If cot a = 1 and sec b = − 5 3 , where  π < α < 3 π 2 a n d π 2 < β < π , then the value of tan(a + b) and the quadrant in which a + b lies, respectively are:

− 1 7 and IVth quadrant

1 7 and Ist quadrant

If cot a = 1 and sec b = − 5 3 , where  π < α < 3 π 2 a n d π 2 < β < π , then the value of tan(a + b) and the quadrant in which a + b lies, respectively are:

− 1 7 and IVth quadrant

1 7 and Ist quadrant

Class 12 Maths Chapter 12 Linear Programming NCERT Solutions PDF

Q: 16. 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 ?.5 and that from a shade is ?.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?

Let the cottage industry manufacture x pedestal lamps and y wooden shades. Therefore, x ≥ 0 and y ≥ 0 The given information can be compiled in a table as follows. 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. Therefore, the constraints are 2x + y ≤ 12 3x + 2y ≤ 20 Total profit, Z = 5x + 3y The mathematical formulation of the given problem is Maximize Z=5x +3y …(1) subject to the constraints, 2x + y ≤ 12 … (2) 3x + 2y ≤ 20 … (3) x, y ≥ 0 … (4) The feasible region determined by the system of constraints is as follows. The corner points are A (6, 0), B (4, 4), and C (0, 10). The values of Z at these corner points are as follows The maximum value of Z is 32 at (4, 4). Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximize his profits.

Q: 12. 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.

Let there be x cakes of first kind and y cakes of second kind. Therefore, x ≥ 0 and y ≥ 0 The given information can be complied in a table as follows. Flour (g) Fat (g) Cakes of first kind, x 200 25 Cakes of second kind, y 100 50 Availability 5000 1000 ∴200x+100y≤5000⇒2x+y≤5025x+50y≤1000⇒x+2y≤40 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+y≤50.......(2)x+2y≤40.......(3)x,y≥0..............(4) The feasible region determined by the system of constraints is as follows The corner points are A (25, 0), B (20, 10), O (0, 0), and C (0, 20). The values of Z at these corner points are as follows. Thus, the maximum numbers of cakes that can be made are 30 (20 of one kind and 10 of the other kind).

Q: 28. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L, respectively. The company has 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 are given in the following table: Distance in (km) From/To A B D 7 3 E 6 4 F 3 2 Assuming that the transportation cost of 10 litres of oil is ?. 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost

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 4500 L. Since x L are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B. Similarly ,(3000–y)Land3500–(7000–x–y)=(x+y–3500) L will be transported from depot B to petrol pumps 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,andx+y≤7000 4500–x≥0,3000–y≥0,andx+y–3500≥0 Then,x≤4500,y≤3000,andx+y≥3500 Cost of transporting 10 L of petrol = Rs. 1 Cost of transporting 1 L of petrol = Rs. 1/10 Hence, the 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: Minimisez=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: Corner Point z = 0.3x + 0.1y + 3950 A (3500, 0) 5000 B (4500, 0) 5300 C (4500, 2500) 5550 D (4000, 3000) 5450 E (500, 3000) 4400 Minimum 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 ?. 4400.

Q: 23. A farmer mixes two brands, P and Q of cattle feed. Brand P, costing ?. 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 ?. 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 for 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?

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 as given below: Vitamin A (units/kg) Vitamin B (units/kg) Vitamin C (units/kg) Cost (Rs/kg) Food P 3 2.5 2 250 Food Q 1.5 11.25 3 200 Requirement (units/kg) 18 45 24 The given problem can be formulated as given below: Minimisez=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 are given below: Corner Point z = 250x + 200y A (18, 0) 4500 B (9, 2) 2650 C (3, 6) 1950 Minimum D (0, 12) 2400 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<1950or5x+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 minimise the cost to ?. 1950.

Q: 20. 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 ?. 6/kg and F2 costs ?. 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?

