# NCERT Solution Class 12 Chapter 12- Linear Programming Exercise 12.1

Mathematics is one of the most crucial and challenging subjects for students in Class 12. Mathematics does not only require a thorough understanding of the concepts but also requires a great deal of practise in solving problems. The best way to score well in Mathematics in the board examinations is to practise a lot of questions. Interestingly, most of the questions in the board examinations are asked from the NCERT textbooks. This makes the NCERT book the most important resource to practise solving questions for the board exams.

Chapter 12 of the NCERT textbook for Mathematics is on Linear Programming. It is a crucial chapter for Class 12 students as questions always appear from this chapter in the exam. This chapter is comparatively briefer than the other chapters in the book. However, the concepts included in it are of utmost importance. Students learn to solve Linear Programming problems in this chapter, which will be very useful for them in the future. Students must solve the questions in the exercises of this chapter to be thorough with the concepts. Class 12 Maths Chapter 12 Exercise 12.1 deals with the basic concepts of Linear Programming. Students are expected to learn to solve Linear Programming problems in which the variables satisfy a set of linear inequalities. These questions require students to follow some steps to form a graph and solve the problem. To help students in this regard, Extramarks provides the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1.

The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 provided by Extramarks include every step required to solve the problems and make it easier for students to arrive at the answer. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 help students apply the concepts taught in the chapter correctly. The NCERT Solutions For Class 12 Maths Chapter 12 Exercise 12.1 are meticulously prepared by highly qualified teachers.These solutions are prepared with the understanding level of Class 12 students in mind. The problems in this exercise have been solved step by step to make it easier to write answers in exams. Examiners always emphasise the importance of writing step-by-step answers in exams. The step-by-step solutions in NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 assist in achieving good scores in the final examinations. Many students struggle to understand and remember the vast number of formulas and processes presented in the NCERT textbook, and as a result, they are unable to apply them to the problems in the exercises. However, practising the step-by-step solutions using the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 makes it easier to remember and apply the formulas correctly. This gives the students an advantage in their exams because they know how to solve such specific problems. Practising the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 multiple times will help students answer such questions quickly and correctly in exams.

## NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming (Ex 12.1) Exercise 12.1

Although NCERT Solutions are available online, it is preferable to have constant access to them. Offline access allows students to learn and practise anywhere they want, at their convenience. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 provided by Extramarks can be downloaded by students for learning whenever they want. Hence, unforeseen network issues and poor internet service will no longer be a barrier to learning. The expert-curated NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 cover all the questions in the textbook, providing an opportunity for learning and exam preparation. Students can download the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 in PDF format from the Extramarks website and mobile application for consistent exam preparation.

### NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming (Ex. 12.1) Exercise 12.1

The CBSE instructs students to follow the textbooks published by the NCERT. Students need to follow the syllabus released by the CBSE while studying. They should also write their answers according to the marking scheme followed by the CBSE to secure maximum marks. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 have been prepared by highly experienced teachers with decades of experience in CBSE board exams. They prepare the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 using the most recent CBSE guidelines. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 are revised and updated frequently following the most recent recommendations. Therefore, there is no reason for students to be concerned about the CBSE’s recent curriculum changes.

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### NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.1

The concepts in the chapter on Linear Programming are a continuation of the concepts learned in previous classes. Students should be thoroughly familiar with the system of linear inequalities that they have studied in Class 11. If they are not thorough with those concepts, they should go through the NCERT textbook for Class 11 and solve the questions on linear inequalities. They can also refer to the NCERT Solutions Class 11 if they are facing difficulties. Students can find that the problems in linear programming are applications of the concept of a system of linear inequalities. There are various types of problems in this chapter where students have to find a maximum profit, a minimum cost, etc. These problems find applications in a variety of fields, such as trade and commerce. While solving the problems of linear programming, students should be careful while plotting graphs. It is essential that students be able to solve the problems in Exercise 12.1 in order to solve the problems in Exercise 12.2. These problems might seem difficult at first, but with repeated practice, they can become easier to solve. Students finding difficulty with the problems can rely on the Extramarks to help them understand the problems better and solve them faster. Students can have access to a variety of educational modules and a wealth of study resources, like NCERT Solutions Class 12 on the platform to meet their needs during preparation for the board exams. For example, the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 on the Extramarks platform focuses on a specific section of a chapter. It is a reliable resource for learning, practising, and answering questions. Another reason to use Extramarks’ NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 is that they are designed and compiled by subject specialists, ensuring that all students have a positive learning experience.

