Chapter 12 Linear Programming Ex 12.2 Solutions
Question - 1 : - 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?
Answer - 1 : -
Let the mixture contain x kg of food P and y kg of food Q.
Hence, x ≥ 0 and y ≥0
The given information can be compiled in a table as given
| 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 11units of vitamin B. Hence, the constraints are
3x + 4y ≥ 8
5x + 2y ≥ 11
Total cost of purchasing food is, Z = 60x + 80y
So, the mathematical formulation of the given problem can bewritten as
Minimise Z = 60x + 80y (i)
Now, subject to the constraints,
3x + 4y ≥ 8 … (2)
5x + 2y ≥ 11 … (3)
x, y ≥0 … (4)
The feasible region determined by the system of constraints isgiven 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 maynot be the minimum value of Z.
For this purpose, we graph the inequality, 60x +80y <160 or 3x +4y <8, and check whether the resulting half plane has points in common with thefeasible region or not
Here, it can be seen that the feasible region has no commonpoint 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
Question - 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?
Answer - 2 : -
Let the first kind of cakes be x and second kind of cakes be y.Hence,
x ≥0 and y ≥0
The given information can be compiled in a table as shown below
| Flour (g) | Fat (g) |
| Cakes of first kind, x | 200 | 25 |
| Cakes of second kind, y | 100 | 50 |
| Availability | 5000 | 1000 |
So, 200x + 100y ≤ 5000
2x + y ≤ 50
25x + 50y ≤ 1000
x + 2y ≤ 40
Total number of cakes Z that can be made are
Z = x + y
The mathematical formulation of the given problem can be writtenas
Maximize Z = x + y (i)
Here, subject to the constraints,
2x + y ≤ 50 (ii)
x + 2y ≤ 40 (iii)
x, y ≥ 0 (iv)
The feasible region determined by the system of constraints isgiven as below
A (25, 0), B (20, 10), O (0, 0) and C (0, 20) are the cornerpoints
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 (20cakes of one kind and 10 cakes of other kind)
Question - 3 : - A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’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.
Answer - 3 : -
Let x and y be the number of rackets and the number of bats tobe made.
Given that the machine time is not available for more than 42hours
Hence, 1.5x + 3y ≤ 42 ……………. (i)
Also, given that the craftsman’s time is not available for morethan 24 hours
Hence, 3x + y ≤ 24 ………… (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 givebelow
| Tennis Racket | Cricket Bat | Availability |
| Machine Time (h) | 1.5 | 3 | 42 |
| Craftsman’s Time (h) | 3 | 1 | 24 |
1.5x + 3y ≤ 42
3x + y ≤ 24
x, y ≥ 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 writtenas
Maximize Z = 20x + 10y ………….. (i)
Subject to the constraints,
1.5x + 3y ≤ 42 …………. (ii)
3x + y ≤ 24 …………….. (iii)
x, y ≥ 0 ………………… (iv)
The feasible region determined by the system of constraints isgiven below
A (8, 0), B (4, 12), C (0, 14) and O (0, 0) are the cornerpoints 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 toits full capacity is Rs 200
Question - 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 maximize his profit, if he operates his machines for at the most 12 hours a day?
Answer - 4 : -
Let the manufacturer produce x package of nuts and y package ofbolts. Hence,
x ≥ 0 and y ≥ 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 ofbolts is Rs 7
Hence, the constraints are
x + 3y ≤ 12
3x + y ≤ 12
Then, total profit, Z = 17.5x + 7y
The mathematical formulation of the given problem can be writtenas follows
Maximize Z = 17.5x + 7y …………. (1)
Subject to the constraints,
x + 3y ≤ 12 …………. (2)
3x + y ≤ 12 ………… (3)
x, y ≥ 0 …………….. (4)
The feasible region determined by the system of constraints isgiven 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 beproduced each day to get the maximum profit of Rs 73.50
Question - 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 maximize his profit? Determine the maximum profit.
Answer - 5 : -
On each day, let the factory manufacture x screws of type A andy screws of type B.
Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as given below
| Screw A | Screw B | Availability |
| Automatic Machine (min) | 4 | 6 | 4 × 60 = 240 |
| Hand Operated Machine (min) | 6 | 3 | 4 × 60 = 240 |
The profit on a package of screws A is Rs 7 and on the packagescrews B is Rs 10
Hence, the constraints are
4x + 6y ≤ 240
6x + 3y ≤ 240
Total profit, Z = 7x + 10y
The mathematical formulation of the given problem can be writtenas
Maximize Z = 7x + 10y …………. (i)
Subject to the constraints,
4x + 6y ≤ 240 …………. (ii)
6x + 3y ≤ 240 …………. (iii)
x, y ≥ 0 ……………… (iv)
The feasible region determined by the system of constraints isgiven below
A (40, 0), B (30, 20) and C (0, 40) are the corner points
The value of Z at these corner points are 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 20packages of screws B to get the maximum profit of Rs 410
Question - 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 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 maximize his profit?
Answer - 6 : -
Let the cottage industry manufacture x pedestal lamps and ywooden shades respectively
Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table is 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 ≤ 12
3x + 2y ≤ 20
Total profit, Z = 5x + 3y …………….. (i)
Subject to the constraints,
2x + y ≤ 12 …………. (ii)
3x + 2y ≤ 20 ………… (iii)
x, y ≥ 0 …………. (iv)
The feasible region determined by the system of constraints isgiven below
A (6, 0), B (4, 4) and C (0, 10) are the corner points
The value of Z at these corner points are 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 point (4, 4)
Therefore, the manufacturer should produce 4 pedestal lamps and4 wooden shades to maximize his profits.
