CBSE Class 12 Maths Revision Notes Chapter 12

Q.1 A manufacturing company makes two models A and B of a product. Each piece of Model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of Model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of Rs 8000 on each piece of model A and Rs 12000 on each piece of Model B. How many pieces of Model A and Model B should be manufactured per week to realise a maximum profit? What is the maximum profit per week?

Ans

Suppose x is the number of pieces of Model A and y is the number of pieces of Model B. Then
Total profit (in Rs) = 8000x + 12000y
Let Z = 8000x + 12000y
Mathematical model for the given problem is as follows:
Maximise Z = 8000 x + 12000 y (1)
subject to the constraints:

9x + 12y ≤ 180 (Fabricating constraint), i.e., 3x + 4y ≤ 60 (2)
x + 3y ≤ 30 (finishing constraint) (3)
x, y ≥ 0 (4)

The feasible region (shaded) OABC determined by the linear inequalities (2) to (4) is shown below.

The company should produce 12 pieces of Model A and 6 pieces of Model B to realise maximum profit and maximum profit then will be Rs 1,68,000.

Q.2 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. Formulate the LPP.

Ans

The mathematical formulation of the given problem is
Maximize Z = 5x + 6y
subject to the constraints,
5x + 8y ≤ 200
10x + 8y ≤ 240, i.e., 5x + 4y ≤ 120

Q.3 A cooperative society of farmers has 50 hectare of land to grow two crops X and Y. The profit from crops X and Y per hectare are estimated as Rs 10,500 and Rs 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at rates of 20 litres and 10 litres per hectare. Further, no more than 800 litres of herbicide should be used in order to protect fish and wild life using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society?

Ans

Let x hectare of land be allocated to crop X and y hectare to crop Y.
Profit per hectare on crop X = Rs 10500
Profit per hectare on crop Y = Rs 9000
Therefore, total profit = Rs (10500x + 9000y)
The mathematical formulation of the problem is as follows:
Maximise Z = 10500
subject to the constraints:
x + y ≤ 50 (constraint related to land) (1)
20x + 10y ≤ 800 (constraint related to herbicide)
i.e., 2x + y ≤ 80 (2)
x, y ≥ 0 (3)

Graph of the system of inequalities (1) to (3). The feasible region OABC is shown below:

Hence, the society will get the maximum profit of Rs 4,95,000 by allocating 30 hectares for crop X and 20 hectares for crop Y.

Q.4 A retired person has Rs 70,000 to invest and two types of bonds are available in the market for investment. First type of bond yields an annual income of 8% on the amount invested and the second type of bonds yields 10% per annum. As per norms, he has to invest a minimum of Rs 10,000 in the first type and not more than Rs 30,000 in the second type. How should he plan his investment so as to get the maximum return, after one year of investment?

Ans

Let investment in first type of bond = Rs x
And investment in second type of bond = Rs y
Maximum return

$\mathrm{Z}=\frac{8}{100}\mathrm{x}+\frac{10}{100}\mathrm{y}=0.08\mathrm{x}+0.1\mathrm{y}$

(1)
Subject to the constraints

x + y ≤ 70,000 (2)
x ≥ 10,000 (3)
y ≤ 30,000 (4)
y ≥ 0 (5)

Graph the inequalities (2) to (4). The feasible region (shaded) is shown below:

Thus, the person will get the maximum return if he invests Rs 40,000 in the first type of bond and Rs 30,000 in the second type of bond.

Q.5 There are two factories, located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below:

How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost?

Ans

Let x units and y units of the commodity be transported from the factory at P to the depots at A and B respectively. Then (8 – x – y) units will be transported to depot at C.

