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NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

Edited By Komal Miglani | Updated on May 12, 2025 03:13 PM IST | #CBSE Class 12th

Linear programming is not just about finding the best solution—it’s about understanding the limits of possibility. Linear programming is a mathematical technique that is used for maximising or minimising a linear objective function, subject to a set of linear constraints. In the NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming, students will learn about the application of the systems of linear inequalities/equations to solve some real-life problems of various types. In real-life situations, a linear programming problem is like getting a maximum profit using limited resources and fulfilling customers' demand without going over budget.

This Story also Contains
  1. NCERT Solution for Class 12 Maths Chapter 12 Solutions: Download PDF
  2. NCERT Class 12 Maths Chapter 12 Question Answer - Important Formulae
  3. NCERT Solutions for Class 12 Maths Chapter 12: Exercise Questions
  4. Class 12 Maths NCERT Chapter 12: Extra Question
  5. Approach to Solve Questions of Linear Programming Class 12
  6. NCERT Solutions for Class 12 Maths: Chapter Wise
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming
LiveCBSE 12th Result 2025 (OUT) LIVE: Class 12 marksheets on Digilocker; school-wise toppersMay 13, 2025 | 9:36 PM IST

In CBSE 12th results 2025, Shiv Nadar School recorded 100%. Praneel Munshi topped the CBSE Board exams 2025 with 98.80% in science stream. 21.10% students scored 95% and above.

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Linear programming teaches us that decisions are not just numbers—they are balanced strategies. The objective of these class 12 NCERT solutions is to provide students with quality study material along with clear explanations. These solutions of NCERT are prepared by experienced careers360 experts following the latest CBSE syllabus.

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NCERT Solution for Class 12 Maths Chapter 12 Solutions: Download PDF

Students who wish to access the Class 12 Maths Chapter 12 NCERT Solutions can click on the link below to download the complete solution in PDF.

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NCERT Class 12 Maths Chapter 12 Question Answer - Important Formulae

Feasible Region: The feasible region, or solution region, of a linear programming problem is the common area determined by all the constraints, including the non-negativity constraints (x ≥ 0, y ≥ 0).

Infeasible Solution: Any point within or on the boundary of the feasible region represents a feasible solution to the constraints. Points outside the feasible region are considered infeasible solutions.

Optimal Solution: An optimal solution is any point within the feasible region that provides the optimal value (maximum or minimum) of the objective function.

Fundamental Theorems in Linear Programming:

Optimality at Corner Points: For a linear programming problem with a feasible region represented as a convex polygon, if the objective function Z = ax + by has an optimal value, this optimal value must occur at one of the corner points (vertices) of the feasible region.

Existence of Maxima and Minima: If the feasible region R is bounded, then the objective function Z has both a maximum and a minimum value on R, and each of these values occurs at a corner point (vertex) of R. If R is unbounded, a maximum or minimum may not exist. However, if it does exist, it must occur at a corner point of R.

Corner Point Method: The corner point method is used to solve a linear programming problem and consists of the following steps:

Find the feasible region of the linear programming problem and determine its corner points (vertices).

Evaluate the objective function Z = ax + by at each corner point. Let M and m represent the largest and smallest values obtained at these points.

If the feasible region is bounded, M and m respectively represent the maximum and minimum values of the objective function.

If the feasible region is unbounded, then:

  • M is the maximum value of the objective function if the open half-plane determined by ax + by > M has no points in common with the feasible region.

  • m is the minimum value of the objective function if the open half-plane determined by ax + by < M has no points in common with the feasible region.

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Theorem 1: Let R be the feasible region (convex polygon) for a linear programming problem and let Z=ax+by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point* (vertex) of the feasible region.

Theorem 2: Let R be the feasible region for a linear programming problem, and let Z=ax+by be the objective function. If R is bounded , then the objective function Z has both a maximum and a minimum value on R, and each of these occurs at a corner point (vertex) of R.

NCERT Solutions for Class 12 Maths Chapter 12: Exercise Questions

Class 12 Maths chapter 12 solutions Exercise: 12.1
Page number: 403-404
Total questions: 10

Question 1: Solve the following Linear Programming Problems graphically: Maximise Z=3x+4y Subject to the constraints x+y4,x0,y0.