Students require well-explained solutions in addition to having access to many questions prior to the board examinations. The majority of the question banks rely heavily on NCERT textbook questions. Having the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 in hand simplifies the process of answering questions. Students are more confident in any examination when they have a quick method of understanding and learning. This also allows students to devote more time to certain other aspects of their studies while keeping their minds cool. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 answer all the questions in the exercise and are error-free. Hence, practising the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 multiple times will help students answer such questions quickly and efficiently in exams.

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Q.1

$\begin{array}{l}\mathrm{Maximise}\mathrm{ }\mathrm{Z}\mathrm{ }=3\mathrm{x}+4\mathrm{y}\\ \mathrm{Subject}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{constraints}\mathrm{ }:\mathrm{x}+\mathrm{y}\le 4,\mathrm{x}\ge 0,\mathrm{y}\ge 0.\end{array}$

Ans

The feasible region determined by the constraints,
x + y ≤ 4 … (i)
x ≥ 0, y ≥ 0 …(ii)
is shown in graph: The corner points of the feasible region are O (0, 0),
A (4, 0), and B (0, 4). The values of Z at these points are as follows.

 Corner point Z = 3x + 4y O(0, 0) 0 A(4, 0) 12 B(0, 4) 16 →Maximum

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

Q.2

$\begin{array}{l}\mathrm{Minimise}\mathrm{ }\mathrm{Z}\mathrm{ }=-3\mathrm{x}+4\mathrm{y}\\ \mathrm{Subject}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{x}+\mathrm{y}\le ,\mathrm{x}+\mathrm{y}\le ,\mathrm{x}\ge 0,\mathrm{y}\ge 0.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{feasible}\mathrm{region}\mathrm{determined}\mathrm{by}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{constraints},\\ \mathrm{x}+2\mathrm{y}\le 8...\left(\mathrm{i}\right)\\ \mathrm{ }3\mathrm{x}+2\mathrm{y}\le 12...\left(\mathrm{ii}\right)\\ \mathrm{x}\ge 0\mathrm{and}\mathrm{y}\ge \mathrm{0}...\left(\mathrm{iii}\right)\mathrm{}\\ \mathrm{is}\mathrm{shown}\mathrm{in}\mathrm{graph}:\end{array}$ The corner points of the feasible region are O (0, 0), A (4, 0), B (2, 3), and C (0, 4).
The values of Z at these corner points are as follows:

 Corner point Z = – 3x + 4 y O (0, 0) 0 A (4, 0) –12 → Minimum B( 2, 3) 6 C(0, 4) 16

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

Q.3

$\begin{array}{l}\mathrm{Maximise}\mathrm{ }\mathrm{Z}\mathrm{ }=5\mathrm{x}+3\mathrm{y}\\ \mathrm{Subject}\mathrm{ }\mathrm{to}\mathrm{ }3\mathrm{x}+\mathrm{y}\le ,\mathrm{x}+\mathrm{y}\le ,\mathrm{x}\ge 0,\mathrm{y}\ge 0.\end{array}$

Ans

The feasible region determined by the system of constraints,
3x + 5y ≤ 15 … (i)
5x + 2y ≤ 10 … (ii)
x ≥ 0, and y ≥ 0 … (iii)
is shown in the given graph: $\begin{array}{l}\mathrm{The}\mathrm{corner}\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{are}\mathrm{O}\left(0, 0\right),\mathrm{A}\left(2, 0\right),\\ \mathrm{B}\left(0, 3\right),\mathrm{and}\mathrm{C}\left(\frac{20}{19}, \frac{45}{19}\right).\\ \mathrm{The}\mathrm{values}\mathrm{of}\mathrm{Z}\mathrm{at}\mathrm{these}\mathrm{corner}\mathrm{points}\mathrm{are}\mathrm{as}\mathrm{follows}:\end{array}$