$\begin{array}{l}\text{Hence, we have}\mathrm{x}\ge 0,\mathrm{y}\ge 0\text{and}8-\mathrm{x}-\mathrm{y}\ge 0\\ \mathrm{or}\mathrm{x}\ge 0,\mathrm{y}\ge 0\text{and}\mathrm{x}+\mathrm{y}\le 8\\ \text{Now}(5-\mathrm{x}),(5-\mathrm{y})\text{and}6-(5-\mathrm{x}+5-\mathrm{y})=\mathrm{x}+\mathrm{y}-4\\ \text{units are to be transported from the factory at Q to the depots}\\ \text{at}\mathrm{A},\mathrm{B}\text{and C respectively.}\\ \text{Thus,}5-\mathrm{x}\ge 0,\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}5}-\mathrm{y}\ge 0,\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}and}\mathrm{x}+\mathrm{y}-4\ge 0\\ \text{or\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{x}\le 5,\mathrm{y}\le 5,\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}and\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{x}+\mathrm{y}\ge 4\\ \text{Total transportation cost}\mathrm{Z}\text{is given by}\\ \mathrm{Z}=160\mathrm{x}+100\mathrm{y}+100(5-\mathrm{x})+120(5-\mathrm{y})+100(\mathrm{x}+\mathrm{y}-4)+150(8-\mathrm{x}-\mathrm{y})\\ =10(\mathrm{x}–7\mathrm{y}+190)\\ \text{Minimise}\mathrm{Z}=10(\mathrm{x}–7\mathrm{y}+190)\\ \text{subject to the constraints:}\\ \mathrm{x}\ge 0,\mathrm{y}\ge 0...\left(1\right)\\ \mathrm{x}+\mathrm{y}\le 8...\left(2\right)\\ \mathrm{x}\le 5...\left(3\right)\\ \mathrm{y}\le 5...\left(4\right)\\ \text{and}\mathrm{x}+\mathrm{y}\ge 4...\left(5\right)\end{array}$

Hence, the optimal transportation strategy will be

Q.6 A catering agency has two kitchens to prepare food at two places A and B. From these places ‘Mid-day Meal’ is to be supplied to three different schools situated at P, Q, R. The monthly requirements of the schools are respectively 40, 40 and 50 food packets. A packet contains lunch for 1000 students. Preparing capacity of kitchens A and B are 60 and 70 packets per month respectively. The transportation cost per packet from the kitchens to schools is given below:

How many packets from each kitchen should be transported to school so that the cost of transportation is minimum? Also find the minimum cost.

Ans

Let x packets and y packets be transported from the kitchen at A to the schools at P and Q respectively. Then (60 – x – y) packets will be transported to school at R.

$\begin{array}{l}\text{Hence, we have}\mathrm{x}\ge 0,\mathrm{y}\ge 0\text{and}60-\mathrm{x}-\mathrm{y}\ge 0\\ \mathrm{or}\mathrm{x}\ge 0,\mathrm{y}\ge 0\text{and}\mathrm{x}+\mathrm{y}\le 60\\ \text{Now}(40-\mathrm{x}),(40-\mathrm{y})\text{and}70-(40-\mathrm{x}+40-\mathrm{y})=\mathrm{x}+\mathrm{y}-10\\ \text{units are to be transported from the kitchen at B to the schools}\\ \text{at}{\mathrm{P}}_{\mathrm{t}}\mathrm{Q}\text{and}\mathrm{R}\text{respectively.}\\ \text{Thus,}40-\mathrm{x}\ge 0,40-\mathrm{y}\ge 0,\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}and}\mathrm{x}+\mathrm{y}-10\ge 0\\ \text{or\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{x}\le 40,\mathrm{y}\le 40,\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}and\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{x}+\mathrm{y}\ge 10\\ \text{Total transportation cost}\mathrm{Z}\text{is given by}\\ \mathrm{Z}=5\mathrm{x}+4\mathrm{y}+4(40-\mathrm{x})+2(40-\mathrm{y})+5(\mathrm{x}+\mathrm{y}-10)+3(60-\mathrm{x}-\mathrm{y})\\ =3\mathrm{x}+4\mathrm{y}+370\\ \text{Minimise}\mathrm{Z}=3\mathrm{x}+4\mathrm{y}+370\\ \text{subject to the constraints:}\\ \mathrm{x}\ge 0,\mathrm{y}\ge 0...\left(1\right)\\ \mathrm{x}+\mathrm{y}\le 60...\left(2\right)\\ \mathrm{x}\le 40...\left(3\right)\\ \mathrm{y}\le 40...\left(4\right)\\ \text{and}\mathrm{x}+\mathrm{y}\ge 10...\left(5\right)\end{array}$

Hence, the optimal transportation strategy is

Q.7 A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires at least 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum amount of vitamin A?

