NCERT Solutions for Class 12 Maths Chapter 12 – Free PDF Download
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming, prepared by the subject experts at BYJU’S, are provided here. We can say that linear programming is a method to achieve the best outcome in a mathematical model whose requirements are represented by linear relationships. In this chapter, students learn about Linear Programming in detail. Solving all the questions from this chapter of the NCERT Solutions for Class 12 Maths repeatedly will help the students in understanding the methods of solving the questions and also aid in acing their board exam.
The Class 12 Maths NCERT Solutions of the chapter Linear Programming are prepared as per the latest CBSE syllabus and are very easy to understand, as the subject experts at BYJUâ€™S ensure that the solutions are given in the simplest form. These NCERT Solutions for Class 12Â Maths Chapter 12 cover all the exercise questions included in the book and are in accordance with the latest CBSE Syllabus 2023-24 and its guidelines.
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming
Access All Exercises of Class 12 Maths Chapter 12
Exercise 12.1 Solutions 10 Questions
Exercise 12.2 Solutions 11 Questions
Miscellaneous Exercise on Chapter 12 Solutions 10 Questions
NCERT Solutions for Class 12 Maths Chapter 12 – Linear Programming
Exercise 12.1 page no: 513
1. Maximise Z = 3xÂ + 4y
Subject to the constraints:
Solution:
The feasible region determined by the constraints, x + y â‰¤ 4, x â‰¥ 0, y â‰¥ 0, is given below.
O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of Z at these points are given below.
Corner point | Z = 3x + 4y | |
O (0, 0) | 0 | |
A (4, 0) | 12 | |
B (0, 4) | 16 | Maximum |
Hence, the maximum value of Z is 16 at the point B (0, 4).
2. Minimise Z = âˆ’3xÂ + 4y
subject to.
Solution:
The feasible region determined by the system of constraints,
is given below.
O (0, 0), A (4, 0), B (2, 3) and C (0, 4) are the corner points of the feasible region.
The values of Z at these corner points are given below.
Corner point | Z = – 3x + 4y | |
O (0, 0) | 0 | |
A (4, 0) | -12 | Minimum |
B (2, 3) | 6 | |
C (0, 4) | 16 |
Hence, the minimum value of Z is â€“ 12 at the point (4, 0).
3. Maximise Z = 5xÂ + 3y
subject to.
Solution:
The feasible region determined by the system of constraints, 3xÂ + 5yÂ â‰¤ 15, 5xÂ + 2yÂ â‰¤ 10,Â xÂ â‰¥ 0, andÂ y â‰¥ 0, is given below.
O (0, 0), A (2, 0), B (0, 3) and C (20 / 19, 45 / 19) are the corner points of the feasible region. The values of Z at these corner points are given below.
Corner point | Z = 5x + 3y | |
O (0, 0) | 0 | |
A (2, 0) | 10 | |
B (0, 3) | 9 | |
C (20 / 19, 45 / 19) | 235 / 19 | Maximum |
Hence, the maximum value of Z is 235 / 19 at the point (20 / 19, 45 / 19).
4. Minimise Z = 3xÂ + 5y
such that.
Solution:
The feasible region determined by the system of constraints, x + 3y â‰¥ 3, x + y â‰¥ 2, and x, y â‰¥ 0, is given below.
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A (3, 0), B (3 / 2, 1 / 2) and C (0, 2).
The values of Z at these corner points are given below.
Corner point | Z = 3x + 5y | |
A (3, 0) | 9 | |
B (3 / 2, 1 / 2) | 7 | Smallest |
C (0, 2) | 10 |
7 may or may not be the minimum value of Z because the feasible region is unbounded.
For this purpose, we draw the graph of the inequality, 3x + 5y < 7 and check whether the resulting half-plane has common points with the feasible region or not.
Hence, it can be seen that the feasible region has no common point with 3x + 5y < 7.
Thus, the minimum value of Z is 7 at point B (3 / 2, 1 / 2).
5. Maximise Z = 3xÂ + 2y
subject to.
Solution:
The feasible region determined by the constraints, xÂ + 2yÂ â‰¤ 10, 3xÂ +Â yÂ â‰¤ 15,Â xÂ â‰¥ 0, andÂ y â‰¥ 0, is given below.
A (5, 0), B (4, 3), C (0, 5) and D (0, 0) are the corner points of the feasible region.
The values of Z at these corner points are given below.
Corner point | Z = 3x + 2y | |
A (5, 0) | 15 | |
B (4, 3) | 18 | Maximum |
C (0, 5) | 10 |
Hence, the maximum value of Z is 18 at points (4, 3).
6. Minimise Z =Â xÂ + 2y
subject to
.
Solution:
The feasible region determined by the constraints, 2xÂ +Â yÂ â‰¥ 3,Â xÂ + 2yÂ â‰¥ 6,Â xÂ â‰¥ 0, andÂ y â‰¥ 0, is given below.
A (6, 0) and B (0, 3) are the corner points of the feasible region.
The values of Z at the corner points are given below.
Corner point | Z = x + 2y |
A (6, 0) | 6 |
B (0, 3) | 6 |
Here, the values of Z at points A and B are same. If we take any other point, such as (2, 2) on line x + 2y = 6, then Z = 6.
Hence, 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.
7. Minimise and Maximise Z = 5xÂ + 10y
subject to.
Solution:
The feasible region determined by the constraints, xÂ + 2yÂ â‰¤ 120,Â xÂ +Â yÂ â‰¥ 60,Â xÂ âˆ’ 2yÂ â‰¥ 0,Â xÂ â‰¥ 0, andÂ y â‰¥ 0, is given below.
A (60, 0), B (120, 0), C (60, 30), and D (40, 20) are the corner points of the feasible region. The values of Z at these corner points are given below.
Corner point | Z = 5x + 10y | |
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).
8. Minimise and Maximise Z =Â xÂ + 2y
subject to.
