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Linear Programming

Class 12th Mathematics Part Ii CBSE Solution
Exercise 12.1
  1. Maximise Z = 3x + 4y subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.…
  2. Minimise Z = - 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.…
  3. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.…
  4. Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
  5. Maximize Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.…
  6. Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. Show that the…
  7. Minimize and Maximize Z = 5x + 10 y subject to x + 2y ≤ 120, x + y ≥ 60, x - 2y…
  8. Minimize and Maximize Z = x + 2y subject to x + 2y ≥ 100, 2x - y ≤ 0, 2x + y ≤…
  9. Maximize Z = - x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥…
  10. Maximize Z = x + y, subject to x - y ≤ -1, -x + y ≤ 0, x, y ≥ 0.
Exercise 12.2
  1. Reshma wishes to mix two types of food P and Q in such a way that the vitamin…
  2. One kind of cake requires 200g of flour and 25g of fat, and another kind of…
  3. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5…
  4. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A…
  5. A factory manufactures two types of screws, A and B. Each type of screw…
  6. A cottage industry manufactures pedestal lamps and wooden shades, each…
  7. A company manufactures two types of novelty souvenirs made of plywood.…
  8. A merchant plans to sell two types of personal computers - a desktop model and…
  9. A diet is to contain at least 80 units of vitamin A and 100 units of minerals.…
  10. There are two types of fertilisers F1 and F2. F1 consists of 10% nitrogen and…
  11. The corner points of the feasible region determined by the following system of…
Miscellaneous Exercise
  1. Refer to Example 9. How many packets of each food should be used to maximize…
  2. A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per…
  3. A dietician wishes to mix together two kinds of food X and Y in such a way that…
  4. A manufacturer makes two types of toys A and B. Three machines are needed for…
  5. An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made…
  6. Two godowns A and B have grain capacity of 100 quintals and 50 quintals…
  7. An oil company has two depots A and B with capacities of 7000 L and 4000 L…
  8. A fruit grower can use two types of fertilizer in his garden, brand P and brand…
  9. Refer to Question 8. If the grower wants to maximise the amount of nitrogen…
  10. A toy company manufactures two types of dolls, A and B. Market tests and…
Exercise 12.1
Question 1.

Maximise Z = 3x + 4y subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.
Answer:

It is given in the question that,

Z = 3x + 4y


We have to subject on the following equation:


x ≥ 0, y ≥ 0, x + y ≤ 4



(x, y) = (0, 4), (4, 0)



We can clearly see that Z is maximum at (0, 4). Hence, maximum value of Z will be 16
Question 2.

Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
Answer:

It is given in the question that,

Z = - 3x + 4y


We have to subject on the following equation:


x +2y ≤ 8, 3x+2y ≤ 12, x ≥ 0, y ≤ 0


3x+2y ≤ 12



(x, y) = (0, 6), (4, 0)


x+2y ≤ 8



(x, y) = (0, 4), (8, 0)




We can clearly see that Z is minimum at (4, 0). Hence, minimum value of Z will be - 12
Question 3.

Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
Answer:

It is given in the question that,

Z = 5x + 3y


We have to subject on the following equation:


3x +5y ≤ 15, 5x+2y ≤ 10, x ≥ 0, y ≤ 0


5x+2y ≤ 10



(x, y) = (0, 5), (2, 0)




(x, y) = (0, 3), (5, 0)




We can clearly see that Z is maximum at . Hence, maximum value of Z will be



Question 4.

Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Answer:

It is given in the question that,

Z = 3x + 5y


We have to subject on the following equation:


x +3y ≥ 3, x+2y ≥ 2, x ≥ 0, y ≤ 0


x+y ≥ 2



(x, y) = (0, 2), (2, 0)


x+3y ≥ 3



(x, y) = (0, 1), (3, 0)



We can clearly see that the feasible region is unbounded and the corner points of the feasible region are



As per the table minimum value of Z is 7 but we can see that the feasible region is unbounded. Thus, 7 may or may not be the minimum value of Z


∴ We will plot the graph of inequality,


Here we will see is the resulting half plane has common points with the feasible region or not. As per the graph we can observe that the feasible region has no points in common with 3x + 5y < 7.


Hence, the minimum value of 7 is at
Question 5.