Let the farmer buy x kg of fertilizer F1 and y kg of fertilizer F2. Therefore, x ≥ 0 and y ≥ 0 The given information can be complied in a table as follows. Nitrogen (%) Phosphoric Acid (%) Cost (Rs/kg) F1 (x) 10 6 6 F2 (y) 5 10 5 Requirement (kg) 14 14 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 y ≥14 x10+y20≥142x+y≥280 F1 consists of 6% phosphoric acid and F2 consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid. ∴6%of x +10%of y ≥14 6x100+10y100≥143x+56y≥700 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 + y ≥ 280 … (2) 3x + 5y ≥ 700 … (3) x, y ≥ 0 … (4) The feasible region determined by the system of constraints is as follows. It can be seen that the feasible region is unbounded. The corner points are A(7003,0),B(100,80),C(0,280) . The values of Z at these points are as follows. As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z. For this, 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. It can be seen that the feasible region has no common point with 6x +5y <1000 Therefore, 100 kg of fertiliser F1 and 80 kg of fertilizer F2 should be used to minimize the cost. The minimum cost is ? 1000.

Updated on Aug 7, 2025 15:31 IST

By Pallavi Pathak, Assistant Manager Content

Class 12 Linear Programming covers problems that need to maximize or minimize profit or cost. These are called optimization problems. Such problems include finding the minimum cost, maximum profit, or minimum use of resources, etc. An important class of optimization problems is called the linear programming problem. These problems have wide applications in commerce, industry, and management science, etc.
Linear Programming Class 12 NCERT Solutions include such problems and their solutions by the graphical method only. These solutions are given in a step-by-step format, which is easy to understand. It helps students to understand the concepts properly and prepare well for the examinations, such as the CBSE Board exam and entrance tests like JEE Mains.
For students who are looking for reliable and accurate key topics and free PDFs of each chapter of Class 11 and Class 12 of Physics, Chemistry, and Mathematics, they can explore here - NCERT Solutions Class 11 and 12.

Table of content

An overview of Linear Programming Class 12
Class 12 Linear Programming: Key Topics, Weightage
Important Formulas of Class 12 Linear Programming
Free PDF: Class 12 Linear Programming NCERT Solutions – Chapter 12
Class 12 Linear Programming Exercise-wise Solutions
Class 12 Linear Programming Exercise 12.1 Solutions
Class 12 Maths NCERT Linear Programming Solutions- FAQs

An overview of Linear Programming Class 12

Here are highlights of the Class 12 Linear Programming:

The linear programming helps in finding the optimal value of a linear function of several variables subject to conditions. The conditions are that the variables satisfy a set of linear inequalities.
Variables are non-negative and sometimes called the decision variables.

Class 12 Linear Programming: Key Topics, Weightage

Class 12 Linear Programming does not have any weightage in JEE Main. It is a small chapter; the following are the topics covered in the Linear Programming Class 12:

Exercise	Topics Covered
12.1	Introduction
12.2	Linear Programming Problem and its Mathematical Formulation

Important Formulas of Class 12 Linear Programming

Class 12 Linear Programming Important Formulae For CBSE and Competitive Exam

1. Objective Function

A function to be maximized or minimized, usually represented as: $Z = ax + by$ where $a$ and $b$ are constants, and $x, y$ are decision variables.

2. Constraints

Linear inequalities that define the feasible region, such as: $a x + b y \leq c, a x + b y \geq c, a x + b y = c$

3. Feasible Region

The common region that satisfies all constraints, including $x \geq 0$ and $y \geq 0$

4. Corner-Point Method: It is used to find the optimal solution:

Identify the corner points of the feasible region vertices.
Substitute each corner point into the objective function
Choose the maximum or minimum value of z, depending on the problem.

Free PDF: Class 12 Linear Programming NCERT Solutions – Chapter 12

Find below the link to the free Linear Programming Class 12 Solutions PDF. Students must download it as it offers detailed and accurate solutions to all the NCERT textbook questions of this chapter.
Class 12 Chapter 12 Linear Programming NCERT Solutions PDF: Free PDF Download

If you are looking for chapter-wise important topics and free PDFs of Class 12 Maths, check Class 12 Maths NCERT Solutions.