Solution:
The feasible region determined by the constraints, xÂ + 2yÂ â‰¥ 100, 2xÂ âˆ’Â yÂ â‰¤ 0, 2xÂ +Â yÂ â‰¤ 200,Â xÂ â‰¥ 0, andÂ y â‰¥ 0, is given below.
A (0, 50), B (20, 40), C (50, 100) and D (0, 200) are the corner points of the feasible region. The values of Z at these corner points are given below.
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 points (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).
9. Maximise Z = âˆ’Â xÂ + 2y, subject to the constraints.
.
Solution:
The feasible region determined by the constraints,
is given below.
Here, 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 given below.
Corner point | Z = – x + 2y |
A (6, 0) | Z = – 6 |
B (4, 1) | Z = – 2 |
C (3, 2) | Z = 1 |
Since the feasible region is unbounded, z = 1 may or may not be the maximum value.
For this purpose, we graph the inequality, – x + 2y > 1, and check whether the resulting half-plane has points in common with the feasible region or not.
Here, the resulting feasible region has points in common with the feasible region.
Hence, z = 1 is not the maximum value.
Z has no maximum value.
10. Maximise Z =Â xÂ +Â y, subject to.
Solution:
The region determined by the constraintsÂ Â is given below.
There is no feasible region, and therefore, z has no maximum value.
Exercise 12.2 page no: 519
- Reshma wishes to mix two types of food, P and Q, in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A, and 11 units of vitamin B. Food P costs Rs 60/kg, and Food Q costs Rs 80/kg. Food P contains 3 units /kg of vitamin A and 5 units /kg of vitamin B while food Q contains 4 units /kg of vitamin A and 2 units /kg of vitamin B. Determine the minimum cost of the mixture?
Solution:
Let the mixture contain x kg of food P and y kg of food Q.
Hence, xÂ â‰¥ 0 andÂ yÂ â‰¥ 0
The given information can be compiled in a table, as given below.
Vitamin A (units/kg) | Vitamin B (units/kg | Cost (Rs/kg) | |
Food P | 3 | 5 | 60 |
Food Q | 4 | 2 | 80 |
Requirement (units/kg) | 8 | 11 |
The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Hence, the constraints are
3xÂ + 4yÂ â‰¥ 8
5xÂ + 2yÂ â‰¥ 11
The total cost of purchasing food is Z = 60x + 80y.
So, the mathematical formulation of the given problem can be written as
Minimise Z = 60x + 80y (i)
Now, subject to the constraints,
3xÂ + 4yÂ â‰¥ 8 â€¦ (2)
5xÂ + 2yÂ â‰¥ 11 â€¦ (3)
x,Â yÂ â‰¥ 0 â€¦ (4)
The feasible region determined by the system of constraints is given below.
Clearly, we can see that the feasible region is unbounded.
A (8 / 3, 0), B (2, 1 / 2) and C (0, 11 / 2)
The values of Z at these corner points are given below.
Corner point | Z = 60x + 80 y | |
A (8 / 3, 0) | 160 | Minimum |
B (2, 1 / 2) | 160 | Minimum |
C (0, 11 / 2) | 440 |
Here, the feasible region is unbounded; therefore, 160 may or may not be the minimum value of Z.
For this purpose, we graph the inequality, 60xÂ + 80yÂ < 160 or 3xÂ + 4y < 8, and check whether the resulting half-plane has points in common with the feasible region or not.
Here, it can be seen that the feasible region has no common point with 3x + 4y < 8.
Hence, at the line segment joining the points (8 / 3, 0) and (2, 1 / 2), the minimum cost of the mixture will be Rs 160.
2. One kind of cake requires 200g flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes?
Solution:
Let the first kind of cake be x and second kind of cakes be y. Hence,
xÂ â‰¥ 0 andÂ yÂ â‰¥ 0
The given information can be compiled in a table, as shown below.
Flour (g) | Fat (g) | |
Cakes of the first kind, x | 200 | 25 |
Cakes of the second kind, y | 100 | 50 |
Availability | 5000 | 1000 |
So, 200x + 100y â‰¤ 5000
2x + y â‰¤ 50
25x + 50y â‰¤ 1000
x + 2y â‰¤ 40
The total number of cakes Z that can be made are
Z = x + y
The mathematical formulation of the given problem can be written as
Maximise Z = x + y (i)
Here, subject to the constraints,
2x + y â‰¤ 50 (ii)
x + 2y â‰¤ 40 (iii)
x, y â‰¥ 0 (iv)
The feasible region determined by the system of constraints is given below.
A (25, 0), B (20, 10), O (0, 0) and C (0, 20) are the corner points.
The values of Z at these corner points are as given below.
Corner point | Z = x + y | |
A (25, 0) | 25 | |
B (20, 10) | 30 | Maximum |
C (0, 20) | 20 | |
O (0,0) | 0 |
Hence, the maximum numbers of cakes that can be made are 30 (20 cakes of one kind and 10 cakes of other kind).
3. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsmanâ€™s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftsmanâ€™s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsmanâ€™s time.
(ii) What number of rackets and bats must be made if the factory is to work at full capacity?
(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.
Solution:
Let x and y be the number of rackets and the number of bats to be made.
Given that the machine time is not available for more than 42 hours.
Hence, 1.5x + 3yÂ â‰¤Â 42 â€¦â€¦â€¦â€¦â€¦. (i)
Also, given that the craftsmanâ€™s time is not available for more than 24 hours.
Hence, 3x + y â‰¤ 24 â€¦â€¦â€¦â€¦ (ii)
The factory is to work at full capacity. Hence,
1.5x + 3y = 42
3x + y = 24
On solving these equations, we get
x = 4 and y = 12
Therefore, 4 rackets and 12 bats must be made.
(i) The given information can be compiled in a table as give below.
