximize Z=x+y(1) subject to the constraints, 2x+y50x+2y40x,y0 NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2

Q3. 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. (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 the maximum profit of the factory when it works at full capacity.

Answer.  (i) Let the number of rackets and the number of bats to be made be x and y respectively.  The machine time is not available for more than 42 hours. 1.5x+3y42 The craftsman's time is not available for more than 24 hours. 3x+y24 The factory is to work at full capacity. Therefore, 1.5x+y243x+y=24  On solving these equations, we obtain x=4 and y=12 Thus, 4 rackets and 12 bats must be made.  (1) The given information can be complied in a table as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2 1.5x+3y423x+y24x,y0  The profit on a racket is Rs 20 and on a bat is Rs 10 . Z=20x+10y The mathematical formulation of the given problem is  Maximize Z=20x+10y (1)  subject to the constraints, 1.5x+3y42 (2)  3x+y24(3)x,y0(4) The feasible region determined by the system of constraints is as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2

Q4. 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 Rs17.50 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 maximise his profit, if he operates his machines for at the most 12 hours a day?

Answer.  Let the manufacturer produce x packages of nuts and y packages of bolts. Therefore, x0 and y0 The given information can be compiled in a table as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2  The profit on a package of nuts is Rs 17.50 and on a package of bolts is Rs 7. Therefore,  the constraints are x+3y123x+y12 Total profit, z=17.5x+7y The mathematical formulation of the given problem is   subject to the constraints, x+3y12(2)3x+y12(3)x,y0(4) The feasible region determined by the system of constraints is as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2

Q5. A factory manufactures 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 maximise his profit? Determine the maximum profit.

Answer.  Let the factory manufacture x screws of type A and y screws of type B on each day.  Therefore, x0 and y0 The given information can be compiled in a table as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2  The profit on a package of screws A is Rs 7 and on the package of screws B is Rs 10 .  Therefore, the constraints are 4x+6y2406x+3y240 Total profit, Z=7x+10y  Maximize z=7x+10y(1) subject to the constraints, 4x+6y2406x+3y2406x+3y240(3)x,y0(4) The feasible region determined by the system of constraints is  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2 The maximum value of z is 410 at (30, 20). Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410.

Q6. 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 maximise his profit?

Answer.  Let the cottage industry manufacture x pedestal lamps and y wooden shades. Therefore, x0 and y0 The given information can be compelled in a table as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2  The profit on a lamp is Rs 5 and on the shades is Rs 3. Therefore, the constraints are 2x+y123x+2y20 Total profit, z=5x+3y The mathematical formulation of the given problem is  Maximize z=5x+3y(1)  Maximize z=5x+3y(1) subject to the constraints, 2x+y12 (2) 3x+2y20x,y0(4) NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2

Q7. 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 maximise the profit?

Answer.  Let the company manufacture x souvenirs of type A and y souvenirs of type B Therefore, x0 and y0 The given information can be complied in a table as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2  The profit on type A souvenirs is Rs 5 and on type B souvenirs is Rs 6. Therefore, the  constraints are 5x+8y20010x+8y240 Total profit, Z=5x+6y  The mathematical formulation of the given problem is  Maximize Z=5x+6y(1) subject to the constraints, 5x+8y2005x+4y120x,y0(4) NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2 Thus, e souvenirs Of type A and 20 souvenirs Of type B should be produced each day to get the maximum profit of Rs 160.

Q8. 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.

Answer.  Let the merchant stock x desktop models and y portable models. Therefore, x0 and y0 The cost of a desktop model is Rs 25000 and of a portable model is Rs 4000. However,  the merchant can invest a maximum of Rs 70 lakhs. 25000x+40000y70000005x+8y1400x+y250  The monthly demand of computers will not exceed 250 units. x+y250 The profit on a desktop model is Rs 4500 and the profit on a portable model is Rs 5000.  Total profit, z=4500x+5000y Thus, the mathematical formulation of the given problem is  Maximum Z=4500x+5000y  Thus, the mathematical formulation of the given problem is  Maximum Z=4500x+5000y subject to the constraints, 5x+8y1400x+y250x+y250x,y0 The feasible region determined by the system of constraints is as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2 The maximum value of Z Is 1150000 at (200, SO). Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs 1150000.

Q9. 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.

Answer.  Let the diet contain x units of food F1 and y units of food F2 . Therefore, x0 and y0 The given information can be complied in a table as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2  The cost of food F1 is Rs 4 per unit and of Food F2 is Rs 6 per unit. Therefore, the  constraints are 3x+6y804x+3y100 x, y0 Total cost of the diet, z=4x+6y  The mathematical formulation of the given problem is  Minimise z=4x+6y(1) subject to the constraints, 3x+6y80(2)4x+3y100(3)x,y0(4) The feasible region determined by the constraints is as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2 A(83,0),B(2,12), and C(0,112) NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2 

Q10. There are two types of fertilisers 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 fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Answer.  Let the farmer buy x kg of fertilizer F1 and y kg of fertilizer F2. Therefore, x0 and y0 The given information can be complied in a table as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2 F1 consists of 10% nitrogen and F2 consists of 5% nitrogen. However, the farmer  requires at least 14 kg of nitrogen. 10% of x+5% of y14 x10+y20142x+y280F1 consists of 6% phosphoric acid and F2 consists of 10% phosphoric acid. However, the  farmer requires at least 14 kg of phosphoric acid.  6x100+10y100143x+56y700 Total cost of fertilizers, Z=6x+5y The mathematical formulation of the given problem is  Minimize z=6x+5y(1) subject to the constraints,   Total cost of fertilizers, z=6x+5y The mathematical formulation of the given problem is  Minimize Z=6x+5y(1) subject to the constraints, 2x+y280(2)3x+5y700(3)x,y0(4) NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2  It can be seen that the feasible region is unbounded.  The corner points are (7003,0),B(100,80), and C(0,280) The values of z at these points are as follows.  NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.2 

Q11. 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

Answer.  The maximum value of Z is unique.  It is given that the maximum value of z occurs at two points, (3,4) and (0,5) .  

Chapter-12 (Linear Programming)

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