Ans

Let x and y be the number of packets of food P and Q respectively. Mathematical formulation of the given problem is as follows:
Minimise Z = 6x + 3y (vitamin A)
subject to the constraints

12x + 3y ≥ 240 (constraint on calcium), i.e., 4x + y ≥ 80 (1)
4x + 20y ≥ 460 (constraint on iron), i.e., x + 5y ≥ 115 (2)
6x + 4y ≤ 300 (constraint on cholesterol), i.e., 3x + 2y ≤ 150 (3)
x, y ≥ 0 (4)

Graph the inequalities (1) to (4). The feasible region (shaded) determined by the constraints (1) to (4) is shown in the graph below:

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

The amount of vitamin A under the constraints given in the problem will be minimum, if 15 packets of food P and 20 packets of food Q are used in the special diet. The minimum amount of vitamin A will be 150 units.

Q.8 Solve the following linear programing problem graphically:
Maximise Z = 60x + 15y
Subject to constraints
x + y ≤ 50
3x + y ≤ 90
x, y ≥ 0

Ans

Maximise Z = 60x + 15y (1)

Subject to constraints
x + y ≤ 50 (2)
3x + y ≤ 90 (3)
x, y ≥ 0 (4)

Graph the inequalities (2), (3) and (4). The feasible region (shaded) is shown below:

The maximum value of Z on the feasible region occurs at x = 30 and y = 0.

Q.9 Solve the following linear programing problem graphically:
Maximise Z = 5x + 3y
Subject to constraints
3x + 5y ≤ 15
5x + 2y ≤ 10
x, y ≥ 0

Ans

Maximise Z = 5x + 3y (1)
Subject to constraints
3x + 5y ≤ 15 (2)
5x + 2y ≤ 10 (3)
x, y ≥ 0 (4)

Graph the inequalities (2) to (4). The feasible region OABC is shown below:

The maximum value of Z on the feasible region occurs at x = 20/19 and y = 45/19.

Q.10 Solve the following linear programing problem graphically:
Minimise Z = 3x + 2y
Subject to constraints
5x + y ≥ 10
x + y ≥ 6
x + 4y ≥ 12
x, y ≥ 0

Ans

Minimise Z = 3x + 2y (1)
Subject to constraints
5x + y ≥ 10 (2)
x + y ≥ 6 (3)
x + 4y ≥ 12 (4)
x, y ≥ 0 (5)

Graph the inequalities (2) to (5). The feasible region (shaded) is shown as below:

Thus, Z is minimum at x = 1 and y = 5.

Q.11 A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 10 units of vitamin C. Food ‘I’ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’. Formulate and solve this problem as a linear programming problem to minimise the cost of such a mixture.

Ans

Minimise cost: Z = 50x + 70y (1)

Equation of vitamin A
2x + y ≥ 8 (2)
Equation of vitamin C
x + 2y ≥ 10 (3)
x, y ≥ 0 (4)

Graph the inequalities (2) to (4). The feasible region determined by the system is shown below:

Hence, the optimal mixing strategy for the dietician would be to mix 2 kg of Food ‘I’ and 4 kg of Food ‘II’, and with this strategy, the minimum cost of the mixture will be Rs 380.

Q.12 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?

Ans

Let x = Number of executive class tickets sold
and y = Number of economy class tickets sold
Maximum profit: Z = 1000x + 600y (1)
20 tickets for executive class are reserved (given)
⇒ x ≥ 20
The number of economy class tickets is at least 4 times of executive class tickets
⇒ y ≥ 4x
Total tickets are 200
⇒ x + y ≤ 200
Points lie on x + y ≤ 200

$\begin{array}{|ccc|}\hline \mathrm{x}& 0& 200\\ \mathrm{y}& 200& 0\\ \hline\end{array}$

Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximize the profit and the maximum profit is Rs 1,36,000.

Q.13 If a young man rides his motorcycle at 25 km/hour, he had to spend Rs 2 per km on petrol. If he rides at a faster speed of 40 km/hour, the petrol cost increases at Rs 5 per km. He has Rs 100 to spend on petrol and wishes to find what is the maximum distance he can travel within one hour. Express this as an LPP and solve it graphically. write one important role of wisely used petrol.

Ans

Let the young man ride with the speed 25 km/hr for x hours and with the speed 40 km/hr for y hours respectively.
Money spent at speed 25 km/hr = 50x
And Money spent at speed 40 km/hr = 200y
Maximum distance

Z = 25x + 40y (1)

subject to the constraints

x + y ≤ 1 (2)
50x + 200y ≤ 100 (3)
x , y ≥ 0 (4)

Graph the inequalities (2) to (4). The feasible region (shaded) is shown as below:

Thus, the maximum distance he can travel is 30 km.
Saving of petrol plays very important role in our life. By using it wisely we can save misuse of fossil fuel and also help in cleaning our environment from the pollution of produced by burning of petrol.