Tennis Racket | Cricket Bat | Availability | |
Machine Time (h) | 1.5 | 3 | 42 |
Craftsmanâ€™s Time (h) | 3 | 1 | 24 |
1.5x + 3y â‰¤Â 42
3x + y â‰¤Â 24
x, y â‰¥ 0
Since the profit on a racket is Rs 20 and Rs 10.
Hence, Z = 20x + 10y
The mathematical formulation of the given problem can be written as
Maximise Z = 20x + 10y â€¦â€¦â€¦â€¦.. (i)
Subject to the constraints,
1.5x + 3y â‰¤Â 42 â€¦â€¦â€¦â€¦. (ii)
3x + y â‰¤Â 24 â€¦â€¦â€¦â€¦â€¦.. (iii)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iv)
The feasible region determined by the system of constraints is given below.
A (8, 0), B (4, 12), C (0, 14) and O (0, 0) are the corner points respectively.
The values of Z at these corner points are given below.
Corner point | Z = 20x + 10y | |
A (8, 0) | 160 | |
B (4, 12) | 200 | Maximum |
C (0, 14) | 140 |
Therefore, the maximum profit of the factory when it works to its full capacity is Rs 200.
4. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs 17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?
Solution:
Let the manufacturer produce x package of nuts and y package of bolts. Hence,
x â‰¥ 0 and y â‰¥ 0
The given information can be compiled in a table, as given below.
Nuts | Bolts | Availability | |
Machine A (h) | 1 | 3 | 12 |
Machine B (h) | 3 | 1 | 12 |
The profit on a package of nuts is Rs 17.50 and on a package of bolts is Rs 7.
Hence, the constraints are
x + 3y â‰¤ 12
3x + y â‰¤ 12
Then, the total profit is Z = 17.5x + 7y
The mathematical formulation of the given problem can be written as follows:
Maximise Z = 17.5x + 7y â€¦â€¦â€¦â€¦. (1)
Subject to the constraints,
x + 3y â‰¤ 12 â€¦â€¦â€¦â€¦. (2)
3x + y â‰¤ 12 â€¦â€¦â€¦â€¦ (3)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦â€¦.. (4)
The feasible region determined by the system of constraints is given below.
A (4, 0), B (3, 3) and C (0, 4) are the corner points.
The values of Z at these corner points are given below.
Corner point | Z = 17.5x + 7y | |
O (0, 0) | 0 | |
A (4, 0) | 70 | |
B (3, 3) | 73.5 | Maximum |
C (0, 4) | 28 |
Therefore, Rs 73.50 at (3, 3) is the maximum value of Z.
Hence, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit of Rs 73.50.
5. A factory manufacturers two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.
Solution:
On each day, let the factory manufacture x screws of type A and y screws of type B.
Hence,
x â‰¥ 0 and y â‰¥ 0
The given information can be compiled in a table, as given below.
Screw A | Screw B | Availability | |
Automatic Machine (min) | 4 | 6 | 4 Ã— 60 = 240 |
Hand Operated Machine (min) | 6 | 3 | 4 Ã— 60 = 240 |
The profit on a package of screws A is Rs 7 and on the package screws, B is Rs 10.
Hence, the constraints are
4x + 6y â‰¤ 240
6x + 3y â‰¤ 240
Total profit, Z = 7x + 10y
The mathematical formulation of the given problem can be written as
Maximise Z = 7x + 10y â€¦â€¦â€¦â€¦. (i)
Subject to the constraints,
4x + 6y â‰¤ 240 â€¦â€¦â€¦â€¦. (ii)
6x + 3y â‰¤ 240 â€¦â€¦â€¦â€¦. (iii)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦â€¦â€¦ (iv)
The feasible region determined by the system of constraints is given below.
A (40, 0), B (30, 20) and C (0, 40) are the corner points.
The value of Z at these corner points is given below.
Corner point | Z = 7x + 10y | |
A (40, 0) | 280 | |
B (30, 20) | 410 | Maximum |
C (0, 40) | 400 |
The maximum value of Z is 410 at (30, 20).
Hence, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410.
6. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shade that he produces, how should he schedule his daily production in order to maximise his profit?
Solution:
Let the cottage industry manufacture x pedestal lamps and y wooden shades, respectively.
Hence,
x â‰¥ 0 and y â‰¥ 0
The given information can be compiled in a table, as given below.
Lamps | Shades | Availability | |
Grinding/Cutting Machine (h) | 2 | 1 | 12 |
Sprayer (h) | 3 | 2 | 20 |
The profit on a lamp is Rs 5 and on the shades is Rs 3. Hence, the constraints are
2x + y â‰¤ 12
3x + 2y â‰¤ 20
Total profit, Z = 5x + 3y â€¦â€¦â€¦â€¦â€¦.. (i)
Subject to the constraints,
2x + y â‰¤ 12 â€¦â€¦â€¦â€¦. (ii)
3x + 2y â‰¤ 20 â€¦â€¦â€¦â€¦ (iii)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦. (iv)
The feasible region determined by the system of constraints is given below.
A (6, 0), B (4, 4) and C (0, 10) are the corner points.
The value of Z at these corner points is given below.
Corner point | Z = 5x + 3y | |
A (6, 0) | 30 | |
B (4, 4) | 32 | Maximum |
C (0, 10) | 30 |
The maximum value of Z is 32 at points (4, 4).
Therefore, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximise his profits.
7. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?
Solution:
Let the company manufacture x souvenirs of type A and y souvenirs of type B, respectively.
Hence,
x â‰¥ 0 and y â‰¥ 0
The given information can be compiled in a table, as given below.