Maximize Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
Answer:

It is given in the question that,

Z = 3x + 2y


We have to subject on the following equation:


x +2y ≤ 10, 3x+y ≤ 15, x, y ≥ 0


3x+y ≤ 15



(x, y) = (0, 15), (5, 0)




(x, y) = (0, 5), (10, 0)




We can clearly see that Z is maximum at (4, 3). Hence, maximum value of Z will be 18
Question 6.

Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. Show that the minimum of Z occurs at more than two points.
Answer:

It is given in the question that,

Z = x + 2y


We have to subject on the following equation:





(x, y) = (0, 3), (6, 0)




(x, y) = (0, 3), (1, 1)




As per the table minimum value of Z is 6 but we can see that the feasible region is unbounded. Thus, 6 may or may not be the minimum value of Z


∴ We will plot the graph of inequality,


Here we will see that there is no common point with the feasible region.


Hence, Z will be minimum on all points joining line (0, 3), (6, 0). Therefore, Z will be minimum on x + 2y = 6
Question 7.

Minimize and Maximize Z = 5x + 10 y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.
Answer:

It is given in the question that,

Minimize and Maximize, Z = 5x + 10y


We have to subject on the following equation:





(x, y) = (0, 60), (120, 0)




(x, y) = (0, 0), (20, 10)




(x, y) = (60, 0), (0, 60)





∴ It is clear that at (60, 0) Z has its minimum value i.e. 300


Also, Z is minimum at two pints (60, 30) and (120, 0)


Hence, the value of Z will be maximum at all points joining (60, 30) and (120, 0)
Question 8.

Minimize and Maximize Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.
Answer:

It is given in the question that,

Minimize and Maximize, Z = x + 2y


We have to subject on the following equation:





(x, y) = (100, 0), (0, 200)




(x, y) = (0, 50), (100, 0)




(x, y) = (0, 0), (50, 100)





∴ It is clear that at (0, 200) Z has its maximum value i.e. 400


Also, Z is minimum at two pints (0, 50) and (20, 40)


Hence, the value of Z will be minimum at all points joining (0, 50) and (20, 40)
Question 9.

Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
Answer:

It is given in the question that,

Z = - x + 2y


We have to subject on the following equation:





(x, y) = (5, 0), (0, 5)




(x, y) = (6, 0), (3, 0)



We can clearly see that the feasible region is unbounded and the corner points of the feasible region are (3, 2), (4, 1) and (6, 0)



As per the table maximum value of Z is 1 but we can see that the feasible region is unbounded. Thus, 1 may or may not be the minimum value of Z


∴ We will plot the graph of inequality,


Here we will see is the resulting half plane has common points with the feasible region or not. As per the graph we can observe that the feasible region have some points in common with - x + 2y > 1.


Hence, there is no maximum value of Z
Question 10.

Maximize Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.
Answer:

It is given in the question that,

Z = x + y


We have to subject on the following equation:





(x, y) = (0, 1), (- 1, 0)




(x, y) = (0, 0), (1, 1)



From the above graph it is clear that, there is no common region or say feasible region. Hence, for the given condition Z has no maximum value
Exercise 12.2
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:

Let the mixture consists of x kg of food P and y kg of food Q. ∴ x ≥ 0 and y ≥ 0


According to given situation in the question, the information can be tabulated as follows:



The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B.


∴ The constraints are:


3x + 4y ≥ 8 and


5x + 2y ≥ 11.


Where x ≥ 0 and y ≥ 0


Total cost of ‘Z’ of purchasing food is Z= 6ox+ 80y


The mathematical formulation of the given problem is: minimize Z= 6ox + 80y…….1


Subject to constraints


3x + 4y ≥ 8 and


5x + 2y ≥ 11.


Where x , y ≥ 0



The graphical representation shows the feasible region is unbounded.


The corner points of the feasible region are


The values of Z at these corner points are



As the feasible region is unbounded


∴ 160 may or may not be the minimum value of Z


But it can be seen that the feasible region has no common oint with 3x+4y <8


So the minimum cost of the mixture will be Rs160 at the line segment joining the points



Question 2.

One kind of cake requires 200g of 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 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.
Answer:

let there be x cakes of first kind and y cakes of second kind.

∴ x≥ 0 and y ≥ 0.


