NCERT Solutions for Exercise 12.2 Class 12 Maths Chapter 12 - Linear Programming

# NCERT Solutions for Exercise 12.2 Class 12 Maths Chapter 12 - Linear Programming

Edited By Ramraj Saini | Updated on Dec 04, 2023 11:08 AM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 12 Exercise 12.2

NCERT Solutions for Exercise 12.2 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 12.2 Class 12 Maths chapter 12 discuss a few types of linear programming problems. There are 11 practice questions given in exercise 12.2 Class 12 Maths. All these problems of NCERT solutions for Class 12 Maths chapter 12 exercise 12.2 are done using graphical methods. Manufacturing problems, diet problems, transportation problems are some of the linear programming problems given in the Class 12 Maths chapter 12 exercise 12.2. According to the given statements, the given objective functions and the constraints are formulated and then solved using graphical methods. Along with the Class 12 Maths chapter 12 exercise 12.2 solutions there are two more exercises.

All these NCERT problems are solved by Mathematics expert faculties and NCERT solutions for Class 12 Maths chapter 12 exercise 12.2 can be used for the preparation of the CBSE Class 12 Board Exam. 12th class Maths exercise 12.2 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: 12.2

JEE Main Highest Scoring Chapters & Topics
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Let mixture contain x kg of food P and y kg of food Q. Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in the table as :

 Vitamin A Vitamin B Cost Food P 3 5 60 Food Q 4 2 80 requirement 8 11
##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

The mixture must contain 8 units of Vitamin A and 11 units of Vitamin B.

Therefore, we have

$3x+4y\geq 8$

$5x+2y\geq 11$

Total cost is Z. $Z=60x+80y$

Subject to constraint,

$3x+4y\geq 8$

$5x+2y\geq 11$

$x\geq 0,y\geq 0$

The feasible region determined by constraints is as follows:

It can be seen that a feasible region is unbounded.

The corner points of the feasible region are $A(\frac{8}{3},0),B(2,\frac{1}{2}),C(0,\frac{11}{2})$

The value of Z at corner points is as shown :

 corner points $Z=60x+80y$ $A(\frac{8}{3},0)$ 160 MINIMUM $B(2,\frac{1}{2})$ 160 minimum $C(0,\frac{11}{2})$ 440

Feasible region is unbounded, therefore 160 may or may not be the minimum value of Z.

For this, we draw $60x+80y< 160\, \, or \, \, \, 3x+4y< 8$ 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. $\, \, 3x+4y< 8$

Hence, Z has a minimum value 160 at line segment joining points $A(\frac{8}{3},0)$ and $B(2,\frac{1}{2})$ .

Let there be x cakes of first kind and y cakes of the second kind.Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in the table as :

 Flour(g) fat(g) Cake of kind x 200 25 Cake of kind y 100 50 Availability 5000 1000

Therefore,

$200x+100y\leq 5000$

$\Rightarrow \, \, \, \, 2x+y\leq 50$

. $\, \, 25x+50y\leq 10000$

$\Rightarrow \, \, x+2y\leq 400$

The total number of cakes, Z. Z=X+Y

Subject to constraint,

$\Rightarrow \, \, \, \, 2x+y\leq 50$

$\Rightarrow \, \, x+2y\leq 400$

$x\geq 0,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of the feasible region are $A(25,0),B(20,10),C(0,20),D(0,0)$

The value of Z at corner points is as shown :

 corner points Z=X+Y $A(25,0)$ 25 $B(20,10)$ 30 maximum $C(0,20)$$D(0,0)$ 20 0 minimum

The maximum cake can be made 30 (20 of the first kind and 10 of the second kind).

(i) What number of rackets and bats must be made if the factory is to work at full capacity?

Let number of rackets be x and number of bats be y.

the machine time availability is not more than 42 hours.

i.e. $1.5x+3y\leq 42$

craftsman’s time availability is 24 hours

i.e. $3x+y\leq 24$

The factory has to work at full capacity.