 Corner point z = 5x + 3y O (0, 0) 0 A (2, 0) 10 B (0, 3) 9 $C\left(\frac{20}{19},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{45}{19}\right)$ $\frac{235}{19}$ → Maximum

$\begin{array}{l}\mathrm{Therefore},\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}\mathrm{Z}\mathrm{is}\frac{235}{19}\mathrm{}\mathrm{at}\mathrm{the}\mathrm{point}\\ \left(\frac{20}{19}, \frac{45}{19}\right).\end{array}$

Q.4

$\begin{array}{l}\mathrm{Minimise}\mathrm{ }\mathrm{Z}\mathrm{ }=3\mathrm{x}+5\mathrm{y}\\ \mathrm{Subject}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{x}+3\mathrm{y}\le ,\mathrm{x}+\mathrm{y}\ge ,\mathrm{ }\mathrm{x},\mathrm{ }\mathrm{y}\ge 0.\end{array}$

Ans

$\begin{array}{l}\text{The feasible region determined by the system of constraints,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{x}+\text{3y}\ge \text{3}\text{}\text{}\dots \left(i\right)\\ \text{}\text{}\text{x}+\text{y}\ge \text{2}\text{}\text{}\dots \left(ii\right)\\ \text{}\text{x,y}\ge \text{0}\text{}\text{}\text{}\dots \left(iii\right)\\ \text{is shown in the following graph:}\end{array}$ $\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{graph},\mathrm{we}\mathrm{see}\mathrm{that}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{is}\mathrm{unbounded}\mathrm{.}\\ \mathrm{The}\mathrm{corner}\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{are}\mathrm{A}\left(3, 0\right),\mathrm{B}\left(\frac{3}{2},\mathrm{ }\frac{1}{2}\right),\\ \mathrm{C}\left(0, 2\right).\mathrm{The}\mathrm{values}\mathrm{of}\mathrm{Z}\mathrm{at}\mathrm{these}\mathrm{corner}\mathrm{points}\mathrm{are}\mathrm{as}\mathrm{follows}:\end{array}$

 Corner point Z = 3x + 5y A (3, 0) 9 $B\left(\frac{3}{2},\frac{1}{2}\right)$ 7 → Smallest C (0, 2) 10

$\begin{array}{l}\mathrm{As}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{is}\mathrm{unbounded},\mathrm{therefore}, 7\mathrm{may}\mathrm{or}\mathrm{may}\\ \mathrm{not}\mathrm{be}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{Z}\mathrm{.}\\ \mathrm{For}\mathrm{this},\mathrm{we}\mathrm{draw}\mathrm{the}\mathrm{graph}\mathrm{of}\mathrm{the}\mathrm{inequality}, 3\mathrm{x}+ 5\mathrm{y}< 7,\mathrm{and}\\ \mathrm{check}\mathrm{whether}\mathrm{the}\mathrm{resulting}\mathrm{half}\mathrm{plane}\mathrm{has}\mathrm{points}\mathrm{in}\mathrm{common}\\ \mathrm{with}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{or}\mathrm{not}.\mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{the}\mathrm{feasible}\\ \mathrm{region}\mathrm{has}\mathrm{no}\mathrm{common}\mathrm{point}\mathrm{with}3\mathrm{x}+ 5\mathrm{y}< 7\mathrm{Therefore},\\ \mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{Z}\mathrm{is}7\mathrm{at}\left(\frac{3}{2},\frac{1}{2}\right)\mathrm{.}\end{array}$

Q.5

$\begin{array}{l}\mathrm{Maximise}\mathrm{ }\mathrm{Z}=3\mathrm{x}+2\mathrm{y}\\ \mathrm{Subject}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{x}+2\mathrm{y}\le ,\mathrm{x}+\mathrm{y}\ge ,\mathrm{ }\mathrm{x},\mathrm{ }\mathrm{y}\ge 0.\end{array}$

Ans

The feasible region determined by the constraints,
x + 2y ≤ 10 …(i)
3x + y ≤ 15 …(ii)
x ≥ 0, y ≥ 0, …(iii)
is shown in the following graph: The corner points of the feasible region are A (5, 0), B(4, 3), and C (0, 5).
The values of Z at these corner points are as follows:

 Corner point Z= 3x + 2y A(5,0) 15 B(4, 3) 18 → Maximum C(0, 5) 10

Therefore, the maximum value of Z is 18 at the point (4, 3).