The information given in the question can be complied in given form



200x + 100y ≤ 5000 ⇒ 2x+ y ≤ 50


& 25x + 50y ≤ 1000 ⇒ x +2y ≤ 40


Let Z be the total number of cakes that can be made


⇒ Z =X=y


Mathematical formulation of the given problem is


Maximize Z =x+y


Subject to constraint 2x+ y ≤ 50 and x +2y ≤ 40 where x, y ≥ 0


The graphical representation shows the feasible region determined by the system of constraints.



The corner points A(25, 0) , B( 20,10), 0(0,0) and C(0,20)


The values of z at these corner points are as follows



Thus, the maximum number of cakes that can be made are 30 (20 of one kind and 10 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 craftman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftman’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.

(i) 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:

Let the number of rackets and the number of bats to be made be x and y respectively.


The machine time available is not more than 42 hours.


∴ 1.5x +3y ≤ 42


The craftsman`s time is not available for more than 24 hours


∴ 3x + y ≤ 24


The factory is to work at full capacity


∴ 1.5x + 3y = 42


3x + y = 24


Where x and y ≥ 0


Solving the two equations we get x= 4 and y = 12


Thus 4 rackets and 12 bats must be made


(i) The given information can be complied in the tables form as follows



The profit on a racket is Rs 20 and on bat is Rs 10


Maximize, Z = 20x +10y ………………1


Subject to constraints


1.5x + 3y = 42


3x + y = 24


x, y ≥ 0


the feasible region determined by the system of constraints is:



The corner points are A(8,0) , B(4, 12) , C(0, 14) and O(0,0)


The values of Z at these corner points are as follows:



Thus the maximum profit of the factory is when it works to its full capacity is Rs200.
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 Rs17.50 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?
Answer:

let the number of packages of nuts produce be x

Let the number of bolts produced be y


The tabular representation of data given is as follows:



The system of constraints is given as


x + 3y ≤ 12


& 3y + x ≤ 12


Where x and y ≥ 0


We need to maximize the profit, so the function here will be maximizing Z


Maximize Z = 17.5x+ 7.5y


Subject to constraints


x + 3y ≤ 12


& 3y + x ≤ 12


Where x and y ≥ 0


The corner points are A(4 , 0) , B( 3,3) , C( 0 , 4) and O( 12, 0)




Hence the profit will be maximum if the company produces


Number of bolts – 3 packages


Number of nuts – 3 packages


Maximum profit is Rs 73.50
Question 5.

A factory manufactures 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.
Answer:

Let the factory manufacture x screws of type A and y screws of type B on each day.


∴ x and y ≥ 0


The tabular form of the given data in the question is



The profit on a package of screws A is Rs 7 and on package of screws B is Rs 10.


∴ the constraints are


4x+ 6y ≤ 240


And 6x + 3y ≤ 240 and x, y ≥ 0


And total profit is Z = 7x+ 10y


The mathematical formulation of the data is


Maximizing Z = 7x+ 10y


Subject to constraints


4x+ 6y ≤ 240


And 6x + 3y ≤ 240 and x, y ≥ 0


The feasible reason is determined by the system of constraints is:



The corner points are A(40, 0) , B( 30 , 20) , C ( 0 , 40)


The value of Z at these points is



The maximum value of Z is at point B(30, 20)


Therefore factory should produce 30 packages of screw A and 20 packages of screw 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 shades that he produces, how should he schedule his daily production in order to maximise his profit?
Answer:

let the cottage industry manufacture x number of pedestal lamps and y number of wooden shades.

∴ x and y ≥ 0


The given information can be represented in the following tabular form:



The profit on lamps is Rs 5 and that on the shades is Rs 3.


∴ The constraints are 2x + y ≤ 12


And 3x + 2y ≤ 20


x, y ≥ 0


Total profit, Z= 5x + 3y


The mathematical formulation of the given problem is


Maximize Z = 5x + 3y


Subject to constraints


2x + y ≤ 12


And 3x + 2y ≤ 20


x, y ≥ 0


the feasible region determined by the system of constraints can be represented graphically as



The corner points are A( 6,0) , B( 4,4) and C( 0,10)


The value of Z at the corner points are as follows



The maximum value of Z is 32 at B(4,4)


Thus the manufacturer should produce 4 pedestal lamps and 4 wooden shades ti maximize his profits of Rs 32.
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:

Let the company manufacturer x souvenirs of type A and y souvenirs of type B

∴ x and y ≥ 0


The tabular representation of the given data is:



The profit on type A souvenirs is Rs 5 and that on type B souvenirs is Rs 6.