Type A | Type B | Availability | |
Cutting (min) | 5 | 8 | 3 Ã— 60 + 20 = 200 |
Assembling (min) | 10 | 8 | 4 Ã— 60 = 240 |
The profit on type A souvenirs is Rs 5 and on type B souvenirs is Rs 6. Hence, the constraints are
5x + 8y â‰¤ 200
10x + 8y â‰¤ 240 i.e.,
5x + 4y â‰¤ 120
Total profit, Z = 5x + 6y
The mathematical formulation of the given problem can be written as
Maximise Z = 5x + 6y â€¦â€¦â€¦â€¦â€¦ (i)
Subject to the constraints,
5x + 8y â‰¤ 200 â€¦â€¦â€¦â€¦â€¦. (ii)
5x + 4y â‰¤ 120 â€¦â€¦â€¦â€¦.. (iii)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦.. (iv)
The feasible region determined by the system of constraints is given below.
A (24, 0), B (8, 20) and C (0, 25) are the corner points.
The values of Z at these corner points are given below.
Corner point | Z = 5x + 6y | |
A (24, 0) | 120 | |
B (8, 20) | 160 | Maximum |
C (0, 25) | 150 |
The maximum value of Z is 200 at (8, 20).
Hence, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of Rs 160.
8. A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25,000 and Rs 40,000, respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4,500 and on the portable model is Rs 5,000.
Solution:
Let the merchant stock x desktop models and y portable models, respectively.
Hence,
x â‰¥ 0 and y â‰¥ 0
Given that the cost of desktop model is Rs 25,000 and of a portable model is Rs 40,000.
However, the merchant can invest a maximum of Rs 70 lakhs.
Hence, 25000x + 40000y â‰¤ 7000000
5x + 8y â‰¤ 1400
The monthly demand of computers will not exceed 250 units.
Hence, x + y â‰¤ 250
The profit on a desktop model is 4500, and the profit on a portable model is Rs 5000.
Total profit, Z = 4500x + 5000y
Therefore, the mathematical formulation of the given problem is
Maximum Z = 4500x + 5000y â€¦â€¦â€¦â€¦ (i)
Subject to the constraints,
5x + 8y â‰¤ 1400 â€¦â€¦â€¦â€¦ (ii)
x + y â‰¤ 250 â€¦â€¦â€¦â€¦.. (iii)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦. (iv)
The feasible region determined by the system of constraints is given below.
A (250, 0), B (200, 50) and C (0, 175) are the corner points.
The values of Z at these corner points are given below.
Corner point | Z = 4500x + 5000y | |
A (250, 0) | 1125000 | |
B (200, 50) | 1150000 | Maximum |
C (0, 175) | 875000 |
The maximum value of Z is 1150000 at (200, 50).
Therefore, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs 11,50,000.
9. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F_{1} and F_{2} are available. Food F_{1} costs Rs 4 per unit food and F_{2} costs Rs 6 per unit. One unit of food F_{1} contains 3 units of vitamin A and 4 units of minerals. One unit of food F_{2} contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for the diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Solution:
Let the diet contain x units of food F_{1} and y units of food F_{2}. Hence,
x â‰¥ 0 and y â‰¥ 0
The given information can be compiled in a table, as given below.
Vitamin A (units) | Mineral (units) | Cost per unit (Rs) | |
Food F_{1} (x) | 3 | 4 | 4 |
Food F_{2} (y) | 6 | 3 | 6 |
Requirement | 80 | 100 |
The cost of food F_{1} is Rs 4 per unit, and of food, F_{2} is Rs 6 per unit.
Hence, the constraints are
3x + 6y â‰¥ 80
4x + 3y â‰¥ 100
x, y â‰¥ 0
The total cost of the diet, Z = 4x + 6y
The mathematical formulation of the given problem can be written as
Minimise Z = 4x + 6y â€¦â€¦â€¦â€¦â€¦ (i)
Subject to the constraints,
3x + 6y â‰¥ 80 â€¦â€¦â€¦â€¦ (ii)
4x + 3y â‰¥ 100 â€¦â€¦â€¦. (iii)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦. (iv)
The feasible region determined by the constraints is given below.
We can see that the feasible region is unbounded.
A (80 / 3, 0), B (24, 4 / 3), and C (0, 100 / 3) are the corner points.
The values of Z at these corner points are given below.
Corner point | Z = 4x + 6y | |
A (80 / 3, 0) | 320 / 3 = 106.67 | |
B (24, 4 / 3) | 104 | Minimum |
C (0, 100 / 3) | 200 |
Here, the feasible region is unbounded, so 104 may or not be the minimum value of Z.
For this purpose, we draw a graph of the inequality, 4x + 6y < 104 or 2x + 3y < 52, and check whether the resulting half-plane has points in common with the feasible region or not.
Here, it can be seen that the feasible region has no common point with 2x + 3y < 52.
Hence, the minimum cost of the mixture will be Rs 104.
10. There are two types of fertilisers, F_{1} and F_{2}. F_{1} consists of 10% nitrogen and 6% phosphoric acid, and F_{2} consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F_{1} costs Rs 6/kg and F_{2} costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Solution:
Let the farmer buy x kg of fertiliser F_{1} and y kg of fertiliser F_{2}. Hence,
x â‰¥ 0 and y â‰¥ 0
The given information can be compiled in a table, as given below.
Nitrogen (%) | Phosphoric Acid (%) | Cost (Rs / kg) | |
F_{1} (x) | 10 | 6 | 6 |
F_{2} (y) | 5 | 10 | 5 |
Requirement (kg) | 14 | 14 |
F_{1} consists of 10% nitrogen, and F_{2} consists of 5% nitrogen.
However, the farmer requires at least 14 kg of nitrogen.
So, 10% of x + 5% of y â‰¥ 14
x / 10 + y / 20 â‰¥ 14
By L.C.M we get
2x + y â‰¥ 280
F_{1} consists of 6% phosphoric acid, and F_{2} consists of 10% phosphoric acid.
However, the farmer requires at least 14 kg of phosphoric acid.