Hence, $1.5x+3y= 42...............1$

$3x+y= 24...............2$

Solving equation 1 and 2, we have

$x=4\, \, and\, \, \, y=12$

Thus, 4 rackets and 12 bats are to be made .

(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.

Let the number of rackets is x and the number of bats is y.

the machine time availability is not more than 42 hours.

craftsman’s time availability is 24 hours

The given information can be repreented in table as shown :

 racket bat availability machine time 1.5 3 42 craftman's time 3 1 24

$1.5x+3y\leq 42$

$3x+y\leq 24$

$x,y\geq 0$

The profit on the bat is 10 and on the racket is 20.

$Z=20x+10y$

The mathematical formulation is :

maximise $Z=20x+10y$

subject to constraints,

$1.5x+3y\leq 42$

$3x+y\leq 24$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points are $A(8,0),B(4,12),C(0,14),D(0,0)$

The value of Z at corner points is as shown :

 CORNER POINTS $Z=20x+10y$ $A(8,0)$ 160 $B(4,12)$ 200 maximum $C(0,14)$ 140 $D(0,0)$ 0

Thus, the maximum profit of the factory when it works at full capacity is 200.

Let packages of nuts be x and packages of bolts be y .Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in table as :

 bolts nuts availability machine A 1 3 12 machine B 3 1 12

Profit on a package of nuts is Rs. 17.5 and on package of bolt is 7.

Therefore, constraint are

$x+3y\leq 12$

$3x+y\leq 12$

$x\geq 0,y\geq 0$

$Z= 17.5x+7y$

The feasible region determined by constraints is as follows:

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

The value of Z at corner points is as shown :

 Corner points $Z= 17.5x+7y$ $A(4,0)$ 70 $B(3,3)$ 73.5 maximum $C(0,4)$ 28 $D(0,0)$ 0

The maximum value of z is 73.5 at $B(3,3)$ .

Thus, 3 packages of nuts and 3 packages of bolts should be manufactured everyday to get maximum profit.

Let factory manufactures screws of type A and factory manufactures screws of type B. Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in the table as :

 screw A screw B availability Automatic machine 4 6 $4\times 60=240$ hand operated machine 6 3 $4\times 60=240$

Profit on a package of screw A is Rs.7 and on the package of screw B is 10.

Therefore, the constraint is

$4x+6y\leq 240$

$6x+3y\leq 240$

$x\geq 0,y\geq 0$

$Z= 7x+10y$

The feasible region determined by constraints is as follows:

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

The value of Z at corner points is as shown :

 Corner points $Z= 7x+10y$ $A(40,0)$ 280 $B(30,20)$ 410 maximum $C(0,40)$ 400 $D(0,0)$ 0

The maximum value of z is 410 at $B(30,20)$ .

Thus, 30 packages of screw A and 20 packages of screw B should be manufactured every day to get maximum profit.

Let the cottage industry manufactures x pedestal lamps and y wooden shades. Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in the table as :

 lamps shades availability machine (h) 2 1 12 sprayer (h) 3 2 20

Profit on a lamp is Rs. 5 and on the shade is 3.

Therefore, constraint is

$2x+y\leq 12$

$3x+2y\leq 20$

$x\geq 0,y\geq 0$

$Z= 5x+3y$

The feasible region determined by constraints is as follows:

The corner points of the feasible region are $A(6,0),B(4,4),C(0,10),D(0,0)$

The value of Z at corner points is as shown :

 Corner points $Z= 5x+3y$ $A(6,0)$ 30 $B(4,4)$ 32 maximum $C(0,10)$ 30 $D(0,0)$ 0

The maximum value of z is 32 at $B(4,4)$ .

Thus, 4 shades and 4 pedestals lamps should be manufactured every day to get the maximum profit.

Let x be Souvenirs of type A and y be Souvenirs of type B .Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in table as :

 Type A Type B availability cutting 5 8 $(3\times 60)+20=200$ asembling 10 8 $4\times 60=240$

Profit on type A Souvenirs is Rs. 5 and on type B Souvenirs is 6.