The constraints here are of the form:


5x+ 8y ≤ 200


And 10x+ 8y ≤ 240 or 5x + 4y ≤ 120


The function is to maximize the profit Z = 5x+ 6y


The mathematical formulation of the data


Maximise Z = 5x+ 6y


Subject to constraints


5x+ 8y ≤ 200


5x + 4y ≤ 120


x and y ≥ 0


The feasible region is determined by the system of constraints is as follows:



The corner points here are A( 24, 0) , B( 8, 20) and C( 0 , 25)


The value of Z at these corner points are :



The maximum value of Z is at the point (8, 20)


Thus the firm should produce 8 number of souvenirs of type A and 20 type B souvenirs each day in order to maximize its profit of Rs160.
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:

Let the merchant stock x number of desktop models and y number of portable models.

∴ x and y ≥ 0


According to the given condition


The cost of a desKtop model is Rs25000 and that of a portable model is Rs 40,000. The merchant can invest up to Rs 7000000


⇒ 25000x + 40000y ≤ 7000000


⇒ 5x + 8y ≤ 1400


The monthly demand of computers will not exceed 250 units.


⇒ x+ y ≤ 250


The profit on a desktop model is Rs 4500 and the profit on a portable model is Rs5000.


Total profit , Z = 4500x + 5000y


Thus the mathematical formulation of the data is


Maximize Z = 4500x + 5000y


Subject to constraints


5x + 8y ≤ 1400


x+ y ≤ 250


x and y ≥ 0


the feasible region by the system of constraints is as follows:



The cornet points are A(250,0) , B( 200,50) and C(0 ,175)


The value of Z at the given corners points are:



The maximum value of Z is Rs1150000 at ( 200,50)


Thus the merchant should stock 200 desktop models and 50 portable models to earn the maximum profit of Rs 11, 50, 000.
Question 9.

A diet is to contain 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 of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Answer:

let the diet contain x units of food F1 and y units of food F2.

∴ x and y ≥ 0


The tabular representation of the data is



The cost of food F1 is Rs 4/unit and the cost of food F2 is Rs 6 per unit.


The constraints here are


3x + 6y ≥ 80


4x+ 3y ≥ 100


x and y ≥ 0


Total cost of the diet Z= 4x+ 6y


The mathematical formulation of the given data is


maximize Z= 4x+ 6y


Subject to constraints


3x + 6y ≥ 80


4x+ 3y ≥ 100


x and y ≥ 0


the feasible region by the system of constraints is as follows:



It can be seen the feasible region is unbounded with


The corner points of the feasible region are A(


The values of Z at the corner points are



As the feasible region is unbounded therefore 104 may or may not be the minimum value of Z.


For this we will draw a graph of inequality 4x+ 6y < 104


It can be seen that the feasible region has no common points with 4x+ 6y < 104


∴ the maximum cost of the mixture will be Rs104.
Question 10.

There are two types of fertilisers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 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, 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?
Answer:

let the farmer buy x kf of fertilizer f1 and y kg of fertilizer f2

∴ x and y ≥ 0


The tabular representation of the data is as follows:



F1 consists of 10% nitrogen and F2 consists of 5% nitrogen. But the farmer requires at least 14 kg of nitrogen.


10% × x + 5% × y ≥ 14



2x+y ≥ 280


food F1 consists of 6% phosphoric acid and food F2 consists of 10% phosphoric acid. But the farmer requires at least 14kg of phosphoric acid


⇒ 6% × x + 10% of y ≥ 14


⇒


3x + 56y ≥ 700


Total cost of fertilizers, Z= 6x+ 5y


The mathematical formulation of the given data is


Minimize, Z = 6x +5y


Subject to constraints


2x+y ≥ 280


3x + 56y ≥ 700


x and y ≥ 0


The feasible region determined by the system of constraint is:



It can be seen from the graph that the feasible region is unbounded


The corner points are


The value of Z at these points are:



As the feasible region is unbounded, 1000 may or may not be the minimum value


For this we will draw graph of inequality


6x+ 5y < 1000


It can be seen there is no common point between feasible region and 6x+ 5y < 1000


∴ 100 KG of fertilizer F1 and 80 kg of fertilizerF2 should be used to minimize the cost and the minimum cost is Rs 1000.
Question 11.