So, 6% of x + 10 % of y â‰¥ 14
6x / 100 + 10y / 100 â‰¥ 14
3x + 5y â‰¥ 700
The total cost of fertilisers, Z = 6x + 5y
The mathematical formulation of the given problem can be written as
Minimise Z = 6x + 5y â€¦â€¦â€¦â€¦.. (i)
Subject to the constraints,
2x + y â‰¥ 280 â€¦â€¦â€¦ (ii)
3x + 5y â‰¥ 700 â€¦â€¦â€¦. (iii)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦. (iv)
The feasible region determined by the system of constraints is given below.
Here, we can see that the feasible region is unbounded.
A (700 / 3, 0), B (100, 80) and C (0, 280) are the corner points.
The values of Z at these points are given below.
Corner point | Z = 6x + 5y | |
A (700 / 3, 0) | 1400 | |
B (100, 80) | 1000 | Minimum |
C (0, 280) | 1400 |
Here, the feasible region is unbounded; hence, 1000 may or may not be the minimum value of Z.
For this purpose, we draw a graph of the inequality, 6x + 5y < 1000, and check whether the resulting half-plane has points in common with the feasible region or not.
Here, it can be seen that the feasible region has no common point with 6x + 5y < 1000
Hence, 100 kg of fertiliser F_{1} and 80 kg of fertiliser F_{2} should be used to minimise the cost. The minimum cost is Rs 1000.
11. The corner points of the feasible region are determined by the following system of linear inequalities:
2x + y â‰¤ 10, x + 3y â‰¤ 15, x, y â‰¥ 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
(A) p = q
(B) p = 2q
(C) p = 3q
(D) q = 3p
Solution:
The maximum value of Z is unique.
Here, it is given that the maximum value of Z occurs at two points, (3, 4) and (0, 5).
Value of Z at (3, 4) = Value of Z at (0, 5)
p (3) + q (4) = p (0) + q (5)
3p + 4q = 5q
3p = 5q â€“ 4q
3p = q or q = 3p
Therefore, the correct answer is option (D).
Miscellaneous Exercise page no: 607
1. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
Solution:
Let the diet contain x and y packets of foods P and Q, respectively. Hence,
x â‰¥ 0 and y â‰¥ 0
The mathematical formulation of the given problem is given below.
Maximise z = 6x + 3y â€¦â€¦â€¦â€¦.. (i)
Subject to the constraints,
4x + y â‰¥ 80 â€¦â€¦â€¦â€¦. (ii)
x + 5y â‰¥ 115 â€¦â€¦â€¦. (iii)
3x + 2y â‰¤ 150 â€¦â€¦â€¦… (iv)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦â€¦ (v)
The feasible region determined by the system of constraints is given below.
A (15, 20), B (40, 15) and C (2, 72) are the corner points of the feasible region.
The values of z at these corner points are given below.
Corner point | z = 6x + 3y | |
A (15, 20) | 150 | |
B (40, 15) | 285 | Maximum |
C (2, 72) | 228 |
So, the maximum value of z is 285 at (40, 15).
Hence, 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 in the diet is 285 units.
2. A farmer mixes two brands, P and Q, of cattle feed. Brand P, costing Rs 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag, contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag. What is the minimum cost of the mixture per bag?
Solution:
Let the farmer mix x bags of brand P and y bags of brand Q, respectively.
The given information can be compiled in a table, as given below.
Vitamin A (units/kg) | Vitamin B (units/kg) | Vitamin C (units/kg) | Cost (Rs/kg) | |
Food P | 3 | 2.5 | 2 | 250 |
Food Q | 1.5 | 11.25 | 3 | 200 |
Requirement (units/kg) | 18 | 45 | 24 |
The given problem can be formulated as given below.
Minimise z = 250x + 200y â€¦â€¦â€¦â€¦ (i)
3x + 1.5y â‰¥ 18 â€¦â€¦â€¦â€¦.. (ii)
2.5x + 11.25y â‰¥ 45 â€¦â€¦â€¦.. (iii)
2x + 3y â‰¥ 24 â€¦â€¦â€¦â€¦.. (iv)
x, y â‰¥ 0 â€¦â€¦â€¦. (v)
The feasible region determined by the system of constraints is given below.
A (18, 0), B (9, 2), C (3, 6) and D (0, 12) are the corner points of the feasible region.
The values of z at these corner points are given below.
Corner point | z = 250x + 200y | |
A (18, 0) | 4500 | |
B (9, 2) | 2650 | |
C (3, 6) | 1950 | Minimum |
D (0, 12) | 2400 |
Here, the feasible region is unbounded; hence, 1950 may or may not be the minimum value of z.
For this purpose, we draw a graph of the inequality, 250x + 200y < 1950 or 5x + 4y < 39, and check whether the resulting half-plane has points in common with the feasible region or not.
Here, it can be seen that the feasible region has no common point with 5x + 4y < 39
Hence, at points (3, 6), the minimum value of z is 1950.
Therefore, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimise the cost to Rs 1950.
3. A dietician wishes to mix together two kinds of food, X and Y, in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food are given below.
Food | Vitamin A | Vitamin B | Vitamin C |
X | 1 | 2 | 3 |
Y | 2 | 2 | 1 |
One kg of food X costs Rs 16, and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet.
Solution:
Let the mixture contain x kg of food X and y kg of food Y, respectively.
The mathematical formulation of the given problem can be written as given below.
Minimise z = 16x + 20y â€¦â€¦.. (i)
Subject to the constraints,
x + 2y â‰¥ 10 â€¦â€¦â€¦â€¦. (ii)
x + y â‰¥ 6 â€¦â€¦â€¦â€¦ (iii)
3x + y â‰¥ 8 â€¦â€¦â€¦â€¦. (iv)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦ (v)
The feasible region determined by the system of constraints is given below.
A (10, 0), B (2, 4), C (1, 5) and D (0, 8) are the corner points of the feasible region.
The values of z at these corner points are given below.
