NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

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NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming are provided here. These NCERT solutions are prepared by the expert team at Careers360 considering the latest syllabus of CBSE 2024-25. In this chapter, we shall study some linear programming problems and their solutions by the graphical method only, though there are many other methods also to solve these problems. In the solutions of Chapter 12 Maths Class 12, we will discuss how to find optimal solutions to problems related to maximizing profit or minimizing cost. Class 12 linear programming ncert solutions will help in formulating these real-life problems into a mathematical model. You should practice linear programming problems to get a command of concepts and an in-depth understanding of this linear programming class 12 chapter.

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NCERT Linear Programming Class 12 Questions And Answers PDF Free Download
NCERT Class 12 Maths Chapter 12 Question Answer - Important Formulae
NCERT Linear Programming Class 12 Questions And Answers (Exercise)
Importance of solving NCERT questions for class 12 Math Chapter 12
NCERT solutions for class 12 subject-wise
Class-wise NCERT Solutions
NCERT Books and NCERT Syllabus

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

In this linear programming class 12 solutions, you are going to deal with problems on linear programming like maximization and minimization of equations, and mathematical and graphical methods to solve problems of linear programming. You can also refer to the linear programming class 12 ncert solutions for a better understanding of concepts. Below we have given complete Class 12 Maths Chapter 12 NCERT solutions. Check all NCERT solutions from class 6 to 12 in a single place to prepare better for exams. Also, read Class 12 Maths Chapter 12 Linear Programming Notes for a better understanding, and exemplar solutions, visit Ncert Exemplar Solutions For Class 12 Maths Chapter 12 Linear Programming.

NCERT Linear Programming Class 12 Questions And Answers PDF Free Download

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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 $Z = a x + b y$ 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 $Z = a x + b y$ 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 $R$ .

Free download NCERT Class 12 Maths Chapter 12 Question Answer for CBSE Exam.

NCERT Linear Programming Class 12 Questions And Answers (Exercise)

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 $Z = 3 x + 4 y$ Subject to the constraints $x + y \leq 4, x \geq 0, y \geq 0.$

Answer:

The region determined by constraints, $x + y \leq 4, x \geq 0, y \geq 0.$ 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 $Z = 3 x + 4 y$

The value of these points at these corner points are :

Corner points	$Z = 3 x + 4 y$
$B (4, 0)$	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 $z = - 3 x + 4 y$ Subject to . $x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0.$

Answer:

The region determined by constraints, $x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0.$ 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	$z = - 3 x + 4 y$
$A (2, 3)$	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 $Z = 5 x + 3 y$ Subject to $3 x + 5 y \leq 15$ , $5 x + 2 y \leq 10$ , $x \geq 0, y \geq 0$

Answer:

The region determined by constraints, $3 x + 5 y \leq 15$ , $5 x + 2 y \leq 10$ , $x \geq 0, y \geq 0$ is as follows :

1627031555890

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 :

Corner points	$Z = 5 x + 3 y$
$A (0, 3)$	9
$B (0, 0)$	0
$C (2, 0)$	10
$D (\frac{20}{19}, \frac{45}{19})$	$\frac{235}{19}$	Maximum

The maximum value of Z is $\frac{235}{19}$ at $D (\frac{20}{19}, \frac{45}{19})$

Question:4 Solve the following Linear Programming Problems graphically: Minimise $Z = 3 x + 5 y$ Such that $x + 3 y \geq 3, x + y \geq 2, x, y \geq 0.$

Answer:

The region determined by constraints $x + 3 y \geq 3, x + y \geq 2, x, y \geq 0.$ is as follows,

1627031646530

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:

Corner points	$Z = 3 x + 5 y$
$A (3, 0)$	9
$B (\frac{3}{2}, \frac{1}{2})$	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 $3 x + 5 y < 7$ 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. $Z = 3 x + 5 y$

Hence, Z has a minimum value of 7 at $B (\frac{3}{2}, \frac{1}{2})$

Question:5 Solve the following Linear Programming Problems graphically: Maximise $Z = 3 x + 2 y$ Subject to $x + 2 y \leq 10, 3 x + y \leq 15, x, y \geq 0$

Answer:

The region determined by constraints, $x + 2 y \leq 10, 3 x + y \leq 15, x, y \geq 0$ 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	$Z = 3 x + 2 y$
$A (5, 0)$	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 $Z = x + 2 y$ Subject to $2 x + y \geq 3, x + 2 y \geq 6, x, y \geq 0.$

Answer:

The region determined by constraints $2 x + y \geq 3, x + 2 y \geq 6, x, y \geq 0.$ 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	$Z = x + 2 y$
$A (6, 0)$	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 $Z = x + 2 y$ , 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 $Z = x + 2 y$ .