The corner points of the feasible region determined by the following system of linear inequalities:

2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 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
Answer:

In the given question the constraints are


2x + y ≤ 10,


x + 3y ≤ 15,


and x, y ≥ 0


The function is to maximize Z = px + qy


The corner points are (0, 0), (5, 0), (3, 4) and (0, 5).


The value of Z at these points are



Since the maximum value of z occurs on (3, 4) and (0, 5)


Hence the value on (3, 4) = Value on (0, 5)


3p+ 4q = 5q


3p = q


∴ Value of Z will be maximum if q =3p


Hence D is the correct answer.
Miscellaneous Exercise
Question 1.

Refer to Example 9. How many packets of each food should be used to maximize the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
Answer:

let x and y be the number of packets of food P and Q respectively. Obviously, x ≥ 0, y ≥ 0. Mathematical formulation of given problem is as follows:


Maximize Z = 6x + 3y ….(1)


Subject to the constraints,


12x + 3y ≥ 240 (constraint on calcium) i.e. 4x + y ≥ 80 …..(2)


4x + 20y ≥ 460 (constraint on iron) i.e. x + 5y ≥ 115 …..(3)


6x + 4y ≤ 300 (constraint on cholesterol) i.e. 3x + 2y ≤ 150 ….(4)


x ≥ 0, y ≥ 0 ….(5)


Now let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown in the fig. Here, we can observe that the feasible region is bounded.



The coordinates of the corner points A(15,20),B(40,15), and C(2,72).



Now, we find the maximum value of Z. According to table the maximum value of Z = 285 at point B (40,15).


Hence, to get the maximum amount of vitamin A in the diet, the packets of food P should be 40 and the packets of food Q should be 15. The maximum amount of vitamin A will be 285.
Question 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?
Answer:

let x and y be the amount of two bands P and Q respectively which are mixed. Obviously, x ≥ 0, y ≥ 0.


The given information can be formulated in a table as:



Mathematical formulation of given problem is as follows:


Maximize Z = 250x + 200y ….(1)


Subject to the constraints,


3x + 1.5y ≥ 18 …..(2)


2.5x + 11.25y ≥ 45 …..(3)


2x + 3y ≥ 24 ….(4)


x ≥ 0, y ≥ 0 ….(5)


Now let us graph the feasible region of the system of inequalities (2) to (5).


The feasible region (shaded) is shown in the fig. Here, we can observe that the feasible region is unbounded.



The coordinates of the corner points A(18,0), B(9,2), C(3,6) and D(0,12).



As, feasible region is unbounded so 1950 may or may not be minimum value of Z.


So, 1950 is the minimum value of Z, if the open half plane is determined by 250x + 200y <1950 has no point in common with the feasible region. Otherwise z has no minimum value.


For this we will draw the graph of inequality 250x + 200y <1950 or 5x + 4y < 39. Here we find that 250x + 200y <1950 has no point in common with the feasible region. Hence the minimum value of Z is 1950 at C(3,6).


So, 3 and 6 be the amount of two bands P and Q respectively which are mixed to get minimum cost to Rs 1950.
Question 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 is given below:



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?
Answer:

let x kg and y kg be amount of food X and Y respectively. Obviously x ≥ 0, y ≥ 0. Mathematical formulation of given problem is as follows:


Minimize Z = 16x + 20y ….(1)


Subject to the constraints,


x + 2y ≥ 10 …..(2)


x + y ≥ 6 …..(3)


3x + y ≥ 8 ….(4)


x ≥ 0, y ≥ 0 ….(5)


Now let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown in the fig. Here, we can observe that the feasible region is unbounded.



The coordinates of the corner points A(10,0),B(2,4), C(1,5)and D(0,8).



As, feasible region is unbounded so 112 may or may not be minimum value of Z.


So, 112 is the minimum value of Z, if the open half plane is determined by 16x + 20y <112 has no point in common with the feasible region. Otherwise z has no minimum value.


For this we will draw the graph of inequality 16x + 20y <112 or 4x + 5y<28. Here we find that 16x + 20y <112 has no point in common with the feasible region. Hence the minimum value of Z is 112 at B(2,4).