Corner point | z = 16x + 20y | |
A (10, 0) | 160 | |
B (2, 4) | 112 | Minimum |
C (1, 5) | 116 | |
D (0, 8) | 160 |
Since the feasible region is unbounded, hence, 112 may or may not be the minimum value of z.
For this purpose, we draw a graph of the inequality, 16x + 20y < 112 or 4x + 5y < 28, and check whether the resulting half-plane has points in common with the feasible region or not.
Here, it can be seen that the feasible region has no common point with 4x + 5y < 28
Hence, the minimum value of z is 112 at (2, 4).
Therefore, the mixture should contain 2 kg of food X and 4 kg of food Y.
The minimum cost of the mixture is Rs 112.
4. A manufacturer makes two types of toys, A and B. Three machines are needed for this purpose, and the time (in minutes) required for each toy on the machines is given below.
Types of Toys | Machines | ||
I | II | III | |
A | 12 | 18 | 6 |
B | 6 | 0 | 9 |
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Solution:
Let x and y toys of type A and type B be manufactured in a day, respectively.
The given problem can be formulated as given below.
Maximise z = 7.5x + 5y â€¦â€¦â€¦â€¦.. (i)
Subject to the constraints,
2x + y â‰¤ 60 â€¦â€¦â€¦â€¦. (ii)
x â‰¤ 20 â€¦â€¦â€¦.. (iii)
2x + 3y â‰¤ 120 â€¦â€¦â€¦.. (iv)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦â€¦. (v)
The feasible region determined by the constraints is given below.
A (20, 0), B (20, 20), C (15, 30) and D (0, 40) are the corner points of the feasible region.
The values of z at these corner points are given below.
Corner point | z = 7.5x + 5y | |
A (20, 0) | 150 | |
B (20, 20) | 250 | |
C (15, 30) | 262.5 | Maximum |
D (0, 40) | 200 |
262.5 at (15, 30) is the maximum value of z.
Hence, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximise the profit.
5. An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Solution:
Let the airline sell x tickets of executive class and y tickets of economy class, respectively.
The mathematical formulation of the given problem can be written as given below.
Maximise z = 1000x + 600y â€¦â€¦â€¦â€¦ (i)
Subject to the constraints,
x + y â‰¤ 200 â€¦â€¦â€¦â€¦â€¦.. (ii)
x â‰¥ 20 â€¦â€¦â€¦â€¦ (iii)
y â€“ 4x â‰¥ 0 â€¦â€¦â€¦â€¦â€¦ (iv)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦â€¦(v)
The feasible region determined by the constraints is given below.
A (20, 80), B (40, 160) and C (20, 180) are the corner points of the feasible region.
The values of z at these corner points are given below.
Corner point | z = 1000x + 600y | |
A (20, 80) | 68000 | |
B (40, 160) | 136000 | Maximum |
C (20, 180) | 128000 |
136000 at (40, 160) is the maximum value of z.
Therefore, 40 tickets for the executive class and 160 tickets for the economy class should be sold to maximise the profit, and the maximum profit is Rs 1,36,000.
6. Two godowns, A and B, have grain capacities of 100 quintals and 50 quintals, respectively. They supply to 3 ration shops, D, E and F, whose requirements are 60, 50 and 40 quintals, respectively. The cost of transportation per quintal from the godowns to the shops is given in the following table:
Transportation cost per quintal (in Rs) | ||
From / To | A | B |
D | 6 | 4 |
E | 3 | 2 |
F | 2.50 | 3 |
How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
Solution:
Let godown A supply x and y quintals of grain to shops D and E.
So, (100 â€“ x â€“ y) will be supplied to shop F.
Since x quintals are transported from godown A, so the requirement at shop D is 60 quintals. Hence, the remaining (60 â€“ x) quintals will be transported from godown B.
Similarly, (50 â€“ y) quintals and 40 â€“ (100 â€“ x â€“ y) = (x + y â€“ 60) quintals will be transported from godown B to shop E and F.
The given problem can be represented diagrammatically, as given below.
x â‰¥ 0, y â‰¥ 0, and 100 â€“ x â€“ y â‰¥ 0
Then, x â‰¥ 0, y â‰¥ 0, and x + y â‰¤ 100
60 â€“ x â‰¥ 0, 50 â€“ y â‰¥ 0, and x + y â€“ 60 â‰¥ 0
Then, x â‰¤ 60, y â‰¤ 50, and x + y â‰¥ 60
Total transportation cost z is given by,
z = 6x + 3y + 2.5 (100 â€“ x â€“ y) + 4 (60 â€“ x) + 2 (50 â€“ y) + 3 (x + y â€“ 60)
= 6x + 3y + 250 – 2.5x â€“ 2.5y + 240 â€“ 4x + 100 â€“ 2y + 3x + 3y â€“ 180
= 2.5x + 1.5y + 410
The given problem can be formulated as given below.
Minimise z = 2.5x + 1.5y + 410 â€¦â€¦â€¦â€¦. (i)
Subject to the constraints,
x + y â‰¤ 100 â€¦â€¦â€¦.. (ii)
x â‰¤ 60 â€¦â€¦â€¦.. (iii)
y â‰¤ 50 â€¦â€¦â€¦. (iv)
x + y â‰¥ 60 â€¦â€¦â€¦ (v)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦..(vi)
The feasible region determined by the system of constraints is given below.
A (60, 0), B (60, 40), C (50, 50) and D (10, 50) are the corner points.
The values of z at these corner points are given below.
Corner point | z = 2.5x + 1.5y + 410 | |
A (60, 0) | 560 | |
B (60, 40) | 620 | |
C (50, 50) | 610 | |
D (10, 50) | 510 | Minimum |
The minimum value of z is 510 at (10, 50)
Hence, the amount of grain transported from A to D, E and F is 10 quintals, 50 quintals and 40 quintals, respectively, and from B to D, E and F is 50 quintals, 0 quintals, 0 quintals, respectively
Thus, the minimum cost is Rs 510.
7. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L, respectively. The company is to supply oil to three petrol pumps, D, E and F, whose requirements are 4500L, 3000L and 3500L, respectively. The distances (in km) between the depots and the petrol pumps are given in the following table:
Distance in (km) | ||
From / To | A | B |
D | 7 | 3 |
E | 6 | 4 |
F | 3 | 2 |
Assuming that the transportation cost of 10 litres of oil is Rs 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Solution:
Let x and y litres of oil be supplied from A to the petrol pumps, D and E. So, (7000 â€“ x â€“ y) will be supplied from A to petrol pump F.
The requirement at petrol pump D is 4500L. Since x L is transported from depot A, the remaining (4500 â€“ x) L will be transported from petrol pump B.
Similarly, (3000 â€“ y)L and 3500 â€“ (7000 â€“ x â€“ y) = (x + y â€“ 3500) L will be transported from depot B to petrol pumps E and F, respectively.
The given problem can be represented diagrammatically, as given below.
x â‰¥ 0, y â‰¥ 0, and (7000 â€“ x â€“ y) â‰¥ 0
Then, x â‰¥ 0, y â‰¥ 0, and x + y â‰¤ 7000
4500 â€“ x â‰¥ 0, 3000 â€“ y â‰¥ 0, and x + y â€“ 3500 â‰¥ 0
Then, x â‰¤ 4500, y â‰¤ 3000, and x + y â‰¥ 3500
Cost of transporting 10L of petrol = Rs 1
Cost of transporting 1L of petrol = Rs 1 / 10
Hence, the total transportation cost is given by
z = (7 / 10) x + (6 / 10) y + 3 / 10 (7000 â€“ x â€“ y) + 3 / 10 (4500 â€“ x) + 4 / 10 (3000 â€“ y) + 2 / 10 (x + y â€“ 3500)
= 0.3x + 0.1y + 3950
The problem can be formulated as given below.
Minimise z = 0.3x + 0.1y + 3950 â€¦â€¦â€¦. (i)
Subject to constraints,
x + y â‰¤ 7000 â€¦â€¦â€¦. (ii)
x â‰¤ 4500 â€¦â€¦â€¦.. (iii)
y â‰¤ 3000 â€¦â€¦. (iv)
x + y â‰¥ 3500 â€¦â€¦â€¦ (v)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦ (vi)
The feasible region determined by the constraints is given below.
A (3500, 0), B (4500, 0), C (4500, 2500), D (4000, 3000) and E (500, 3000) are the corner points of the feasible region.
The values of z at these corner points are given below.
Corner point | z = 0.3x + 0.1y + 3950 | |
A (3500, 0) | 5000 | |
B (4500, 0) | 5300 | |
C (4500, 2500) | 5550 | |
D (4000, 3000) | 5450 | |
E (500, 3000) | 4400 | Minimum |
The minimum value of z is 4400 at (500, 3000).
Hence, the oil supplied from depot A is 500 L, 3000 L and 3500 L and from depot B is 4000 L, 0 L and 0 L to petrol pumps D, E and F, respectively.
Therefore, the minimum transportation cost is Rs 4400.
8. A fruit grower can use two types of fertiliser 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.
If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added to the garden?
Kg per bag | ||
Brand P | Brand Q | |
Nitrogen | 3 | 3.5 |
Phosphoric acid | 1 | 2 |
Potash | 3 | 1.5 |
Chlorine | 1.5 | 2 |
Solution:
Let the fruit grower use x bags of brand P and y bags of brand Q, respectively.
The problem can be formulated as given below.
Minimise z = 3x + 3.5y â€¦â€¦â€¦â€¦. (i)
Subject to the constraints,
x + 2y â‰¥ 240 â€¦â€¦â€¦.. (ii)
x + 0.5y â‰¥ 90 â€¦â€¦.. (iii)
1.5x + 2y â‰¤ 310 â€¦â€¦â€¦.. (iv)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦. (v)
The feasible region determined by the system of constraints is given below.
A (240, 0), B (140, 50) and C (20, 140) are the corner points of the feasible region.
The value of z at these corner points is given below.
Corner point | z = 3x + 3.5y | |
A (140, 50) | 595 | |
B (20, 140) | 550 | |
C (40, 100) | 470 | Minimum |
The maximum value of z is 470 at (40, 100).
Therefore, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimise the amount of nitrogen.
Hence, the minimum amount of nitrogen added to the garden is 470 kg.
9. Refer to Question 8. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
Solution:
Let the fruit grower use x bags of brand P and y bags of brand Q, respectively.
The problem can be formulated as given below.
Maximise z = 3x + 3.5y â€¦â€¦â€¦.. (i)
Subject to the constraints,
x + 2y â‰¥ 240 â€¦â€¦â€¦.. (ii)
x + 0.5y â‰¥ 90 â€¦â€¦â€¦.. (iii)
1.5x + 2y â‰¤ 310 â€¦â€¦â€¦.. (iv)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦. (v)
The feasible region determined by the system of constraints is given below.
A (140, 50), B (20, 140) and C (40, 100) are the corner points of the feasible region.
The values of z at these corner points are given below.
Corner point | z = 3x + 3.5y | |
A (140, 50) | 595 | Maximum |
B (20, 140) | 550 | |
C (40, 100) | 470 |
The maximum value of z is 595 at (140, 50).
Hence, 140 bags of brand P and 50 bags of brand Q should be used to maximise the amount of nitrogen.
Thus, the maximum amount of nitrogen added to the garden is 595 kg.
10. A toy company manufactures two types of dolls, A and B. Market research and available resources have indicated that the combined production level should not exceed 1200 dolls per week, and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll, respectively, on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
Solution:
Let x and y be the number of dolls of type A and B, respectively, that are produced in a week.
