# NCERT Solutions for Class 12 Maths Linear Programming

Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get the * NCERT Solutions for Class 12 Maths Linear Programming Exercise 12.2*.

As we know that all the schools affiliated from CBSE follow the NCERT books for all subjects. You can check the **CBSE NCERT Syllabus**. Thus, * NCERT Solutions* helps the students to solve the exercise questions as given in

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**NCERT Solutions for Class 12 Maths Linear Programming Exercise 12.2**## NCERT Solutions for Class 12 Maths Linear Programming

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 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.

**NCERT Solutions**

Let Reshma mix x kg of food P and y kg of food Q. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the mixture must contain atleast 8 units of vitamin A and 11 units of vitamin B.

Thus, we have constraints as 3x + 4y >= 8 and 5x + 2y >= 11.

Now we have total cost of purchasing x kg of food P and y kg of food Q is Z = 60x + 80y.

Then, we need to minimize Z = 60x + 80y by the constraints as 3x + 4y >= 8 and 5x + 2y >= 11.

Now, let the line l_{1} : 3x + 4y = 8 and l_{2} : 5x + 2y = 11 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is unbounded, we need to use Corner Point Method to minimize Z = 60x + 80y.

Now, on solving the equations of lines, we get that the coordinates A(8/3, 0), E(2, 1/2), D(0, 11/2).

Thus, minimum value of Z is 160 at the point (8/3, 0) and (2, 1/2).

Hence, the required minimum cost of mixture is Rs 160.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 2. **

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 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.

**NCERT Solutions**

Let the number of first kind of cake is x and the number of second kind of cake is y. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the maximum number of cakes can be made from 5 kg of flour ans 1 kg of fat.

Thus, we have constraints as 200x + 100y <= 5000 or 2x + y <= 50 and 25x + 50y <= 1000 or x + 2y <= 40.

Now we have total number of cakes Z = x + y.

Then, we need to maximize Z = x + y by the constraints as 2x + y <= 50 and x + 2y <= 40.

Now, let the line l_{1} : 2x + y = 50 and l_{2} : x + 2y = 40 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = x + y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), A(25, 0), E(20, 10) and B(0, 20).

Thus, maximum value of Z is 30 at the point (20, 10).

Hence, the required maximum value of Z = x + y is 30 when 20 cakes of one kind and 10 cakes of another kind is made of ingredients.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 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 hour 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 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.

**NCERT Solutions**

Let the number of tennis rackets is x and number of cricket bat is y. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the availability in the factory is not more than 42 hours of machine time and 24 craftsman time.

Thus, we have constraints as 1.5x + 3y <= 42 or x + 2y <= 28 and 3x + y <= 24.

(i) Total number of bats and rackets made in a factory is Z = x + y.

Then, we need to maximize Z = x + y by the constraints as x + 2y <= 28 and 3x + y <= 24.

Now, let the line l_{1} : x + 2y = 28 and l_{2} : 3x + y = 24 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = x + y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), C(8, 0), E(4, 12) and B(0, 14).

Thus, maximum value of Z is 16 at the point (4, 12).

Hence, the required maximum value of Z = x + y is 16 when 4 tennis rackets and 12 cricket bats must be made so that the factory works at full capacity.

(ii) Now the profit function is Z = 20x + 10y

Since the maximum value of Z occurred at the point (4, 12),

then Z_{max} = 20(4) + 10(12) = 80 + 120 = Rs. 200 when 4 tennis rackets and 12 cricket bats are made.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 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 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?

**NCERT Solutions**

Let the number of packets of nuts of cake is x and the number of packets of bolts is y. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the machine is operated for atmost 12 hours a day.

Thus, we have constraints as x + 3y <= 12 and 3x + y <= 12.

Now, Total profit Z earned on package of nuts and bolts is Z = 17.5x + 7y.

Then, we need to maximize Z = 17.5x + 7y by the constraints as x + 3y <= 12 and 3x + y <= 12.