So, 2kg and 4kg be the amount of two foods P and Q respectively which are mixed to get minimum cost to Rs 112.
Question 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:



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.
Answer:

let x and y be the number of toys manufactured per day of type A and B respectively. Obviously x ≥ 0, y ≥ 0. Mathematical formulation of given problem is as follows:


Maximize Z = 7.5x + 5y ….(1)


Subject to the constraints,


2x + y ≤ 60 …..(2)


x ≤ 60 …..(3)


2x + 3y ≤ 120 ….(4)


x ≥ 0, y ≥ 0 ….(5)


Now let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown in the fig. Here, we can observe that the feasible region is bounded.



The coordinates of the corner points A(20,0),B(20,20), C(15,30)and D(0,40).



Now, we find the maximum value of Z. According to table the maximum value of Z = 262.5 at point C (15,30).


Hence, 15 and 20 be the number of toys manufactured per day of type A and B respectively to get the maximum profit.
Question 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 the 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?
Answer:

let x and y be the number of tickets of executive class and economy class respectively. Obviously x ≥ 0, y ≥ 0. Mathematical formulation of given problem is as follows:


Maximize Z = 1000x + 600y ….(1)


Subject to the constraints,


x + y ≤ 200 …..(2)


x ≥ 20 …..(3)


y - 4x ≥ 0 ….(4)


x ≥ 0, y ≥ 0 ….(5)


Now let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown in the fig. Here, we can observe that the feasible region is bounded.



The coordinates of the corner points A(20,80),B(40,160) and C(20,180).



Now, we find the maximum value of Z. According to table the maximum value of Z = 136000 at point B (40,160).


Hence, 40 and 160 be the number of tickets of executive class and economy class respectively to get the maximum profit.
Question 6.

Two godowns A and B have grain capacity 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 are given in the following table:



How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
Answer:

let x and y quintal of grain supplied by A to shop d and E respectively. Then 100 - x - y will be supplied to shop F. Obviously x ≥ 0, y ≥ 0.


Since requirement at shop D is 60 quintal but supplied x quintal hence, 60 - x is sent form B to D.


Similarly requirement at shop E is 50 quintal but supplied y quintal hence, 50 - y is sent form B to E.


Similarly requirement at shop E is 40 quintal but supplied 100 - x - y quintal hence, 40 - (100 - x - y) i.e. x + y - 60 is sent form B to E.


Diagrammatically it can be explained as:



As x ≥ 0, y ≥ 0 and 100 - x - y ≥ 0 ⇒ x + y ≤ 100


60 - x ≥ 0, 50 - y ≥ 0 and x + y - 60 ≥ 0


⇒ x ≤ 60, y ≤ 50 and x + y ≥ 60


Total transportation cost Z can be calculated as:


Z = 6x + 3y + 2.5(100 - x - y) + 4(60 - x) + 2(50 - y) + 3(x + y - 60)


= 6x + 3y + 250 - 2.5x - 2.5y + 240 - 4x + 100 - 2y + 3x + 3y - 180


Z = 2.5x + 1.5y + 410


Mathematical formulation of given problem is as follows:


Minimize Z = 2.5x + 1.5y + 410 ….(1)


Subject to the constraints,


x + y ≤ 100 …..(2)


x ≤ 60 …..(3)


y ≤ 50 ….(4)


x + y ≥ 60 …(5)


x ≥ 0, y ≥ 0 ….(6)


Now let us graph the feasible region of the system of inequalities (2) to (6). The feasible region (shaded) is shown in the fig. Here, we can observe that the feasible region is bounded.



The coordinates of the corner points A(60,0),B(60,40),C(50,50) and D(10,50).



Now, we find the minimum value of Z. According to table the minimum value of Z = 510 at point D (10,50).


Hence, the amount of grain delivered from A to D, E and F is 10 quintal, 50 quintals and 40 quintals respectively and from B to D,E and F is 50 quintal, 0 quintal and 0 quintal respectively.
Question 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 is given in the following table:



Assuming that the transportation cost of 10 liters of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Answer:

let x and y liters of oil be supplied by A to shop D and E respectively. Then 7000 - x - y will be supplied to F. Obviously x ≥ 0, y ≥ 0.


Since requirement at D is 4500L but supplied x quintal hence, 4500 - x is sent form B to D.