Q.14 Define optimisation problems.

Ans

Certain type problems that seek to maximise (or minimise) profit (or cost) form a general class of problems called optimisation problems.

Q.15 Define linear constraints.

Ans

A set of linear inequalities is called linear constraints.

Q.16 What do you mean by the word ‘programming’ in ‘linear programming’?

Ans

The term programming refers to the method of determining a particular programme or a plan of action.

Q.17 Define an objective function.

Ans

Linear function Z = ax + by, where a, b are constraints that have to be maximised or minimised, is called a linear objective function.

Q.18 What are decision variables?

Ans

In linear function Z = ax + by, x and y are called decision variables.

Q.19 Define overriding conditions.

Ans

The linear inequalities, equations or restrictions on the variables of a linear programming problem are called constraints.

Q.20 Define feasible region.

Ans

The common region determined by all the constraints including non-negative constraints x, y ≥ 0, of a linear programming problem is called feasible region.

Q.21 A firm makes items A and B. The total number of items it can make in a day is 24. It takes one hour to make an item of A and only half an hour to make an item of B. The maximum time available per day is 16 hours. The profit on an item of A is Rs. 300 and on one item of B is Rs.160. How many items of each type should be produced to maximise the profit? Solve the problem graphically.

Ans

Let x items of A and y items of B are produced per day.
According to the given condition,
x ≥ 0, y ≥ o, x + y ≤ 24 and x + ½ y ≤ 16 ⇒ 2x + y ≤ 32
Max. profit Z = 300 x + 160 y
Plotting these inequations on the graph, we get

shaded portion as the optimum solution.
Maximum profit can be at the points A(16,0), B(8,16) and C(0,24).

$\begin{array}{|lll|}\hline \text{Points}& \text{Z = 300x + 160y}& \text{Value}\\ \text{A(16, 0)}& 4800+0& \text{4800}\\ \text{B(8, 16)}& \text{2400 + 2560}& \text{4960 [Maximum]}\\ \text{C(0, 24)}& 0+3840& 3840\\ \hline\end{array}$

Profit is maximum at point B (8, 16), i.e., x = 8, y = 16.
Hence, 8 items at A and 16 items of B must be produced to get maximum profit.

Q.22 Name three important linear programming problems.

Ans

(i) Diet Problems
(ii) Manufacturing Problems
(iii) Transportation Problems

Q.23 What is infeasible solution to the problem?

Ans

Any point outside the feasible region is called an infeasible solution to the problem.

Q.24 Define feasible solution to the problem.

Ans

Every point of feasible region is called feasible solution to the problem.

Q.25 A manufacturer manufactures two types of steel trunks. He has two machines – A and B. For completing the first type of trunk, it requires 3 hours on machine A and 1 hour on machine B. Machine A can work for 18 hours and B for 8 hours only per day. There is a profit of Rs.30 on the first type of trunk and Rs. 48 on the second type of trunk. How many trunks of each type should be manufactured every day to earn maximum profit? Solve graphically.

Ans

Let the number of A type steel trunk be x per day and number of B type steel trunk be y per day. According to the given conditions, L.P.P. is
x ≥ 0, y ≥ 0 … (1)
3x + 3y ≤ 48 ⇒ x + y ≤ 6 … (ii)
x + 2y ≤ 8 … (iii)
and to maximise Z = 30x + 48y
Plotting these inequations on the graph, we get

We get a shaded portion as the optimum solution. Maximum profit can be at the points A(6, 0), B(4, 2) and C(0, 4).

$\begin{array}{|lll|}\hline \text{Points}& \text{Z = 30x + 48y}& \text{Value}\\ \text{A(6, 0)}& 180+0& \text{180}\\ \text{B(4, 2)}& \text{120 + 96}& \text{216 [Maximum]}\\ \text{C(0, 4)}& 0+192& 192\\ \hline\end{array}$

We notice maximum profit is at B (4, 2), i.e., x = 4, y = 2.
Hence, 4 trunks of type I and 2 trunks of type II must be manufactured each day to get a maximum profit of Rs. 216.

