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Linear Programming-Students understand a series of Mathematical methods applied to optimize various situations of real life. This article elaborates how students can maximize or minimize any expression subject to certain constraints. Concepts like linear inequalities, feasible region, objective function, optimal solution, graph based approach to solve various LPPs will cover and also will familiarize the students with real applications of mathematics in various fields of commerce, finance and management, etc. Compiled with due care by subject matter experts at Careers360 based on the updated syllabus of CBSE,
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these NCERT Solutions for class 12 Mathematics chapter provide error-free solutions in step by step form for every question mentioned in the textbook. The NCERT Solutions for class 12 help students not only to conceptualise graphically but also helps to solve word problems that often require a thorough understanding of Mathematical modeling. This chapter is crucial from exam perspective like JEE Main, JEE Advanced and consistent practice ensures good grasp of conceptual as well as problem-solving skills.
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.
Here are the NCERT Class 12 Maths Chapter 12 Linear Programming question answers with clear and detailed solutions.
| Linear Programming Class 12 Question Answers Exercise: 12.1 Page number: 403-404 Total questions: 10 |
Answer:
The region determined by constraints, $x+y\leq 4,x\geq 0,y\geq 0.$ is as follows,

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 :
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Corner points
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$Z = 3x + 4y$
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$B(4,0)$
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12
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$C(0,0)$
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0
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$D(0,4)$
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16
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maximum
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The maximum value of Z is 16 at $D(0,4)$
Answer:
The region determined by constraints, $x+2y\leq 8,3x+2y\leq 12,x\geq 0,y\geq 0.$ is as follows,

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:
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Corner points
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$z=-3x+4y$
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$A(2,3)$
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6
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$B(4,0)$
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-12
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Minimum
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$C(0,0)$
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0
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$D(0,4)$
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16
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The minimum value of Z is -12 at $B(4,0)$
Answer:
The region determined by constraints, $3x + 5y \leq 15$ , $5x+2y\leq 10$ , $x\geq 0,y\geq 0$ is as follows :

The corner points of feasible region are $A(0,3),B(0,0),C(2,0),D(\frac{20}{19},\frac{45}{19})$
The value of these points at these corner points are :
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Corner points
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$Z = 5x + 3y$
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$A(0,3)$
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9
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$B(0,0)$
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0
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$C(2,0)$
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10
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$D(\frac{20}{19},\frac{45}{19})$
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$\frac{235}{19}$
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Maximum
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The maximum value of Z is $\frac{235}{19}$ at $D(\frac{20}{19},\frac{45}{19})$
Answer:
The region determined by constraints $x+3y\geq 3,x+y\geq 2,x,y\geq 0.$ is as follows,

The feasible region is unbounded, as shown.
The corner points of the feasible region are $A(3,0),B(\frac{3}{2},\frac{1}{2}),C(0,2)$
The value of these points at these corner points is:
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Corner points
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$Z = 3x + 5y$
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|
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$A(3,0)$
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9
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|
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$B(\frac{3}{2},\frac{1}{2})$
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7
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Minimum
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$C(0,2)$
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10
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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 $B(\frac{3}{2},\frac{1}{2})$
Answer:
The region determined by constraints, $x+2y\leq 10,3x+y\leq 15,x,y\geq 0$ is as follows,

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 :
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Corner points
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$Z = 3x + 2y$
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|
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$A(5,0)$
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15
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$B(4,3)$
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18
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Maximum
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$C(0,5)$
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10
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The maximum value of Z is 18 at $B(4,3)$
Answer:
The region determined by constraints $2x+y\geq 3,x+2y\geq 6,x,y\geq 0.$ is as follows,

The corner points of the feasible region are $A(6,0),B(0,3)$
The value of these points at these corner points are :
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Corner points
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$Z = x + 2y$
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$A(6,0)$
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6
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$B(0,3)$
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6
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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 $Z = x + 2y$.
Answer:
The region determined by constraints, $x+2y\leq 120,x+y\geq 60,x-2y\geq 0,x,y\geq 0$ is as follows,

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:
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Corner points
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$z=5x+10y$
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$A(40,20)$
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400
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$B(60,30)$
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600
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Maximum
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$C(60,0)$
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300
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Minimum
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$D(120,0)$
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600
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maximum
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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 $D(120,0)$
Answer:
The region determined by constraints $x+2y\geq 100,2x-y\leq 0,2x+y\leq 200,x,y,\geq 0$ is as follows,

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 :
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Corner points
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$z=x+2y$
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|
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$A(0,50)$
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100
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Minimum
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$B(20,40)$
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100
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Minimum
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$C(50,100)$
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250
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$D(0,200)$
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400
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Maximum
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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 $D(0,200)$.
Answer:
The region determined by constraints $x\geq 3,x+y\geq 5,x+2y\geq 6,y\geq 0.$ is as follows,

