NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming: We often come across problems where we seek to maximize profit or minimize cost. NCERT solutions for class 12 maths chapter 12 linear programming is helpful to find the most optimal solution for such problems. In the NCERT solutions for class 12 maths chapter 12 linear programming you will learn to formulate these real-life problems into a mathematical model. In class 11th you have already learnt linear inequalities and their solutions by the graphical method and systems of linear equations and their applications in daily life problems. In this chapter, you are going to deal with problems on linear programming like maximization and minimization of equations, mathematical and graphical methods to solve problems of linear programming. In NCERT solutions for class 12 maths chapter 12 linear programming article, questions from all these topics are covered. Check all NCERT solutions from class 6 to 12 at a single place, which will help you get a better understanding of concepts in a much easy way. Here you will get NCERT solutions for class 12 also.
Generally, one question( 5 marks) from this chapter is asked in the 12th board final examination. In this chapter, there are 2 exercises with 21 questions. In this article, you will find NCERT solutions for class 12 maths chapter 12 linear programming which are prepared and explained in a detailed manner. It will help you to score well in the board exam as well as in the competitive exams.
Let's take an NCERT problem - A furniture dealer deals in only two items–chairs and tables. Has storage space of at most 60 pieces and He has Rs 50,000 to invest. A chair costs Rs 500 and A table costs Rs 2500. He estimates that from the sale of one chair, he can make a profit of Rs 75 and that from the sale of one table a profit of Rs 250. He wants to know how many chairs and tables he should buy from the available money so as to maximize his total profit, assuming that he can sell all the items which he buys.
That type of problem which involves- minimize profit or maximize cost is called optimization problems. Linear programming problems are a very important class of optimization problems. The above-stated problem is an example of linear programming. This chapter is an important chapter because of its wide applicability in industry, management science, commerce, etc. In this chapter, you will learn mathematical and graphical methods to solve problems of linear programming.
Topics of NCERT class 12 maths chapter-12 Linear Programming
12.1 Introduction
12.2 Linear Programming Problem and its Mathematical Formulation
12.2.1 Mathematical formulation of the problem
12.2.2 Graphical method of solving linear programming problems
12.3 Different Types of Linear Programming Problems
NCERT solutions for class 12 maths chapter 12 Linear Programming- Exercise Questions
NCERT solutions for class 12 maths chapter 12 linear programming-Exercise: 12.1
Answer:
The region determined by constraints, is as follows,
The region A0B represents the feasible region
The corner points of the feasible region are
Maximize
The value of these points at these corner points are :
Corner points | ||
12 | ||
0 | ||
16 | maximum |
The maximum value of Z is 16 at
Answer:
The region determined by constraints, is as follows,
The corner points of feasible region are
The value of these points at these corner points are :
Corner points | ||
6 | ||
-12 | Minimum | |
0 | ||
16 |
The minimum value of Z is -12 at
Answer:
The region determined by constraints, , , is as follows :
The corner points of feasible region are
The value of these points at these corner points are :
Corner points | ||
9 | ||
0 | ||
10 | ||
Maximum |
The maximum value of Z is at
Answer:
The region determined by constraints is as follows,
The feasible region is unbounded as shown.
The corner points of the feasible region are
The value of these points at these corner points are :
Corner points | ||
9 | ||
7 | Minimum | |
10 | ||
The feasible region is unbounded, therefore 7 may or may not be the minimum value of Z .
For this, we draw 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.
Hence, Z has a minimum value of 7 at
Answer:
The region determined by constraints, is as follows,
The corner points of feasible region are
The value of these points at these corner points are :
Corner points | ||
15 | ||
18 | Maximum | |
10 | ||
The maximum value of Z is 18 at
Question:6 Solve the following Linear Programming Problems graphically: Minimise Subject to
Show that the minimum of Z occurs at more than two points.
Answer:
The region determined by constraints is as follows,
The corner points of the feasible region are
The value of these points at these corner points are :
Corner points | |
6 | |
6 |
Value of Z is the same at both points.
If we take any other point like on line , 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 .
Answer:
The region determined by constraints, is as follows,
The corner points of feasible region are
The value of these points at these corner points are :
Corner points | ||
400 | ||
600 | Maximum | |
300 | Minimum | |
600 | maximum |
The minimum value of Z is 300 at and maximum value is 600 at all points joing line segment and
Answer:
The region determined by constraints is as follows,
The corner points of the feasible region are
The value of these points at these corner points are :
Corner points | ||
100 | Minimum | |
100 | Minimum | |
250 | ||
400 | Maximum |
The minimum value of Z is 100 at all points on the line segment joining points and .
The maximum value of Z is 400 at .