Similarly requirement at E is 3000L but supplied y quintal hence, 3000 - y is sent form B to E.


Similarly requirement at E is 3500 quintal but supplied 7000 - x - y quintal hence, 3500 - (7000 - x - y) i.e. x + y - 3500 is sent form B to E.


Diagrammatically it can be explained as:



As x ≥ 0, y ≥ 0 and 7000 - x - y ≥ 0 ⇒ x + y ≤ 7000


4500 - x ≥ 0, 3000 - y ≥ 0 and x + y - 3500 ≥ 0


⇒ x ≤ 4500, y ≤ 3000 and x + y ≥ 3500


Cost of delivering 10L of petrol = Re.1


Then Cost of delivering 1L of petrol = Rs.1/10


Total transportation cost Z can be calculated as:


Z =


Z = 0.3x + 0.1y + 3950


Mathematical formulation of given problem is as follows:


Minimize Z = 0.3x + 0.1y + 3950 ….(1)


Subject to the constraints,


x + y ≤ 7000 …..(2)


x ≤ 4500 …..(3)


y ≤ 3000 ….(4)


x + y ≥ 3500 …(5)


x ≥ 0, y ≥ 0 ….(6)


Now let us graph the feasible region of the system of inequalities (2) to (6). The feasible region (shaded) is shown in the fig. Here, we can observe that the feasible region is bounded.



The coordinates of the corner points A(3500,0),B(4500,0),C(4500,2500), D(4000,3000) and E(500,3000) .



Now, we find the minimum value of Z. According to table the minimum value of Z = 4400 at point E (500,3000).


Hence, the amount of oil delivered from A to D, E and F is 500L, 3000L and 3500L quintals respectively and from B to D,E and F is 4000L, 0 L and 0 L respectively.
Question 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.

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 in the garden?
Answer:

let x and y be the bags of brands P and Q respectively. Obviously x ≥ 0, y ≥ 0. Mathematical formulation of given problem is as follows:


Minimize Z = 3x + 3.5y ….(1)


Subject to the constraints,


x + 2y ≤ 240 …..(2)


3x + 1.5y ≤ 270 ⇒ x + 0.5y=90 …..(3)


1.5x + 2y ≤ 310 ….(4)


x ≥ 0, y ≥ 0 ….(5)


Now let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown in the fig. Here, we can observe that the feasible region is bounded.



The coordinates of the corner points A(140,50),B(20,40) and C(40,100).



Now, we find the minimum value of Z. According to table the minimum value of Z = 470 at point C (40,100).


Hence, 40 and 100 be be the bags of brands P and Q respectively to be added to the garden to minimize the amount of nitrogen.
Question 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?
Answer:

let x and y be the bags of brands P and Q respectively. Obviously x ≥ 0, y ≥ 0. Mathematical formulation of given problem is as follows:


Maximize Z = 3x + 3.5y ….(1)


Subject to the constraints,


x + 2y ≤ 240 …..(2)


3x + 1.5y ≤ 270 ⇒ x + 0.5y=90 …..(3)


1.5x + 2y ≤ 310 ….(4)


x ≥ 0, y ≥ 0 ….(5)


Now let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown in the fig. Here, we can observe that the feasible region is bounded.



The coordinates of the corner points A(140,50),B(20,40) and C(40,100).



Now, we find the maximum value of Z. According to table the maximum value of Z = 595 at point A (140,50).


Hence, 140 and 50 be be the bags of brands P and Q respectively to be added to the garden to maximize the amount of nitrogen.
Question 10.

A toy company manufactures two types of dolls, A and B. Market tests 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?
Answer:

let x and y be the number of dolls of type A and B respectively. Obviously x ≥ 0, y ≥ 0. Mathematical formulation of given problem is as follows:


Maximize Z = 12x + 16y ….(1)


Subject to the constraints,


x + y ≤ 1200 …..(2)


y ≤ x/2 ⇒ 2y ≤ x …..(3)


x – 3y ≤ 600 ….(4)


x ≥ 0, y ≥ 0 ….(5)


Now let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown in the fig. Here, we can observe that the feasible region is bounded.



The coordinates of the corner points A(600,0), B(1050,150) and C(800,400).



Now, we find the maximum value of Z. According to table the maximum value of Z = 16000 at point C (800,400).


Hence, 800 and 400 be the number of dolls of type A and B respectively.