Answer:

The region determined by constraints, x+y4,x0,y0. is as follows,

1627031435613

The region A0B represents the feasible region

The corner points of the feasible region are B(4,0),C(0,0),D(0,4)

Maximize Z=3x+4y

The value of these points at these corner points are :

Corner points
Z=3x+4y

B(4,0)
12

C(0,0)
0

D(0,4)
16
maximum

The maximum value of Z is 16 at D(0,4)

Question 2: Solve the following Linear Programming Problems graphically: Minimise z=3x+4y Subject to . x+2y8,3x+2y12,x0,y0.

Answer:

The region determined by constraints, x+2y8,3x+2y12,x0,y0. is as follows,

1627031511366

The corner points of feasible region are A(2,3),B(4,0),C(0,0),D(0,4)

The value of these points at these corner points are :

Corner points
z=3x+4y

A(2,3)
6

B(4,0)
-12
Minimum
C(0,0)
0

D(0,4)
16

The minimum value of Z is -12 at B(4,0)

Question 3: Solve the following Linear Programming Problems graphically: Maximise Z=5x+3y Subject to 3x+5y15 , 5x+2y10 , x0,y0

Answer:

The region determined by constraints, 3x+5y15 , 5x+2y10 , x0,y0 is as follows :

1627031555890

The corner points of feasible region are A(0,3),B(0,0),C(2,0),D(2019,4519)

The value of these points at these corner points are :

Corner points
Z=5x+3y

A(0,3)
9

B(0,0)
0

C(2,0)
10

D(2019,4519)
23519
Maximum

The maximum value of Z is 23519 at D(2019,4519)

Question 4: Solve the following Linear Programming Problems graphically: Minimise Z=3x+5y Such that x+3y3,x+y2,x,y0.

Answer:

The region determined by constraints x+3y3,x+y2,x,y0. is as follows,

1627031646530

The feasible region is unbounded as shown.

The corner points of the feasible region are A(3,0),B(32,12),C(0,2)

The value of these points at these corner points is:

Corner points
Z=3x+5y

A(3,0)
9

B(32,12)
7
Minimum
C(0,2)
10




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

For this, we draw 3x+5y<7 and check whether the resulting half plane has a point in common with the feasible region or not.

We can see that a feasible region has no common point with. Z=3x+5y

Hence, Z has a minimum value of 7 at B(32,12)

Question 5: Solve the following Linear Programming Problems graphically: Maximise Z=3x+2y Subject to x+2y10,3x+y15,x,y0

Answer:

The region determined by constraints, x+2y10,3x+y15,x,y0 is as follows,

1627031733350

The corner points of feasible region are A(5,0),B(4,3),C(0,5)

The value of these points at these corner points are :

Corner points
Z=3x+2y

A(5,0)
15

B(4,3)
18
Maximum
C(0,5)
10




The maximum value of Z is 18 at B(4,3)

Question 6: Solve the following Linear Programming Problems graphically: Minimise Z=x+2y Subject to 2x+y3,x+2y6,x,y0.

Answer:

The region determined by constraints 2x+y3,x+2y6,x,y0. is as follows,

1627031776022

The corner points of the feasible region are A(6,0),B(0,3)

The value of these points at these corner points are :

Corner points
Z=x+2y
A(6,0)
6
B(0,3)
6

The value of Z is the same at both points. A(6,0),B(0,3)

If we take any other point like (2,2) on line Z=x+2y, then Z=6.

Thus, the minimum value of Z occurs at more than 2 points.

Therefore, the value of Z is minimum at every point on the line Z=x+2y.