Q.26 A firm deals with two kinds of fruit juices – pineapple and orange juice. These are mixed and two mixtures are sold as soft drinks A and B. One tin of A requires 4 litres of pineapple and 1 litre of orange juice. One tin of B requires 2 litres of pineapple and 3 litres of orange juice. The firm has only 46 litres of pineapple juice and 24 litres of orange juice. Each tin of A and B is sold at a profit of Rs. 4 and Rs. 3 respectively. How many tins of each type should the firm produce to maximise the profit? Solve the problem graphically.

Ans

Let the number of tins of A be x and that of B be y.
According to the given conditions, the L.P.P. is
x ≥ 0, y ≥ 0 … (1)
4x + 2y ≤ 46 ⇒ 2x + y ≤ 23 … (2)
x + 3y ≤ 24 … (3)
and to maximise Z = 4x + 3y.
Plotting the inequations (i), (ii) and (iii) on the graph,

We get the shaded portion as the optimum solution. Possible points for maximum profit are A(23/2,0), B(9,5) and C(0,8)

$\begin{array}{|lll|}\hline \text{Points}& \mathrm{Z}=4\mathrm{x}+3\mathrm{y}& \text{Value}\\ \text{A}\left(\frac{23}{2},0\right)& 46+0& 46\\ \text{B(9, 5)}& 36+15& \text{51 [maximum]}\\ \text{C(0, 8)}& 0+24& 24\\ \hline\end{array}$

We notice profit is maximum at B (9,5), i.e., x = 9, y = 5.
Hence, 9 tins of A and 5 tins of B should be produced.

Q.27 Two tailors A and B earn Rs. 15 and Rs. 20 per day respectively. A can stitch 6 shirts and 4 pants per day, while B can stitch 10 shirts and 4 pants per day. How many days shall each work if it is desired to produce at least 60 shirts and 32 pants at a minimum labour cost? Solve the problem graphically.

Ans

Let the tailors A and B work for x and y days respectively. The required L.P.P. is
6x + 10y ≥ 60 or 3x + 5y ≥ 30 … (1)
4x + 4y ≥ 32 or x + y ≥ 8 … (2)
x ≥ 0, y ≥ 0
To minimise Z = 15 x + 20 y
Plotting the inequations (i), (ii) and (iii) on the graph,

Now, drawing the lines 3x + 5y = 30, which is the line AB and of line x + y = 8, which is the line CD.
These lines intersect at E (5, 3).
The shaded unbounded region is the feasible region.

$\begin{array}{|lll|}\hline \text{Points}& \text{Z = 15x + 20y}& \text{Value}\\ \text{A(10, 0)}& 150+0& \text{150}\\ \text{E(5, 3)}& \text{75 + 60}& \text{135 [Minimum]}\\ \text{D(0, 8)}& 0+160& 160\\ \hline\end{array}$

Hence, Z is minimum at E(5, 3), i.e., x = 5, y = 3 and the minimum cost is Rs. 135.
So, tailor A should work for 5 days and tailor B should work for 3 days.

Q.28 A man has Rs. 1,500 for purchase of rice and wheat. A bag of rice and a bag of wheat cost Rs. 180 and Rs. 120 respectively. He has a storage capacity of 10 bags only. He earns a profit of Rs. 11 and Rs. 9 respectively per bag of rice and wheat. Formulate it as a linear programming problem and solve it graphically for maximum profit.

Ans

Let the man purchases x bags of rice and y bags of wheat. So, the objective function is to minimise Z = 11x + 9y
Subject to constraints,
x + y ≤ 10 … (i)
180x + 120y ≤ 1500
i.e., 3x + 2y ≤ 25 … (ii)
x ≥ 0, y ≥ 0 … (iii)
Plotting the inequations (i), (ii) and (iii) on the graph,

The feasible region of L.P.P. is the shaded region.
Maximum profit can be at the points A(25/2,0), P(5,5) and B(0,10).
The value of the objective function at these points is given:

$\begin{array}{|ll|}\hline \text{Points}& \text{Value of the objective function Z = 11x + 9y}\\ \text{A(25/3, 0)}& 275/3+0=91.66\\ \text{P(5, 5)}& 55+45=100\left[\mathrm{Maximum}\right]\\ \text{P(0, 10)}& 0+90=90\\ \hline\end{array}$

Z is maximum at x = 5 and y = 5.
Hence, for the maximum profit, he will purchase 5 bags of rice and 5 bags of wheat.