Question:7 Solve the following Linear Programming Problems graphically: Minimise and Maximise $z = 5 x + 10 y$ Subject to $x + 2 y \leq 120, x + y \geq 60, x - 2 y \geq 0, x, y \geq 0$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, $x + 2 y \leq 120, x + y \geq 60, x - 2 y \geq 0, x, y \geq 0$ 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	$z = 5 x + 10 y$
$A (40, 20)$	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 joing line segment $B (60, 30)$ and $D (120, 0)$

Question:8 Solve the following Linear Programming Problems graphically: Minimise and Maximise $z = x + 2 y$ Subject to $x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200, x, y, \geq 0$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200, x, y, \geq 0$ 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	$z = x + 2 y$
$A (0, 50)$	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 $Z = - x + 2 y$ Subject to the constraints: $x \geq 3, x + y \geq 5, x + 2 y \geq 6, y \geq 0.$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $x \geq 3, x + y \geq 5, x + 2 y \geq 6, y \geq 0.$ 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	$Z = - x + 2 y$
$A (6, 0)$	- 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 $- x + 2 y > 1$ 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 $x - y \leq - 1, - x + y \leq 0, x, y, \geq 0.$ . Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $x - y \leq - 1, - x + y \leq 0, x, y, \geq 0.$ is as follows,

1627032109317

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

If you are looking for ncert exercise solutions of linear programming class 12 then they are listed below.

Linear Programming Class 12 Exercise 12.1

NCERT Solutions for Class 12 Maths: Chapter Wise

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions

NCERT solutions for class 12 maths chapter 3 Matrices

NCERT solutions for class 12 maths chapter 4 Determinants

NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability

NCERT solutions for class 12 maths chapter 6 Application of Derivatives

NCERT solutions for class 12 maths chapter 7 Integrals

NCERT solutions for class 12 maths chapter 8 Application of Integrals

NCERT solutions for class 12 maths chapter 9 Differential Equations

NCERT solutions for class 12 maths chapter 10 Vector Algebra

NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry

NCERT solutions for class 12 maths chapter 12 Linear Programming

NCERT solutions for class 12 maths chapter 13 Probability

Importance of solving NCERT questions for class 12 Math Chapter 12

Linear programming in Class 12 is important because it equips students with a powerful tool for optimizing resource allocation and solving complex problems, with applications in various fields like business, economics, and engineering. It helps find the most efficient solutions by maximizing or minimizing a linear objective function subject to constraints. Doing practice by NCERT questions and solutions ensures a foundation for students preparing for both board exams and higher classes studies in mathematics.

Class 12 Maths Chapter 12 NCERT solutions make it very easy for you to understand the concepts as they are explained in a step-by-step manner.
NCERT Class 12 Maths Solutions Chapter 12 will give you some new insight into the concepts.
Experts with their experience and knowledge to create solutions of NCERT for class 12 maths chapter 12 linear programming that will certainly help you improve your scores in the board exam.
You should solve the miscellaneous exercise also, to develop a grip on the concepts. Here, you will get solutions for miscellaneous exercises too.

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NCERT solutions for class 12 subject-wise

Here, you can find the NCERT Solutions for other subjects as well.

Class-wise NCERT Solutions

Here, you can find the NCERT Solutions for classes 9 to 12.

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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Questions related to CBSE Class 12th

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Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

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Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

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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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

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Consider Professional Help:
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Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
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Seek Support:
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Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
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Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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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NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

NCERT Linear Programming Class 12 Questions And Answers PDF Free Download

NCERT Class 12 Maths Chapter 12 Question Answer - Important Formulae

NEET/JEE Coaching Scholarship

JEE Main Important Mathematics Formulas

NCERT Linear Programming Class 12 Questions And Answers (Exercise)

Importance of solving NCERT questions for class 12 Math Chapter 12

NCERT solutions for class 12 subject-wise

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