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 :
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Corner points
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$Z = -x+2y$
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$A(6,0)$
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- 6
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minimum
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$B(4,1)$
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-2
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$C(3,2)$
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1
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maximum
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|
|
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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 whether the resulting half-plane has a point in common with a feasible region or not.
We can see that 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.
Answer:
The region determined by constraints $x-y\leq -1,-x+ y\leq 0,x,y,\geq 0.$ is as follows,

There is no feasible region, and thus, Z has no maximum value.
Check the NCERT Solutions of Linear Programming Class 12 Maths Chapter 12 exercise from the following link
Question: In an LPP, if the objective function $z=a x+b y$ has the same maximum value on two corner points of the feasible region, then the number of points at which $z_{\max }$ 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".
Here is the list of important topics that are covered in Class 12 Chapter 12 Linear Programming.
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.
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.
Theorem 1: Let R be the feasible region (convex polygon) for a linear programming problem and let $\mathrm{Z}=a x+b y$ 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 $\mathrm{Z}=a x+b y$ 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 $R$.
Using these approaches, students can tackle the Linear Programming Class 12 Chapter 12 Question Answers with greater confidence.
This chapter helps you understand how to make the best possible decisions using maths. It shows how real-life problems can be solved by finding maximum or minimum values. These Class 12 Maths chapter 12 Linear Programming question answers make these ideas easier to learn and apply through step-by-step examples. Here are some more points on why these question answers are important.
This chapter includes an elaboration of various concepts of Linear Programming by using graphical techniques. NCERT Solutions explain the problems in an easy manner with detailed descriptions.The students practice various problems based on application related to 10 questions included in 1 textbook exercises that aid in improving concept and numericals skills of the students and building their confidence and accuracy. Practice of the questions in this chapter improves interpretation skills, drawing of graphs, logic application ability and graphic thinking ability that helps in their exams. Through this chapter the student is able to solve various problems on the optimisation in practical form.
This is the most applicable chapter in Class 12 Maths, since the concept helps in solving real-time optimization problems by means of modeling. With a clear understanding of how graphical methods and feasible regions are solved, you can solve the questions of linear programming even faster and more accurately. Preparing graphical representation in an adept manner and learning solutions for every question available in the NCERT textbook would enable you to achieve accuracy, efficiency and get prepared for board examinations as well as competitive exams such as JEE Main and Advanced. These were the suggestions from our seasoned Mathematics faculty members of Careers360.
Given below is the chapter-wise list of the NCERT Class 12 Maths solutions with their respective links:
Also, read,
Here, you can find the NCERT Solutions for other subjects as well.
Here, you can find the NCERT Solutions for classes 9 to 11.
Here, you can find the NCERT books and syllabus for class 12.
Also, check,
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Frequently Asked Questions (FAQs)
Linear Programming is a technique used to find the optimal (maximum or minimum) value of an objective function under the condition of some constraints.
Linear programming is important because students get to know the real world applications of optimization.
Topics covered in the linear programming chapter includes linear inequalities, feasible regions, objective functions, optimization problems, and graphically solving problems.
NCERT solutions of chapter 12 can ease the work of students in solving graphical and application based problems due to step-by-step approach.
Absolutely, the essential basics of linear programming can be useful for both the general and board examinations and various similar competition examinations
Students should put most practice into finding the feasible region and plotting the optimal solution on a graph.
By practicing NCERT questions and accurately plotting the graph students can definitely secure a higher score.
Mistakes typically made by the students while dealing with linear programming includes graphing, identifying feasible regions and choosing the correct vertex as the optimal solution.
Yes, NCERT solutions provide thorough explanations and solutions of every textbook question.
On Question asked by student community
Hello Ananya,
Please specify the class for which you need the question papers. I am providing Class 10 and 12 papers.
Here are the links to the CBSE Half-yearly Question Papers (2025-2026).
Hello Ananya,
Please specify the class for which you need the question papers. I am providing Class 10 and 12 papers.
Here are the links to the CBSE Half-yearly Question Papers (2025-2026).
Hello Pawan,
CBSE Class 10 Mathematics 2026 and previous year question paper:
https://school.careers360.com/boards/cbse/cbse-class-10-question-paper-2026
CBSE Class 12 Mathematics 2026 and previous year question paper:
https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers-class-12-maths
Hello Dharani,
Check the link below to download NCERT Class 12 previous year question papers in PDF format for all subjects.
https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers-class-12
Hello Vipin,
Check the link below to download CBSE Class 12 question papers in PDF format for all subjects, including Mathematics.
https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers-class-12
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