Q.29 Solve the following linear programming problem graphically:
Maximise Z = 60x+15y
Subject to constraints
x + y ≤ 50
3x + y ≤ 90
x, y ≥ 0

Ans

We have,
Maximise Z = 60x + 15y
Subject to constraints,
x + y ≤ 50 … (i)
3x + y ≤ 90 … (ii)
x ≥ 0, y ≥ 0 … (iii)
Plotting the inequations (i), (ii) and (iii) on the graph,

From the graph, it is clear that the shaded portion is the optimum solution. Possible points for maximum Z are A(30,0), B(20,30) and C(0,50).

$\begin{array}{|lll|}\hline \text{Points}& \text{Z = 60x + 15y}& \text{Value}\\ \text{A(30, 0)}& 1800+0& \text{1800 [Maximum]}\\ \text{B(20, 30)}& \text{1200 + 450}& 1650\\ \text{C(0, 50)}& 0+750& 750\\ \hline\end{array}$

Hence, Z is maximum at A(30, 0), i.e., x = 30, y = 0.

Q.30 A manufacturer produces two types of steel trunks. He has two machines A and B. The first type of trunk requires 3 hours on machine A and 3 hours on machine B. The second type of trunk requires 3 hours on machine A and 2 hours on machine B. Machines A and B run daily for 18 hours and 15 hours respectively. There is a profit of Rs. 30 on the first type of trunk and Rs. 25 on the second type of trunk. How many trunks of each type should be produced and sold to make a maximum profit?

Ans

Let x trunks of type I and y trunks of type II are produced.
According to the given condition,
Maximise Z = 30x + 25y
3x + 3y ≤ 18
i.e., x + y ≤ 6 … (i)
3x + 2y ≤ 15 … (ii)
x ⇒ 0, y ⇒ 0 … (iii)
Put (0,0) in x + y ≤ 6 ⇒ 0 ≤ 6, true.
So, shaded portion contains (0,0).
Put (0,0) in 3x + 2y ≤ 15 ⇒ 0 ≤ 15, true.
So, shaded portion contains (0,0).
Plotting the inequations (i), (ii) and (iii) on the graph,

Since the shaded portion represents the optimum solution, the possible points for maximum Z are A(5,0), B(3,3) and C(0,6).

$\begin{array}{|lll|}\hline \text{Point}& \text{Z = 30x + 25y}& \text{Value}\\ \text{A(5, 0)}& 150+0& 150\\ \text{B(3, 3)}& \text{90 + 75}& \text{165 [Maximum]}\\ \text{C(0, 6)}& 0+150& 150\\ \hline\end{array}$

We notice profit is maximum at B(3,3), i.e., x = 3, y = 3.
Hence, 3 trunks of each type should be produced to get a maximum profit of Rs. 165.

Q.31 A dealer wishes to purchase a number of fans and sewing machines. He has only Rs. 5,760 to invest and has space for at the most 20 items. A fan and sewing machine cost Rs. 360 and Rs. 240 respectively. He can sell a fan at a profit of Rs. 22 and sewing machine at a profit of Rs. 18. Assuming that he can sell whatever he buys, how should he invest his money in order to maximise his profit? Translate the problem into LPP and solve it graphically.

Ans

Let a man purchases x fans and y sewing machines.
According to given condition, L.P.P. is
To maximise Z= 22x + 18y
Subject to constraints,
x ≥ 0, y ≥ 0 … (i)
x + y ≤ 20 … (ii)
360x + 240y ≤ 5760
i.e., 3x + 2y ≤ 48 …(iii)
Let the equation be x + y = 0.
Some points on graph are:

$\begin{array}{|cccc|}\hline \mathrm{x}& 0& 20& 10\\ \mathrm{y}& 20& 0& 10\\ \hline\end{array}$

Let the equation be 3x + 2y = 48
y = ( 48 – 3x)/2
Some points on graph are:

$\begin{array}{|cccc|}\hline \mathrm{x}& 0& 16& 8\\ \mathrm{y}& 24& 0& 12\\ \hline\end{array}$

Substituting (0,0) in x + y ≤ 20 and 3x + 2y ≤ 48, we notice both inequations are true.
On plotting the above information on graph,

We notice that the shaded portion represents the optimum solution. Possible points for maximum Z are A (16,0), B(8,12) and C(0,20).