Therefore, constraint are

$5x+8y\leq 200$

$10x+8y\leq 240$

$x\geq 0,y\geq 0$

$Z=5x+6y$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(24,0),B(8,20),C(0,25),D(0,0)$

The value of Z at corner points is as shown :

 Corner points $Z=5x+6y$ $A(24,0)$ 120 $B(8,20)$ 160 maximum $C(0,25)$ 150 $D(0,0)$ 0

The maximum value of z is 160 at $B(8,20)$ .

Thus,8 Souvenirs of type A and 20 Souvenirs of type B should be manufactured everyday to get maximum profit.

Let merchant plans has personal computers x desktop model and y portable model

.Thus, $x\geq 0,y\geq 0$ .

The cost of desktop model is cost Rs 25000 and portable model is Rs 40000.

Merchant can invest Rs 70 lakhs maximum.

$25000x+40000y\leq 7000000$

$5x+8y\leq 1400$

the total monthly demand of computers will not exceed 250 units.

$x+y\leq 250$

profit on the desktop model is Rs 4500 and on portable model is Rs 5000.

Total profit = Z , $Z=4500x+5000y$

The mathematical formulation of given problem is :
$5x+8y\leq 1400$

$x+y\leq 250$

$x\geq 0,y\geq 0$

$Z=4500x+5000y$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(250,0),B(200,50),C(0,175),D(0,0)$

The value of Z at corner points is as shown :

 Corner points $Z=4500x+5000y$ $A(250,0)$ 1125000 $B(200,50)$ 1150000 maximum $C(0,175)$ 875000 $D(0,0)$ 0

The maximum value of z is 1150000 at $B(200,50)$ .

Thus, merchant should stock 200 desktop models and 50 portable models to get maximum profit.

Let diet contain x unit of food F1 and y unit of foof F2 .Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in table as :

 Vitamin minerals cost per unit foof F1 3 4 4 food F2 6 3 6 80 100

Cost of food F1 is Rs 4 per unit and Cost of food F2 is Rs 6 per unit

Therefore, constraint are

$3x+4y\geq 4$

$6x+3y\geq 6$

$x\geq 0,y\geq 0$

$Z= 4x+6y$

The feasible region determined by constraints is as follows:

We can see feasible region is unbounded.

The corner points of feasible region are $A(\frac{80}{3},0),B(24,\frac{4}{3}),C(0,\frac{100}{3})$

The value of Z at corner points is as shown :

 Corner points $Z= 4x+6y$ $A(\frac{80}{3},0)$ 106.67 $B(24,\frac{4}{3}),$ 104 minimum $C(0,\frac{100}{3})$ 200 maximum

Feasible region is unbounded , therefore 104 may or may not be minimum value of Z .

For this we draw $4x+6y< 104$ or $2x+3y< 52$ and check whether resulting half plane has point in common with feasible region or not.

We can see feasible region has no common point with $2x+3y< 52$ .

Hence , Z has minimum value 104.

Let farmer buy x kg of fertilizer F1 and y kg of F2 .Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in table as :

 Nitrogen phosphoric acid Cost F1 10 6 6 F2 5 10 5 requirement 14 14

F1 contain 10% nitrogen and F2 contain 5% nitrogen .Farmer requires atleast 14 kg of nitrogen

$10\%x+5\%y\geq 14$

$\frac{x}{10}+\frac{y}{20}\geq 14$

$2x+y\geq 280$

F1 contain 6% phophoric acid and F2 contain 10% phosphoric acid .Farmer requires atleast 14 kg of nitrogen

$6\%x+10\%y\geq 14$

$\frac{6x}{100}+\frac{y}{20}\geq 14$

$3x+56y\geq 700$

Total cost is Z . $Z=6x+5y$

Subject to constraint ,

$2x+y\geq 280$

$3x+56y\geq 700$

$x\geq 0,y\geq 0$

$Z=6x+5y$

The feasible region determined by constraints is as follows:

It can be seen that feasible region is unbounded.