Question - 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 maximize the profit?
Answer - 7 : -
Let the company manufacture x souvenirs of type A and ysouvenirs of type B respectively
Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table is given below
| Type A | Type B | Availability |
| Cutting (min) | 5 | 8 | 3 × 60 + 20 = 200 |
| Assembling (min) | 10 | 8 | 4 × 60 = 240 |
The profit on type A souvenirs is Rs 5 and on type B souvenirsis Rs 6. Hence, the constraints are
5x + 8y ≤ 200
10x + 8y ≤ 240 i.e.,
5x + 4y ≤ 120
Total profit, Z = 5x + 6y
The mathematical formulation of the given problem can be writtenas
Maximize Z = 5x + 6y …………… (i)
Subject to the constraints,
5x + 8y ≤ 200 ……………. (ii)
5x + 4y ≤ 120 ………….. (iii)
x, y ≥ 0 ………….. (iv)
The feasible region determined by the system of constraints isgiven 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 shouldbe produced each day to get the maximum profit of Rs 160
Question - 8 : - A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 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 4500 and on portable model is Rs 5000.
Answer - 8 : -
Let the merchant stock x desktop models and y portable modelsrespectively.
Hence,
x ≥ 0 and y ≥ 0
Given that the cost of desktop model is Rs 25000 and of aportable model is Rs 40000.
However, the merchant can invest a maximum of Rs 70 lakhs
Hence, 25000x + 40000y ≤ 7000000
5x + 8y ≤ 1400
The monthly demand of computers will not exceed 250 units.
Hence, x + y ≤ 250
The profit on a desktop model is 4500 and the profit on aportable model is Rs 5000
Total profit, Z = 4500x + 5000y
Therefore, the mathematical formulation of the given problem is
Maximum Z = 4500x + 5000y ………… (i)
Subject to the constraints,
5x + 8y ≤ 1400 ………… (ii)
x + y ≤ 250 ………….. (iii)
x, y ≥ 0 …………. (iv)
The feasible region determined by the system of constraints isgiven 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 50portable models to get the maximum profit of Rs 1150000.
Question - 9 : - A diet is tocontain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs 4 per unit food and F2 costs Rs 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit offood F2 contains 6 units of vitaminA and 3 units of minerals. Formulate this as a linear programming problem. Findthe minimum cost for diet that consists of mixture of these two foods and alsomeets the minimal nutritional requirements.
Answer - 9 : -
Let the diet contain x units of food F1 andy units of food F2.Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table is given below
| Vitamin A (units) | Mineral (units) | Cost per unit (Rs) |
| Food F1 (x) | 3 | 4 | 4 |
| Food F2 (y) | 6 | 3 | 6 |
| Requirement | 80 | 100 | |
The cost of food F1 is Rs 4 per unit and of food F2 isRs 6 per unit
Hence, the constraints are
3x + 6y ≥ 80
4x + 3y ≥ 100
x, y ≥ 0
Total cost of the diet, Z = 4x + 6y
The mathematical formulation of the given problem can be writtenas
Minimise Z = 4x + 6y …………… (i)
Subject to the constraints,
3x + 6y ≥ 80 ………… (ii)
4x + 3y ≥ 100 ………. (iii)
x, y ≥ 0 …………. (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 cornerpoints
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 theminimum 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 haspoints in common with the feasible region or not
Here, it can be seen that the feasible region has no commonpoint with 2x + 3y < 52
Hence, the minimum cost of the mixture will be Rs 104
Question - 10 : - There are two typesof fertilizers F1 and F2. F1 consists of 10% nitrogenand 6% phosphoric acid and F2 consists of 5% nitrogen and10% phosphoric acid. After testing the soil conditions, a farmer finds that sheneeds atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6 / kg and F2 costs Rs 5 / kg, determinehow much of each type of fertilizer should be used so that nutrientrequirements are met at a minimum cost. What is the minimum cost?
Answer - 10 : -
Let the farmer buy x kg of fertilizer F1 andy kg of fertilizer F2.Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table is given below
| Nitrogen (%) | Phosphoric Acid (%) | Cost (Rs / kg) |
| F1 (x) | 10 | 6 | 6 |
| F2 (y) | 5 | 10 | 5 |
| Requirement (kg) | 14 | 14 | |
F1 consistsof 10% nitrogen and F2 consistsof 5% nitrogen.
However, the farmer requires at least 14 kg of nitrogen
So, 10% of x + 5% of y ≥ 14
x / 10 + y / 20 ≥ 14
By L.C.M we get
2x + y ≥ 280
F1 consistsof 6% phosphoric acid and F2 consists of 10% phosphoric acid.
However, the farmer requires at least 14 kg of phosphoric acid
So, 6% of x + 10 % of y ≥ 14
6x / 100 + 10y / 100 ≥ 14
3x + 5y ≥ 700
Total cost of fertilizers, Z = 6x + 5y
The mathematical formulation of the given problem can be writtenas
Minimize Z = 6x + 5y ………….. (i)
Subject to the constraints,
2x + y ≥ 280 ……… (ii)
3x + 5y ≥ 700 ………. (iii)
x, y ≥ 0 …………. (iv)
The feasible region determined by the system of constraints isgiven 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 maynot 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 withthe feasible region or not.
Here, it can be seen that the feasible region has no commonpoint with 6x + 5y < 1000
Hence, 100 kg of fertilizer F1 and 80 kg of fertilizer F2 should be used to minimize the cost. The minimum cost isRs 1000