Answer:
The region determined by constraints is as follows,
The corner points of the feasible region are
The value of these points at these corner points are :
Corner points | ||
- 6 | minimum | |
-2 | ||
1 | maximum | |
The feasible region is unbounded, therefore 1 may or may not be the maximum value of Z.
For this, we draw 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.
Answer:
The region determined by constraints is as follows,
There is no feasible region and thus, Z has no maximum value.
NCERT solutions for class 12 maths chapter 12 linear programming-Exercise: 12.2
Answer:
Let mixture contain x kg of food P and y kg of food Q. Thus, .
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 |
The mixture must contain 8 units of Vitamin A and 11 units of Vitamin B.
Therefore, we have
Total cost is Z.
Subject to constraint,
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
The value of Z at corner points is as shown :
corner points | ||
160 | MINIMUM | |
160 | minimum | |
440 |
Feasible region is unbounded, therefore 160 may or may not be the minimum value of Z.
For this, we draw 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.
Hence, Z has a minimum value 160 at line segment joining points and .
Answer:
Let there be x cakes of first kind and y cakes of the second kind.Thus, .
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,
.
The total number of cakes, Z. Z=X+Y
Subject to constraint,
The feasible region determined by constraints is as follows:
The corner points of the feasible region are
The value of Z at corner points is as shown :
corner points | Z=X+Y | |
25 | ||
30 | maximum | |
| 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?
Answer:
Let number of rackets be x and number of bats be y.
the machine time availability is not more than 42 hours.
i.e.
craftsman’s time availability is 24 hours
i.e.
The factory has to work at full capacity.
Hence,
Solving equation 1 and 2, we have
Thus, 4 rackets and 12 bats are to be made .
Answer:
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 |
The profit on the bat is 10 and on the racket is 20.
The mathematical formulation is :
maximise
subject to constraints,
The feasible region determined by constraints is as follows:
The corner points are
The value of Z at corner points is as shown :
CORNER POINTS | ||
160 | ||
200 | maximum | |
140 | ||
0 |
Thus, the maximum profit of the factory when it works at full capacity is 200.
Answer:
Let packages of nuts be x and packages of bolts be y .Thus, .
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
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
Corner points | ||
70 | ||
73.5 | maximum | |
28 | ||
0 |
The maximum value of z is 73.5 at .
Thus, 3 packages of nuts and 3 packages of bolts should be manufactured everyday to get maximum profit.
Answer:
Let factory manufactures screws of type A and factory manufactures screws of type B. Thus, .
The given information can be represented in the table as :
screw A | screw B | availability | |
Automatic machine | 4 | 6 | |
hand operated machine | 6 | 3 | |
Profit on a package of screw A is Rs.7 and on the package of screw B is 10.
Therefore, the constraint is
The feasible region determined by constraints is as follows:
The corner points of the feasible region are
The value of Z at corner points is as shown :
Corner points | ||
280 | ||
410 | maximum | |
400 | ||
0 |
The maximum value of z is 410 at .
Thus, 30 packages of screw A and 20 packages of screw B should be manufactured every day to get maximum profit.
Answer:
Let the cottage industry manufactures x pedestal lamps and y wooden shades. Thus, .
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
The feasible region determined by constraints is as follows:
The corner points of the feasible region are
The value of Z at corner points is as shown :
Corner points | ||
30 | ||
32 | maximum | |
30 | ||
0 |
The maximum value of z is 32 at .
Thus, 4 shades and 4 pedestals lamps should be manufactured every day to get the maximum profit.
Answer:
Let x be Souvenirs of type A and y be Souvenirs of type B .Thus, .
The given information can be represented in table as :
Type A | Type B | availability | |
cutting | 5 | 8 | |
asembling | 10 | 8 | |
Profit on type A Souvenirs is Rs. 5 and on type B Souvenirs is 6.
Therefore, constraint are
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
Corner points | ||
120 | ||
160 | maximum | |
150 | ||
0 |
The maximum value of z is 160 at .
Thus,8 Souvenirs of type A and 20 Souvenirs of type B should be manufactured everyday to get maximum profit.
Answer:
Let merchant plans has personal computers x desktop model and y portable model
.Thus, .
The cost of desktop model is cost Rs 25000 and portable model is Rs 40000.
Merchant can invest Rs 70 lakhs maximum.
the total monthly demand of computers will not exceed 250 units.
profit on the desktop model is Rs 4500 and on portable model is Rs 5000.
Total profit = Z ,
The mathematical formulation of given problem is :
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
Corner points | ||
1125000 | ||
1150000 | maximum | |
875000 | ||
0 |
The maximum value of z is 1150000 at .