$\begin{array}{|lll|}\hline \text{Point}& \text{Z = 22x + 18y}& \text{Value}\\ \text{A(16, 0)}& 352+0& 352\\ \text{B(8, 12)}& \text{176 + 216}& \text{392 [Maximum]}\\ \text{C(0, 20)}& 0+360& 360\\ \hline\end{array}$

We notice Z is maximum for (8, 12), i.e., x = 8, y = 12.
Hence, man should purchase 8 fans and 12 sewing machines for a maximum profit of Rs. 392.

Q.32 A shopkeeper sells only tables and chairs. He has only Rs. 6,000 to invest and has space for at the most 20 items. A table costs him Rs. 400 and a chair Rs. 250. He can sell the table at a profit of Rs. 25 and the chair for a profit of Rs. 40. Suppose, he can sell whatever he buys. Formulate the problem as LPP and solve it graphically for maximum profit.

Ans

Let the shopkeeper can sell ‘x’ tables and ‘y’ chairs. According to the given condition,
L.P.P. is to maximise Z = 25x + 40y
Subject to constraints,
x ≥ 0,y ≥ 0 … (i)
x + y ≤ 20 … (ii)
400x + 250y ≤ 6000
⇒ 8x + 5y ≤ 120 … (iii)
On plotting the above information on graph,

We notice that the shaded portions represent the optimum solution. Possible points for maximum Z are A(15,0), B(20/3,40/3) and C(0,20).

$\begin{array}{|lll|}\hline \text{Points}& \text{Z = 25x + 40y}& \text{Value}\\ \text{A(15, 0)}& 375+0& 375\\ \text{B(20/3, 40/3)}& \text{(500/3) + (1600/3)}& \text{700}\\ \text{C(0, 20)}& 0+800& \text{800 [Maximum]}\\ \hline\end{array}$

Hence, the shopkeeper must buy only 20 chairs and no tables for a maximum profit of Rs.800.Z is maximum at C (0,20), i.e., x = 0, y = 20.

Q.33 If a young man rides his motorcycle at 25 km/hr, he has to spend Rs. 2 per km on petrol. If he rides at a faster speed of 40 km/hr, the petrol cost increases at Rs. 5 per km. He has Rs. 100 to spend on petrol and wishes to find the maximum distance that he can travel in one hour. Express this as an LPP and solve it graphically.

Ans

Let the young man travels x km at 25 km/hr and y km at 40 km/hr. According to the given conditions, L.P.P. is to maximise Z = x + y,
Subject to constraints,
x ≥ 0,y ≥ 0 … (i)
x/25 + y/40 ≤ 1 … (ii)
2x + 5y ≤ 100 …(iii)
On plotting the above information on graph,

Since the shaded portion is the optimum solution, the possible points for maximum distance are A(25, 0); B( 50/3, 40/3); C(0, 20).

$\begin{array}{|lll|}\hline \text{Points}& \text{Z = 25x + 40y}& \text{Value}\\ \text{A(25, 0)}& 25+0& 25\\ \text{B(50/3, 40/3)}& \text{(50/3) + (40/3)}& \text{30 [Maximum]}\\ \text{C(0, 20)}& 0+20& 20\\ \hline\end{array}$

Distance is maximum at B( 50/3, 40/3), i.e., x = 50/3 and y = 40/3. Hence, the young man rides 50/3 km at 25 km/hr and 40/3 km at 40 km/hr for maximum distance of 30 km.

Q.34 A factory owner purchases two types of machines – A and B for his factory. The requirements and the limitations for the machines are as follows:

$\begin{array}{|cccc|}\hline \text{Machine}& \begin{array}{l}\text{Area}\\ \text{occupied}\end{array}& \begin{array}{l}\text{Labour}\\ \text{force}\end{array}& \begin{array}{l}\text{Daily output}\\ \text{(in units)}\end{array}\\ \mathrm{A}& 1000{\mathrm{m}}^2& 12\mathrm{men}& 60\\ \mathrm{B}& 1200{\mathrm{m}}^2& 8\mathrm{men}& 40\\ \hline\end{array}$

He has maximum area of 9000 m² available, and 72 skilled labourers who can operate both the
machines. How many machines of each type should he buy to maximise the daily output?