Q.6

$\begin{array}{l}\mathrm{Minimise}\mathrm{ }\mathrm{Z}=\mathrm{x}+2\mathrm{y}\\ \mathrm{Subject}\mathrm{ }\mathrm{to}\mathrm{ }2\mathrm{x}+\mathrm{y}\ge ,\mathrm{x}+2\mathrm{y}\ge ,\mathrm{x},\mathrm{y}\ge 0.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{feasible}\mathrm{region}\mathrm{determined}\mathrm{by}\mathrm{the}\mathrm{constraints},\\ 2\mathrm{x}+\mathrm{y}\ge 3\mathrm{}\dots \left(\mathrm{i}\right)\\ \mathrm{x}+2\mathrm{y}\ge 6\dots \left(\mathrm{ii}\right)\\ \mathrm{}\mathrm{x}\ge 0,\mathrm{y}\ge 0,\mathrm{}\dots \left(\mathrm{iii}\right)\mathrm{}\\ \mathrm{}\mathrm{is}\mathrm{shown}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{graph}:\end{array}$ The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, and y≥ 0, is as follows:

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

Here, the value of Z at points A and B is same. If we take any other point such as (2, 2) on line
x + 2y = 6, then Z = 6.
Thus, the minimum value of Z occurs for more than 2 points.
Therefore, the value of Z is minimum at every point on the line, x + 2y = 6.
Show that the minimum of Z occurs at more than two points.

Q.7

$\begin{array}{l}\mathrm{Minimise}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{Maximise}\mathrm{ }\mathrm{Z}=5\mathrm{x}+10\mathrm{y}\\ \mathrm{Subject}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{x}+2\mathrm{y}\le 120,\mathrm{x}+\mathrm{y}\ge ,\mathrm{x}-2\mathrm{y}\ge 0;\mathrm{x},\mathrm{y}\ge 0.\end{array}$

Ans

The feasible region determined by the constraints,
x + 2y ≤ 120 … (i)
x + y ≥ 60 …. (ii)
x−2y≥ 0 ….. (iii)
x, y ≥ 0 …… (iv)
is shown in the following graph: The corner points of the feasible region are A (60, 0), B(120, 0), C (60, 30), and D (40, 20).
The values of Z at these corner points are as follows:

 Corner Point Z = 5x + 10 y A(60, 0) 300 → Minimum B(120, 0) 600 → Maximum C(60, 30) 600 → Maximum D(40, 20) 400

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

Q.8

$\begin{array}{l}\mathrm{Minimise}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{Maximise}\mathrm{ }\mathrm{Z}=\mathrm{x}+2\mathrm{y}\\ \mathrm{Subject}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{x}+2\mathrm{y}\le 100,2\mathrm{x}-\mathrm{y}\le ,2\mathrm{x}+\mathrm{y}\le 00;\mathrm{x},\mathrm{y}\ge 0.\end{array}$

Ans

The feasible region determined by the constraints,
x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤ 200, x ≥ 0, and y ≥ 0, is as follows: The corner points of the feasible region are A(0, 50), B(20, 40), C(50, 100), and D(0, 200).
The values of Z at these corner points are as follows:

 Corner point Z = x + 2y A(0, 50) 100 → Minimum B(20, 40) 100 → Minimum C(50, 100) 250 D(0, 200) 400 → Maximum

The maximum value of Z is 400 at (0, 200) and the minimum value of Z is 100 at all the points on the line segment joining the points (0, 50) and (20, 40)