Question 7: Solve the following Linear Programming Problems graphically: Minimise and Maximise z=5x+10y Subject to x+2y120,x+y60,x2y0,x,y0 Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, x+2y120,x+y60,x2y0,x,y0 is as follows,

1627031823610

The corner points of feasible region are A(40,20),B(60,30),C(60,0),D(120,0)

The value of these points at these corner points are:

Corner points
z=5x+10y

A(40,20)
400

B(60,30)
600
Maximum
C(60,0)
300
Minimum
D(120,0)
600
maximum

The minimum value of Z is 300 at C(60,0) and the maximum value is 600 at all points joining line segment B(60,30) and D(120,0)

Question 8: Solve the following Linear Programming Problems graphically: Minimise and Maximise z=x+2y Subject to x+2y100,2xy0,2x+y200,x,y,0 Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints x+2y100,2xy0,2x+y200,x,y,0 is as follows,

1627031869634

The corner points of the feasible region are A(0,50),B(20,40),C(50,100),D(0,200)

The value of these points at these corner points are :

Corner points
z=x+2y

A(0,50)
100
Minimum
B(20,40)
100
Minimum
C(50,100)
250

D(0,200)
400
Maximum

The minimum value of Z is 100 at all points on the line segment joining points A(0,50) and B(20,40).

The maximum value of Z is 400 at D(0,200).

Question 9: Solve the following Linear Programming Problems graphically: Maximise Z=x+2y Subject to the constraints: x3,x+y5,x+2y6,y0. Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints x3,x+y5,x+2y6,y0. is as follows,

1627032034587

The corner points of the feasible region are A(6,0),B(4,1),C(3,2)

The value of these points at these corner points are :

Corner points
Z=x+2y

A(6,0)
- 6
minimum
B(4,1)
-2

C(3,2)
1
maximum



The feasible region is unbounded, therefore, 1 may or may not be the maximum value of Z.

For this, we draw x+2y>1 and check whetherthe resulting half-plane has a point in common with a feasible region or not.

We can see the resulting feasible region has a common point with the feasible region.

Hence, Z =1 is not the maximum value; Z has no maximum value.

Question 10: Solve the following Linear Programming Problems graphically: Maximise Z=x+y, Subject to xy1,x+y0,x,y,0.. Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints xy1,x+y0,x,y,0. is as follows,

1627032109317

There is no feasible region, and thus, Z has no maximum value.

Also, read,

Class 12 Maths NCERT Chapter 12: Extra Question

Question: In an LPP, if the objective function z=ax+by has the same maximum value on two corner points of the feasible region, then the number of points at which zmax occurs is:

Solution:
In an LPP, if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points gives the same maximum value. Hence, the number of points at which Zmax occurs is infinite.

Hence, the correct answer is "infinite".

Approach to Solve Questions of Linear Programming Class 12

  • Determine whether the problem requires maximisation or minimisation.
  • Check if variables are negative or non-negative. If they are non-negative, then they will satisfy a set of linear constraints.
  • Plot all the constraints carefully on the graph paper based on their inequalities. Find and shade the feasible region—bounded or unbounded—that satisfies all constraints simultaneously.
  • Shortcut tricks:
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- Label all the axes and lines clearly in the graph paper to understand the representation.

- Be aware of the mistakes made during the plotting or solving of intersections.

- If the region is unbounded, the optimal value may not exist.

NCERT Solutions for Class 12 Maths: Chapter Wise

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NCERT solutions for class 12 Subject-wise

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Class-wise NCERT Solutions

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NCERT Books and NCERT Syllabus

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Frequently Asked Questions (FAQs)

1. What are the real-life applications of Linear Programming in Class 12 Maths?

In Class 12 Maths, Linear Programming finds real-life applications in optimizing resource allocation, such as maximizing profits or minimizing costs in scenarios like production planning, diet planning, and transportation problems.

2. What is the difference between feasible solution and optimal solution in LPP?

In Linear Programming Problems (LPP), a feasible solution satisfies all constraints, while an optimal solution is a feasible solution that either maximizes or minimizes the objective function.

3. What are the types of linear programming problems in NCERT Class 12?

Linear equations and linear inequalities are the types of linear programming problems in NCERT Class 12. The problems in LPP basically consist of the problems that include the calculation of the minimum or maximum value.

4. How many questions are there in NCERT Class 12 Maths Chapter 12?

There are 10 questions in NCERT Class 12 Maths Chapter 12.

5. What is the graphical method in linear programming?

Graphical Method: Owing to the importance of linear programming models in various industries, many types of algorithms have been developed over the years to solve them. Some famous mentions include the Simplex method, the Hungarian approach, and others

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Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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