The given problem can be formulated as given below.
Maximise z = 12x + 16y â€¦â€¦â€¦.. (i)
Subject to the constraints,
x + y â‰¤ 1200 â€¦â€¦â€¦â€¦ (ii)
y â‰¤ x / 2 or x â‰¥ 2y â€¦â€¦â€¦. (iii)
x â€“ 3y â‰¤ 600 â€¦â€¦â€¦â€¦. (iv)
x, y â‰¥ 0 â€¦â€¦â€¦â€¦â€¦ (v)
The feasible region determined by the system of constraints is given below.
A (600, 0), B (1050, 150) and C (800, 400) are the corner points of the feasible region.
The values of z at these corner points are given below.
Corner point | z = 12x + 16y | |
A (600, 0) | 7200 | |
B (1050, 150) | 15000 | |
C (800, 400) | 16000 | Maximum |
The maximum value of z is 16000 at (800, 400).
Hence, 800 and 400 dolls of type A and type B should be produced, respectively, to get the maximum profit of Rs 16000.
Also AccessÂ |
NCERT Exemplar for Class 12 Maths Chapter 12 |
CBSE Notes for Class 12 Maths Chapter 12 |
Beneath are the theoretical concepts covered in Class 12 Maths Chapter 12 Linear Programming
12.1 Introduction
This section recollects the discussion of linear equations, linear inequalities, and applications of linear inequalities in previous grades. It introduces the concept of optimisation problems and a special case of optimisation problem called linear programming problem, using an example. An ideal example of optimisation would be maximising the profit and minimising the cost of a production unit.Â
12.2 Linear Programming Problem and its Mathematical Formulation
This section consists of an example of a furniture dealer who is trying to maximise profits by choosing and experimenting with different combinations of buying chairs and tables.
12.2.1 Mathematical formulation of the problem
- This section further explains the formulation of the above-mentioned mathematical problem. It defines the non-negative constraints, objective function, and decision variables.
- The optimal value of a linear function can be defined as an objective function.
- A linear programming problem is finding the optimal value [maximum or minimum] of a linear function of variables, which are subjected to certain conditions and satisfying a set of linear constraints.
- The variables involved in the objective function are called decision variables.
- The constraints are the restrictions on the variables.
12.2.2 Graphical method of solving linear programming problems
- This section comprises the definition of the feasible region, feasible solution and infeasible solution, optimal solution, and bounded and unbounded region of feasible solution. It briefs about the Corner Point Method, which is used to solve linear programming problems with solved examples.
- The region obtained by the constraints [including non-negative constraints] can be termed as the feasible region.
- The points within the feasible region are called feasible solutions.
- The points outside the feasible region are called infeasible solutions.
- The point in the feasible region which gives the optimal value of the objective function is called an optimal solution.
12.3 Different Types of Linear Programming Problems
In this section, the different types of linear programming problems are discussed.
12.3.1 Manufacturing problems
- These problems can be seen in the manufacturing sector in order to optimise production by maximising profits. The profits can be a function of the number of workers, working hours, materials required, the value of the product in the market, the demand for the product, the supply of the product etc.
12.3.2 Diet problems
- Such problems involve the optimisation of the amount of intake of different types of foods, which are required by the body to obtain necessary nutrients. The agenda of a diet problem will be selecting those foods with the required nutrient at a lesser cost.
12.3.3 Transportation problems
- These problems involve the transportation of manufactured goods efficiently to different places, such that the transportation cost is minimum. For big companies, the analysis of transportation costs is very much important, as it caters to a wide area.Â
Students can utilise the NCERT Solutions for Class 12 Maths Chapter 12 for any quick reference to comprehend complex topics.
Access NCERT Solutions for Class 12 Maths Chapter 12 exercise-wise from the below-given links
Exercise 12.1 SolutionsÂ – 10 Questions
Exercise 12.2 SolutionsÂ – 11 Questions
Miscellaneous Exercise on Chapter 12 SolutionsÂ – 10 Questions
A few points on NCERT Class 12 Maths Chapter 12 Linear Programming
The chapter Linear Programming makes up a whole unit that carries five marks of the total eighty marks. There are two exercises, along with a miscellaneous exercise, in this chapter to help the students understand the concepts related to Linear Programming thoroughly. Some of the topics discussed in Chapter 12 of NCERT Solutions for Class 12 Maths are as follows:
1. A linear programming problem is concerned with finding the optimal value (maximum or minimum) of a linear function of several variables (called objective function), subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities (called linear constraints). Variables are sometimes called decision variables and are non-negative.
2. A few important linear programming problems are
- Diet problems
- Manufacturing problems
- Transportation problems
3. The common region determined by all the constraints, including the non-negative constraints x â‰¥ 0, y â‰¥ 0 of a linear programming problem, is called the feasible region (or solution region) for the problem.
4. Points within and on the boundary of the feasible region represent feasible solutions to the constraints.
- Any point outside the feasible region is an infeasible solution.
- Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution.
5.Â If the feasible region is unbounded, then a maximum or a minimum may not exist. However, if it exists, it must occur at a corner point of R.
6. The corner point method is used in solving a linear programming problem.
7. If two corner points of the feasible region are both optimal solutions of the same type, i.e., both produce the same maximum or minimum, then any point on the line segment joining these two points is also an optimal solution of the same type.
Studying the Linear Programming of Class 12 usingÂ NCERT SolutionsÂ enables the students to understand the following:
Introduction, related terminology, such as constraints, objective function, optimisation, different types of linear programming (L.P.) problems, mathematical formulation of L.P. problems, graphical method of solution for problems in two variables, feasible and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints). Furthermore, NCERT Solutions improve the confidence among students while facing other exams in the future.
Disclaimer –Â
Dropped Topics –Â 12.3 Different Types of Linear Programming Problems andÂ Page 528-529 Summary Points 2-9
Comments