Class 12 Linear Programming Exercise-wise Solutions

Chapter 12 Linear Programming solutions are designed to help students understand key concepts like formulating linear programming problems, identifying feasible regions, and using the graphical method to find optimal solutions. Exercise 12.1 deals with Basics of Linear programming such as Introduction, formulating the objective function, and setting up constraints based on real-life scenarios. Exercise 12.2 focuses on solving LPP using the graphical method, identifying feasible regions, and applying the corner-point method. Miscellaneous Exercises covers various problems based on te earlier discussed topics. Students can check the complete exercise-wise solutions below;

Class 12 Linear Programming Exercise 12.1 Solutions

Class 12 Linear Programming Exercise 12.1 focuses on the foundational problems of Linear Programming, where students can learn to frame linear inequalities, represent feasible regions graphically, and find optimal solutions using corner points and other methods. Linear Programming Exercise 12.1 Solutions consists of 10 Questions. Students can access the complete solution of Exercise 12.1 below;

Class 12 Linear Programming Exercise 12.1 Solutions

Solve the following Linear Programming Problems graphically:

Q1. Maximize Z = 3x + 4y subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.

A.1. Maximise $z = 3 x + 4 y$

Subject to the constraints: $x + y \leq 4, x \geq 0, y \geq 0$

The corresponding equation of the above inequality are

$\begin{array}{l} x + y = 4 \\ x = 0, y = 0 \end{array}$

$\Rightarrow \frac{x}{4} + \frac{y}{4} = 1$

$x = 0, y = 0$

The graph of the given inequalities.

The shaded region OAB is the feasible region which is bounded.

The corresponding of the corner point of the feasible region are O(0,0),A(4,0), and B(0,4).

The value of Z at these points are as follows,

Corner point $z = 3 x + 4 y$

O(0,0) 0

A(4,0) 12

B(0,4) 16

Therefore the maximum value of Z is 16 at the point B(0,4).

Q2. Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

A.2. Minimize $z = - 3 x + 4 y$

Subject to $x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0$

The corresponding equation of the given inequalities are

$\begin{array}{l} x + 2 y = 8 \\ 3 x + 2 y = 12 \\ x = 0, y = 0 \end{array}$

$\Rightarrow \frac{x}{8} + \frac{y}{4} = 1$

$\begin{array}{l} \frac{x}{4} + \frac{y}{6} = 1 \\ x = 0, y = 0 \end{array}$

The graph is shown below.

The bounded region OABC is the feasible region with the corner points O(0,0),A(4,0),B(2,3), and C(0,4

The value of Z at these points are

Therefore, the minimum value of Z is -12 at (4,0).

Q3. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

A.3. Maximise $z = 5 x + 3 y$

Subject to $2 x + 5 y \leq 15, 5 x + 2 y \leq 10, x \geq 0, y \geq 0$

The corresponding equation of the above linear inequalities are

$\begin{array}{l} 3 x + 5 y = 15 \\ 5 x + 2 y = 10 & x = 0, y = 0 \end{array}$

$\Rightarrow \frac{x}{5} + \frac{y}{3} = 1$

$\begin{array}{l} \frac{x}{2} + \frac{y}{5} = 1 \\ x = 0, y = 0 \end{array}$

The graph of its given inequalities.

The shaded region OABC is the feasible region which is bounded with the corner points

$O (0, 0), A (2, 0), B (\frac{20}{19}, \frac{45}{19}) & C (0, 3)$

The values of Z at these points are

Therefore the maximum value of Z is $\frac{235}{19} a t, B (\frac{20}{19}, \frac{45}{19})$

Q4. Minimise Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

A.4. Minimize $z = 3 x + 5 y$

Such that $x + 3 y \geq 3, x + y \geq 2, x, y \geq 0$

The corresponding equation of the given inequalities are

$\begin{array}{l} x + 3 y = 3 \\ x + y = 2 \\ x, y = 0 \end{array}$

$\begin{array}{l} \Rightarrow \frac{x}{3} + \frac{y}{1} = 1 \\ \frac{x}{2} + \frac{y}{2} = 1 \\ x, y = 0 \end{array}$

The graph of the given inequalities is

The feasible region is unbounded. The corner points are $A (3, 0), B (\frac{3}{2}, \frac{1}{2}) & C (0, 2)$

The values of Z at these corner points as follows.