Thus, merchant should stock 200 desktop models and 50 portable models to get maximum profit.
Answer:
Let diet contain x unit of food F1 and y unit of foof F2 .Thus, .
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
The feasible region determined by constraints is as follows:
We can see feseable region is unbounded.
The corner points of feasible region are
The value of Z at corner points is as shown :
Corner points | ||
106.67 | ||
104 | minimum | |
200 | maximum | |
Feasible region is unbounded , therefore 104 may or may not be minimum value of Z .
For this we draw or 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 .
Hence , Z has minimum value 104.
Answer:
Let farmer buy x kg of fertilizer F1 and y kg of F2 .Thus, .
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
F1 contain 6% phophoric acid and F2 contain 10% phosphoric acid .Farmer requires atleast 14 kg of nitrogen
Total cost is Z .
Subject to constraint ,
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
The value of Z at corner points is as shown :
corner points | ||
1400 | ||
1000 | minimum | |
1400 |
Feasible region is unbounded , therefore 1000 may or may not be minimum value of Z .
For this we draw 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 .
Hence , Z has minimum value 1000 at point
Question:11 The corner points of the feasible region determined by the following system of linear inequalities:
are and . Let where Condition on p and q so that the maximum of Z occurs at both and is
Answer:
The maximum value of Z is unique.
It is given that maximum value of Z occurs at two points .
Value of Z at =value of Z at
Hence, D is correct option.
NCERT solutions for class 12 maths chapter 12 linear programming-Miscellaneous Exercise
Question:1 Reference of Example 9 (Diet problem): A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol.
Answer:
Let diet contain x packets of food P and y packets of food Q. Thus, .
The mathematical formulation of the given problem is as follows:
Total cost is Z .
Subject to constraint,
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
corner points | ||
150 | MINIMUM | |
285 | maximum | |
228 |
Hence, Z has a maximum value of 285 at the point .
to maximise the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used. The maximum amount of vitamin A is 285 units.
Answer:
Let farmer mix x bags of brand P and y bags of brand Q. Thus, .
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 |
The given problem can be formulated as follows:
Therefore, we have
Subject to constraint,
The feasible region determined by constraints is as follows:
The corner points of the feasible region are
The value of Z at corner points is as shown :
corner points | ||
4500 | ||
2650 | ||
1950 | minimum | |
2400 |
Feasible region is unbounded, therefore 1950 may or may not be a minimum value of Z. For this, we draw 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 .
Hence, Z has a minimum value 1950 at point .
Answer:
Let mixture contain x kg of food X and y kg of food Y.
Mathematical formulation of given problem is as follows:
Minimize :
Subject to constraint ,
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
corner points | ||
160 | ||
112 | minimum | |
116 | ||
160 |
The feasible region is unbounded , therefore 112 may or may not be minimum value of Z .
For this we draw 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 .
Hence , Z has minimum value 112 at point
Answer:
Let x and y toys of type A and type B.
Mathematical formulation of given problem is as follows:
Minimize :
Subject to constraint ,
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
corner points | ||
150 | ||
250 | ||
262.5 | maximum | |
200 |
Therefore 262.5 may or may not be maximum value of Z .
Hence , Z has maximum value 262.5 at point
Answer:
Let airline sell x tickets of executive class and y tickets of economy class.
Mathematical formulation of given problem is as follows:
Minimize :
Subject to constraint ,
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
corner points | ||
68000 | ||
136000 | maximum | |
128000 | ||
therefore 136000 is maximum value of Z .
Hence , Z has maximum value 136000 at point
Answer:
Let godown A supply x and y quintals of grain to shops D and E respectively. Then , (100-x-y) will be supplied to shop F. Requirements at shop D is 60 since godown A supply x .Therefore remaining (60-x) quintals of grain will be transported from godown B.
Similarly, (50-y) quintals and 40-(100-x-y)=(x+y-60) will be transported from godown B to shop E and F respectively. The problem can be represented diagrammatically as follows:
and
and
Total transportation cost z is given by ,
Mathematical formulation of given problem is as follows:
Minimize :
Subject to constraint ,
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
corner points | ||
560 | ||
620 | ||
610 | ||
510 | minimum |
therefore 510 may or may not be minimum value of Z .
Hence , Z has miniimum value 510 at point
Answer:
Let x and y litres of oil be supplied from A to petrol pump,D and E. Then , (7000-x-y) will be supplied from A to petrol pump F.
Requirements at petrol pump D is 4500 L. since x L A are transported from depot A,remaining 4500-x L will be transported from petrol pump B
Similarly, (3000-y)L and 3500-(7000-x-y)=(x+y-3500) L will be transported from depot B to petrol E and F respectively.