The corner points of feasible region are $A(\frac{700}{3},0),B(100,80),C(0,280)$

The value of Z at corner points is as shown :

 corner points $Z=6x+5y$ $A(\frac{700}{3},0)$ 1400 $,B(100,80)$ 1000 minimum $C(0,280)$ 1400

Feasible region is unbounded , therefore 1000 may or may not be minimum value of Z .

For this we draw $6x+5y< 1000$ and check whether resulting half plane has point in common with feasible region or not.

We can see feasible region has no common point with $6x+5y< 1000$ .

Hence , Z has minimum value 1000 at point $,B(100,80)$

$2x+y \leq 10,x+3y \leq 15,x,y\geq 0$ are $(0,0),(5,0),(3,4)$ and $(0,5)$ . Let $Z=px+qy,$ where $p,q > 0.$ Condition on p and q so that the maximum of Z occurs at both $(3,4)$ and $(0,5)$ is

The maximum value of Z is unique.

It is given that maximum value of Z occurs at two points $(3,4)\, \, and\, \, \, (0,5)$ .

$\therefore$ Value of Z at $(3,4)$ =value of Z at $(0,5)$

$\Rightarrow \, \, \, p(3)+q(4)=p(0)+q(5)$

$\Rightarrow \, \, \, 3p+4q=5q$

$\Rightarrow \, \, \, q=3p$

Hence, D is correct option.

## More about NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.2

Before exercise 12.2 Class 12 Maths, there are 3 solved examples. The first examples are diet, allocation and manufacturing problems. The Class 12 Maths chapter 12 exercise 12.2 have 11 problems out of which 1 is a multiple-choice question. After exercise 12.2 there are miscellaneous examples and exercises for practice.

Also Read| Linear Programming Class 12th Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.2

• For CBSE board exams students can expect one similar question as discussed in NCERT book exercise 12.2 Class 12 Maths
• These solutions will help students to improve their conceptual understanding

## Key Features Of NCERT Solutions for Exercise 12.2 Class 12 Maths Chapter 12

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 12.2 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 12.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 12.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 12.2 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 12.2 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 12.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

## Subject Wise NCERT Exemplar Solutions

1. What are the benefits of solving exercise 12.2 Class 12 Maths?

By solving the Class 12 Maths NCERT syllabus chapter 12 exercise 2 problems students will be able to clarify their doubts in the concepts covered in the chapter and will be helpful for exams also

2. List out a few types of linear programming problems?

Some of the linear programming problems are diet problems, manufacturing problems, allocation problems and transportation problems

3. What is the manufacturing problems discussed in the Class 12 NCERT Maths chapter 12?

The number of products which should be produced and sold is determined here subjected to some constraints like manpower, machines, labour etc

4. What is diet problems in linear programming?

Diet problems minimise the cost of diet subjected to the amount of different nutrients required

5. What is the significance of linear programming problems in the CBSE board exam point of view?

Students may get up to 5 marks questions from the Class 12 NCERT chapter linear programming

6. Can we solve Class 12 maths chapter 12 exercise 12.2 directly without solving exercise 12.1 Class 12 Maths?

No, exercise 12.1 explains the method of solving linear programming problems using graphs. Once students are through with the first exercise they can move to the second one. The concepts covered in the first exercise are used to solve the second exercise.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

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.

Possible steps:

• Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
• Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
• Practice Regularly: Consistent practice is key to mastering chemistry.
2. Consider Professional Help:

• Tutoring: A tutor can provide personalized guidance and support.
• Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
3. 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.
4. Focus on NEET 2025 Preparation:

• Stay Dedicated: Continue your NEET preparation with renewed determination.
• Utilize Resources: Make use of study materials, online courses, and mock tests.
5. Seek Support:

• Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
• Join Study Groups: Collaborating with peers can create a supportive learning environment.

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.

Hi,

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Hope you find this useful!

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