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

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

Edited By Ramraj Saini | Updated on Dec 04, 2023 11:02 AM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 12 Exercise 12.1

NCERT Solutions for Exercise 12.1 Class 12 Maths Chapter 12 Linear Programming are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 12.1 Class 12 Maths chapter 12 gives practice questions to understand linear programming problems. There are 10 questions explained in exercise 12.1 Class 12 Maths. There are many methods to solve linear programming problems. Here in this Class 12 NCERT Mathematics chapter, only graphical methods for solving linear programming problems are discussed. In the NCERT Solutions for class 12 maths chapter 12 exercise 12.1 all the 10 questions are solved using graphs. Class 12 Maths chapter 12 exercise 12.1 solves problems related to maximising or minimising linear functions subjected to certain constraints. These constraints are a set of linear inequalities.

12th class Maths exercise 12.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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## Linear Programming Class 12 Chapter 12-Exercise: 12.1

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 :

 Corner points $Z = 3x + 4y$ $B(4,0)$ 12 $C(0,0)$ 0 $D(0,4)$ 16 maximum
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The maximum value of Z is 16 at $D(0,4)$

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 :

 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 $B(4,0)$

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 :

 Corner points $Z = 5x + 3y$ $A(0,3)$ 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 $D(\frac{20}{19},\frac{45}{19})$

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 are :

 Corner points $Z = 3x + 5y$ $A(3,0)$ 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 $3x + 5y< 7$ and check whether resulting half plane has a point in common with the feasible region or not.

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

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

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 :

 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 $B(4,3)$

Show that the minimum of Z occurs at more than two points.

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 :

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

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 $Z = x + 2y$ .

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 :

 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 maximum value is 600 at all points joing line segment $B(60,30)$ and $D(120,0)$

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 :

 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 $D(0,200)$ .

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 :

 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 whether 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 a feasible region.

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

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.

## More About NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1

In order to understand the concepts well it is important to solve the NCERT syllabus exercise questions. The NCERT solutions for Class 12 Maths chapter 12 exercise 12.1 helps in solving the first exercise of the chapter linear programming. Students can make use of Class 12 Maths chapter 12 exercise 12.1 solutions for preparation of board exams as well as engineering entrance exam like JEE Mains.

Also Read| Linear Programming Class 12th Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1

• Class 12th Maths chapter 12 exercise 12.1 are prepared by the best faculties of Mathematics
• All main topics are covered and exercise 12.1 Class 12 Maths gives answers to all the questions and are in detail
• Students can use Class 12 Maths chapter 12 exercise 12.1 to prepare for CBSE exams.

## Key Features Of NCERT Solutions for Exercise 12.1 Class 12 Maths Chapter 12

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 12.1 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 12.1, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 12.1 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 12.1 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 12.1 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 12.1 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

## Subject Wise NCERT Exemplar Solutions

1. What is the number of exercises in the NCERT chapter linear programming?

There are three exercises including miscellaneous.

2. How many questions are there in exercise 12.1 Class 12 Maths?

Ten questions are explained in the NCERT Class 12 chapter exercise 1

3. What number of solved examples are given before the NCERT Class 12 chapter linear programming exercise 12.1?

There are 5 solved examples before exercise 12.1

4. Why to solve NCERT exercises?

Solving NCERT exercise give more conceptual understanding and students will be able to clear their doubts and can understand the are where they have to improve.

5. How to use NCERT Solutions for Class 12 Maths chapter 12 exercise 12.1?

First understand the concepts and practice solved example. Then move on to the exercise and try to solve it yourself. If you have any doubts look in to the Class 12 Maths chapter 12 exercise 12.1 solutions.

6. What is the importance of the Class 12 NCERT Maths chapter 12 linear programming for CBSE Class 12 board exams?

For CBSE Class 12 Maths exam one question of 5 marks is expected from the chapter linear programming.

7. Give the pattern of linear programming questions of Class 12 NCERT exercise 12.1 .

The linear programming questions will have an objective function. Either maximise or minimize the it according to the given constrains.

8. What method is adopted in the class 12 NCERT chapter 12 problems?

Graphical method is used to solve the problems in Class 12 chapter 12

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

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.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

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) Option 2) Option 3) Option 4)

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) Option 2) Option 3) Option 4)

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

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, 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