Ans

Let number of A type machine = x
and number of B type machine = y
Area required for A type machine = 1000 m²
Area required for A type machine = 1200 m²
Then, L.P.P. of given conditions are
Maximum Z = 60x + 40y
Subject to constraints,
x ≥ 0,y ≥ 0 … (i)

1000 x + 1200 y ≤ 9000
⇒ 5x + 6y ≤ 45 … (ii)
12x+8y ≤ 72
⇒ 3x +2y ≤ 18 … (iii)
Plotting the inequations (i), (ii) and (iii) on the graph,

Since shaded portion is the optimum solution, the possible points for maximum output are A(6, 0), B( 9/4, 45/8) and C(0, 15/2).

$\begin{array}{|lll|}\hline \text{Points}& \mathrm{Z}=60\mathrm{x}+40\mathrm{y}& \text{Value}\\ \text{A(6, 0)}& 360+0& \text{360 [maximum]}\\ \text{B(9/4, 45/8)}& 135+225& \text{360 [maximum]}\\ \text{C(0, 15/2)}& 0+300& 300\\ \hline\end{array}$

Maximum output will be at A( 6, 0) or B(9/4, 45/8). Hence, 6 machines of type A or 9/4 ~ 2 machines of type A and 45/8 ~ 6 machines of type B should be purchased.

Q.35 A mobile company make two models (Beta-10 and Beta-20) of mobile phones. Each piece of model Beta-10 requires 6 hours of assembling and 1 hour of packaging. Each piece of model Beta-20 requires 7 hours of assembling and 1.5 hours of packaging. In a week, for assembling and packaging, the maximum working hours available are 150 and 40 respectively. The company makes a profit of ₹5000 on each piece of model Beta-10 and ₹4000 on each piece of model beta-20. How many pieces of model Beta-10 and Beta-20 should be manufactured per week to realize a maximum profit? What is the maximum profit per week?

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{be}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{models}\mathrm{of}\mathrm{Beta}-10\mathrm{and}\\ \mathrm{Beta}-20\mathrm{mobile}\mathrm{phones}\mathrm{are}\mathrm{made}\mathrm{respectively}.\\ \mathrm{Profit}=\; 5000\mathrm{x}+4000\mathrm{y}\\ \mathrm{Let}\mathrm{this}\mathrm{profit}\mathrm{is}\mathrm{represented}\mathrm{by}\mathrm{Z}.\\ \mathrm{So}\mathrm{the}\mathrm{objective}\mathrm{function}\mathrm{is}:\mathrm{Z}=\; 5000+4000\mathrm{y}.\dots \dots \left(1\right)\\ \mathrm{Therefore},\mathrm{Z}\mathrm{is}\mathrm{to}\mathrm{be}\mathrm{maximised}\mathrm{subject}\mathrm{to}\mathrm{the}\mathrm{following}\\ \mathrm{constraints}:\\ \mathrm{Assembling}:\\ 6\mathrm{x}+7\mathrm{y}\le 150.\dots .\left(2\right)\\ \mathrm{Packaging}:\\ \mathrm{x}+1.5\mathrm{y}\le 40.\dots ..\left(3\right)\\ \mathrm{Note}\mathrm{that}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{cannot}\mathrm{be}\mathrm{negative}.\\ \mathrm{So},\mathrm{x}\ge 0,\mathrm{y}\ge .\dots ..\left(4\right)\\ \mathrm{Plot}\mathrm{all}\mathrm{the}\mathrm{inequalities}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{set}\mathrm{of}\mathrm{axes}.\end{array}$

$\begin{array}{l}\mathrm{The}\mathrm{feasible}\mathrm{region}\mathrm{is}\mathrm{OAD},\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{bounded}\mathrm{region}.\\ \mathrm{Cordinates}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{A}:\left(0,\mathrm{\hspace{0.33em}}\frac{150}{7}\right)\\ \mathrm{CordinatesofthepointD}:\left(0,\mathrm{\hspace{0.33em}}25\right)\\ \mathrm{Now}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{the}\mathrm{objective}\mathrm{function}\mathrm{Z}\mathrm{at}\mathrm{corner}\\ \mathrm{points}\mathrm{are}\mathrm{as}\mathrm{follows}:\end{array}$

$\begin{array}{l}\text{Clearly, objective function}\left(Z\right)\text{\hspace{0.33em}has maximum 125,000}\\ \text{value at D\hspace{0.33em}}\left(25,0\right).\text{\hspace{0.33em}Therefore, the company should make 25 sets}\\ \text{of Beta-10 and no set of Beta-20 in order to gain maximum}\\ \text{profit of 125,000.}\end{array}$