The problem can be represented diagrammatically as follows:
and
and
Cost of transporting 10 L petrol =Re 1
Cost of transporting 1 L petrol
Total transportation cost z is given by ,
Mathematical formulation of given problem is as follows:
Minimize :
Subject to constraint ,
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
corner points | ||
5000 | ||
5300 | ||
5550 | ||
4400 | minimum | |
5450 |
Hence , Z has miniimum value 4400 at point
Answer:
Let fruit grower use x bags of brand P and y bags of brand Q.
Mathematical formulation of given problem is as follows:
Minimize :
Subject to constraint ,
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
corner points | ||
595 | ||
550 | ||
470 | minimum | |
Therefore 470 is minimum value of Z .
Hence , Z has minimum value 470 at point
Question:9 Reference of Que 8 : A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.
Kg per bag | ||
Brand A | Brand P | |
Nitrogen | 3 | 3.5 |
Phosphoric Acid | 1 | 2 |
Potash | 3 | 1.5 |
Chlorine | 1.5 | 2 |
Answer:
Let fruit grower use x bags of brand P and y bags of brand Q.
Mathematical formulation of given problem is as follows:
Maximize :
Subject to constraint ,
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
corner points | ||
595 | maximum | |
550 | ||
470 | minimum | |
therefore 595 is maximum value of Z .
Hence , Z has minimum value 595 at point
Answer:
Let x and y be number of dolls of type A abd B respectively that are produced per week.
Mathematical formulation of given problem is as follows:
Maximize :
Subject to constraint ,
The feasible region determined by constraints is as follows:
The corner points of feasible region are
The value of Z at corner points is as shown :
corner points | ||
7200 | ||
15000 | ||
16000 | Maximum | |
Therefore 16000 is maximum value of Z .
Hence , Z has minimum value 16000 at point
NCERT solutions for class 12 maths - Chapter wise
chapter 1 | NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions |
chapter 2 | NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions |
chapter 3 | |
chapter 4 | |
chapter 5 | NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability |
chapter 6 | NCERT solutions for class 12 maths chapter 6 Application of Derivatives |
chapter 7 | |
chapter 8 | NCERT solutions for class 12 maths chapter 8 Application of Integrals |
chapter 9 | NCERT solutions for class 12 maths chapter 9 Differential Equations |
chapter 10 | NCERT solutions for class 12 maths chapter 10 Vector Algebra |
chapter 11 | NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry |
chapter 12 | NCERT solutions for class 12 maths chapter 12 Linear Programming |
chapter 13 |
NCERT solutions for class 12 subject wise
NCERT Solutions class wise
- NCERT solutions for class 12
- NCERT solutions for class 11
- NCERT solutions for class 10
- NCERT solutions for class 9
Benefits of NCERT solutions
- NCERT solutions are prepared and explained in a step-by-step manner, so it is very easy for you to understand the concepts.
- These NCERT solutions will give you new ways of solving the problems.
- NCERT solutions for class 12 maths chapter 12 linear programming will help you to score good marks in the exam as these questions are answered by the experts who know how best to answer the questions in the board exam.
- Miscellaneous exercise is very important if you wish to develop a grip on the concepts. In NCERT solutions for class 12 maths chapter 12 linear programming, you will get solutions for miscellaneous exercise too.
Happy learning !!!
Frequently Asked Question (FAQs) - NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming
Question: How does NCERT solutions are helpful in CBSE board exam ?
Answer:
As most the questions in CBSE board exam are asked directly from NCERT textbook, so one must know the NCERT well but only knowing the answer is not guarantee to score good marks in the exam. One should know how to answer in board exam in order to get good marks. NCERT solutions are provided by the experts who knows how best to write answer in the board exam in order to get good marks.
Question: Does the chapter linear programming has practical application in real-life ?
Answer:
Yes, linear programming is useful in formulate these real-life problems into a mathematical model and than solve them. Some optimisation problems like maximizing the profits and minimizing the cost can be solved by linear programming.
Question: Does CBSE provides the solutions of NCERT for class 12 maths ?
Answer:
No, CBSE doesn’t provided NCERT solutions for any class or subject.
Question: Where can I find the complete solutions of NCERT for class 12 maths ?
Answer:
A Here you will get the detailed NCERT solutions for class 12 maths by clicking on the link.
Question: What is the weightage of the chapter Linear Programming for CBSE board exam ?
Answer:
Generally, one question of 5 marks is asked from this chapter in the CBSE 12th board final exam.
Question: What are the important topics in chapter Linear Programming ?
Answer:
Linear programming problem and its mathematical formulation, graphical method of solving linear programming problems are the important topics of this chapter.
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