As the feasible region is unbounded, 7 may or may not be minimum value of Z.

We draw the graph of inequality $3 x + 5 y < 7$ .

The feasible region has no common point with $3 x + 5 y < 7$ .

Therefore minimum value of Z in 7 at $B (\frac{3}{2}, \frac{1}{2})$

Commonly asked questions

16. 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 ?.5 and that from a shade is ?.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?

12. 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.

28. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L, respectively. The company has 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 are given in the following table:

Distance in (km)
From/To	A	B
D	7	3
E	6	4
F	3	2

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

23. A farmer mixes two brands, P and Q of cattle feed. Brand P, costing ?. 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 ?. 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 for 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?

20. 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 ?. 6/kg and F2 costs ?. 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?

6. Minimise Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.

7. Minimise and Maximise Z = 5x + 10 y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.

29. A fruit grower can use two types of fertilisers 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, 270 kg of potash and at most 310 kg of chlorine.

If the grower wants to minimise 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 to the garden?

Kg per Bag
	Brand P	Brand Q
Nitrogen	3	3.5
Phosphoric acid	1	2
Potash	3	1.5
Chlorine	1.5	2

18. 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 ? 70 lakhs and if his profit on the desktop model is ?. 4500 and on portable model is ?. 5000.

19. 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 ?. 4 per unit food and F2 costs ?. 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.

27. Two godowns, A and B, have grain capacities 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 is given in the following table:

Transportation Cost per Quintal (in Rs)
From/To	A	B
D	6	4
E	3	2
F	2.50	3

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?

Solve the following Linear Programming Problems graphically:

1. Maximize Z = 3x + 4y subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.

14. 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 ?. 17.50 per package on nuts and ?. 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?

24. 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 of food are given below:

Food	Vitamin A	Vitamin B	Vitamin C
X	1	2	3
Y	2	2	1

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

31. 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 types by at most 600 units. If the company makes a profit of ?. 12 and ?. 16 per doll on dolls A and B, respectively, how many of each should be produced weekly in order to maximise the profit?

15. 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 ?. 7 and screws B at a profit of ?. 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.

17. 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 ?.5 each for type A and ?.6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

26. An aeroplane can carry a maximum of 200 passengers. A profit of ?. 1000 is made on each executive class ticket, and a profit of ?. 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 executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?

11. 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 ?. 60/kg and Food Q costs ?. 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.

13. 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.

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

10. Maximise Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.

8. Minimise and Maximise Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

30. Refer to Question 8. If the grower wants to maximise 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?

2. Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

3. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

4. Minimise Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

5. Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.

9. Maximise Z = – x + 2y, subject to the constraints x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

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

$2 x + y \leq 10, x + 3 y \leq 15, x, y \geq 0$ are (0, 0), (5, 0), (3, 4) and (0, 5). Let $Z = p x + q y$ , 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

25. 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 given below:

Types of Toys	Machines
	I	II	III
A	12	18	6
B	6	0	9

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

Class 12 Maths NCERT Linear Programming Solutions- FAQs

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Maths Ncert Solutions class 12th Exam

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Important Formulas of Class 12 Linear Programming

Class 12 Linear Programming Important Formulae For CBSE and Competitive Exam

Free PDF: Class 12 Linear Programming NCERT Solutions – Chapter 12

Class 12 Linear Programming Exercise-wise Solutions

Class 12 Linear Programming Exercise 12.1 Solutions

Class 12 Linear Programming Exercise 12.1 Solutions

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Class 12 Maths NCERT Linear Programming Solutions- FAQs

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