Now, let the line l_{1} : x + 3y = 12 and l_{2} : 3x + y = 12 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = 17.5x + 7y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), C(4, 0), E(3, 3) and B(0, 4).

Thus, maximum value of Z is 73.5 at the point (3, 3).

Hence, the required maximum value of Z = 17.5x + 7y is 73.5 when the factory produces 3 packages of nuts and 3 packages of bolts.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 5. **

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.

**NCERT Solutions**

Let the number of packages of screws A is x and the number of packages of screws B is y. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the machine is available for the at most 4 hours a day.

Thus, we have constraints as

4x + 6y <= 240 or 2x + 3y <= 120

and 6x + 3y <= 240 or 2x + y <= 80.

Now, Total profit Z earned on packages of screws A and B is Z = 7x + 10y.

Then, we need to maximize Z = 7x + 10y by the constraints as 2x + 3y <= 120 and 2x + y <= 80.

Now, let the line l_{1} : 2x + 3y = 120 and l_{2} : 2x + y = 80 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = 7x + 10y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), C(40, 0), E(30, 20), B(0, 40).

Thus, maximum value of Z is 410 at the point (30, 20).

Hence, the required maximum value of profit Z = 7x + 10y is Rs 410 when the factory produces 30 packages of screws A and 20 packages of screws B.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 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 maximise his profit?.

**NCERT Solutions**

Let the cottage industry manufacturer x pedestal lamp and y wooden shades. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the cutting machine is available for atmost 12 hours and sprayer is available for atmost 20 hours.

Thus, we have constraints as 2x + y <= 12 and 3x + 2y <= 20.

Now, Total profit Z earned is Z = 5x + 3y.

Then, we need to maximize Z = 5x + 3y by the constraints as 2x + y <= 12 and 3x + 2y <= 20.

Now, let the line l_{1} : 2x + y = 12 and l_{2} : 3x + 2y = 20 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = 5x + 3y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), A(3, 0), E(4, 4), C(0, 10).

Thus, maximum value of Z is 32 at the point (4, 4).

Hence, the required maximum value of profit Z = 5x + 3y is Rs 32 when the manufacturer sell 4 lamps and 4 shades.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 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?

**NCERT Solutions**

Let the company manufacture x souvenirs of type A and y souvenirs of type B. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the cutting machine is available for 200 minutes and for assembling is 240 minutes.

Thus, we have constraints as 5x + 8y <=200 and 10x + 8y <= 240 or 5x + 4y <= 120.

Now, Total profit Z earned is Z = 5x + 6y.

Then, we need to maximize Z = 5x + 6y by the constraints as 5x + 8y <=200 and 5x + 4y <= 120.

Now, let the line l_{1} : 5x + 8y = 200 and l_{2} : 5x + 4y = 120 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = 5x + 6y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), C(24, 0), E(8, 20), B(0, 25).

Thus, maximum value of Z is 160 at the point (8, 20).

Hence, the required maximum value of profit Z = 5x + 6y is Rs 160 when the company manufacture 8 souvenirs of type A and 20 souvenirs of type B.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 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.

**NCERT Solutions**

Let the the merchant stock x desktop computers and y portable computers. Then, x >= 0 and y >= 0.

According to question,

x + y <=250 and 25000x + 40000y <=7000000 or 5x + 8y <= 1400.

Now, Total profit Z earned is Z = 4500x + 5000y.

Then, we need to maximize Z = 4500x + 5000y by the constraints as

x + y <=250, 5x + 8y <= 1400 and x >= 0 and y >= 0.

Now, let the line l_{1} : x + y = 250 and l_{2} : 5x + 8y = 1400 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = 4500x + 5000y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), A(250, 0), E(200, 50), D(0, 175).

Thus, maximum value of Z is 1150000 at the point (200, 50).

Hence, the required maximum value of profit Z = 4500x + 5000y is Rs 1150000 when the company stocks 200 units of desktop model and 50 units of portable model.