Q.9

$\begin{array}{l}\mathrm{Maximise}\mathrm{ }\mathrm{Z}\mathrm{ }=-\mathrm{ }\mathrm{x}+2\mathrm{y},\mathrm{ }\mathrm{subject}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{contraints}:\\ \mathrm{x}\ge ,\mathrm{x}+\mathrm{y}\ge ,\mathrm{x}+2\mathrm{y}\ge ;\mathrm{y}\ge 0.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{feasible}\mathrm{region}\mathrm{determined}\mathrm{by}\mathrm{the}\mathrm{constraints},\\ \mathrm{ }\mathrm{x}\ge 3...\left(\mathrm{i}\right)\\ \mathrm{}\mathrm{x}+\mathrm{y}\ge 5...\left(\mathrm{ii}\right)\\ \mathrm{}\mathrm{x}+2\mathrm{y}\le \mathrm{6}...\left(\mathrm{iii}\right)\\ \mathrm{ }\mathrm{y}\ge 0\mathrm{}...\left(\mathrm{iv}\right)\\ \mathrm{is}\mathrm{shown}\mathrm{in}\mathrm{graph}:\end{array}$ .

It can be seen that the feasible region is unbounded. The values of Z at corner points A (6, 0), B (4, 1), and C (3, 2) are as follows:

 Corner point Z = – x + 2y A(6, 0) – 6 B(4, 1) – 2 C(3, 2) 1

As the feasible region is unbounded, therefore, Z = 1 may or may not be the maximum value.
For this, we graph the inequality, −x + 2y > 1, and check whether the resulting half plane has points in common with the feasible region or not.
The resulting feasible region has points in common with the feasible region.
Therefore, Z = 1 is not the maximum value. Z has no maximum value.

Q.10

$\begin{array}{l}\mathrm{Maximise}\mathrm{ }\mathrm{Z}\mathrm{ }=\mathrm{ }\mathrm{x}+\mathrm{y},\mathrm{ }\mathrm{subject}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{contraints}:\\ \mathrm{x}-\mathrm{y}\le -,-\mathrm{x}+\mathrm{y}\le ;\mathrm{x},\mathrm{y}\ge 0.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{region}\mathrm{determined}\mathrm{by}\mathrm{the}\mathrm{constraints},\\ \mathrm{x}-\mathrm{y}\le -1...\left(\mathrm{i}\right)\\ -\mathrm{x}+\mathrm{y}\le 0...\left(\mathrm{ii}\right)\\ \mathrm{x},\mathrm{y}\ge 0...\left(\mathrm{iii}\right)\\ \mathrm{is}\mathrm{show}\mathrm{in}\mathrm{graph}:\end{array}$ There is no feasible region and thus, Z has no maximum value.

## 1. Where can students find the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1?

Students can find the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 on the Extramarks website and mobile app. Academic experts curate the solutions according to the CBSE guidelines, making them a trustworthy source of information. To provide students with an immersive experience, the solutions are explained step by step and cross-checked with the answers mentioned in the NCERT textbooks.

## 2. Do the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 and other exercises help students understand the concepts presented in the various chapters?

Understanding the concepts of the NCERT solutions can help students progress in their studies. The chapter’s concepts can be confusing and difficult at times, but having NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 can be a useful learning aid. Students must comprehend the steps required to arrive at the answer. The clarity of the steps and procedures can help students develop the ability to answer any question that arises. Using these learning resources can hence make the concepts of Linear Programming easier to understand.

## 3. What are the benefits of referring to the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 during preparation for the board exams?

The NCERT textbook on Mathematics is widely regarded as the most important book for CBSE board exams. Questions are frequently posed directly from NCERT exercises and examples. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 provide answers to these questions correctly and simply. It is also essential that students practice the NCERT textbook on a daily basis to complete the question paper in the time allotted. One of the most common reasons for student’s failure to finish the exam in the time allowed is a lack of practice. Having access to NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1 teaches students how to structure their answers to complete the question paper on time for the examination.

## 4. What are the concepts covered in the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.1?

The NCERT solutions for Exercise 12.1 cover the basic concepts of Linear Programming. It provides the correct step-by-step procedure for solving the problems of linear inequalities. Students should be thorough with the concept of linear inequalities taught in Class 11. It will help them solve problems in linear programming.