Plus Two Maths Chapter Wise Questions and Answers Chapter 12 Linear Programming

are part of Plus Two Maths Chapter Wise Questions and Answers. Here we have given Plus Two Maths Chapter Wise Questions and Answers Chapter 12 Linear Programming.

Board | SCERT, Kerala |

Text Book | NCERT Based |

Class | Plus Two |

Subject | Maths Chapter Wise Questions |

Chapter | Chapter 12 |

Chapter Name | Linear Programming |

Number of Questions Solved | 29 |

Category | Kerala Plus Two |

## Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 12 Linear Programming

**Short Answer Type Questions (Score 3)**

Question 1.

Food X contains 6 units of vitamin A and 7 units of vitamin B per gram and it costs 12 paise per gram. Food Y contains 8 units of vitamin A andl2 units of vitamin B per gram and it costs 20 paise per gram. Daily minimum requirements of vitamin A and vitamin B are 100 units and 120 units respectively. Formulate the LPP mathematically so that the cost is to be minimized. **(3)**

Answer.

Let x gram of food X and y gram of food Y be taken. Clearly x ≥ 0 , y ≥ 0

Let Z denote the total cost. Then

Z = 12x + 20y

By the given conditions, 6x + 8 y ≥ 100 7x + 12 y ≥ 120 .

Hence the required LPP is :

Minimise Z = 12x + 20y

Subject to: 6x + 8 y ≥ 100 ,

7x + 12 y ≥ 120 and x, y ≥0

Maximize P = 6x + 8y subject to

4x + 2y≤ 60, 2x + 4y ≤48, x, y≥ 60,

P is maximum at (12, 6) and maximum value of P = 120.

Question 2.

Manu has Rs. 36,000 for purchase of rice and wheat. A bag of rice and a bag of wheat cost Rs 180 and Rs. 120 respectively. He has storage capacity for 250 bags only. He earns a profit of Rs. 11 and Rs. 9 per bag of rice and wheat respectively.

i. Formulate an LPP to maximize the profit. **(1)**

ii. Solve the LPP. **(2)**

Answer.

L Number of rice bags be x, of wheat bags be y.

So maximize Z = 11x + 9y subject to

180x +120y ≤ 36000 or 3x + 2y ≤ 600,

x + y ≤ 250 and x ≥ 0, y ≥ 0.

ii. The feasible region is obtained from the following graph

Question 3.

i. Draw the graph of 5.x + 10y = 50,

x + y = 1, y= 4, x,y = 0. 2 **(2)**

ii. Solve graphically the linear programming problem: Minimize

Z = 2x + y subject to 5x+10y ≤ 50, x+y ≥1,y≤4 and x,y ≥0. **(1)**

Answer.

ii. The corner points of the feasible region are A(0, 4), B (0,1), C(1,0), D(10, 0) and E (2, 4). The corresponding z values are 4,1,2, 20 and 8 respectively. The minimum value of z is 1 at x = 0, y = 1.

Question 4.

A manufacture make two types of furniture, chairs and tables. Both the products are processed on three machines A_{1}, A_{2} and A_{3}. Machine A_{2} requires 3 hours for a chair and 3 hours for a table, machine A_{2} requires5 hours for a hair and 2 hours for a table and machine A_{3} requires 2 hours for a chair and 6 hours for a table. Maximum time available on machine A_{1}, A_{2} and A_{3} is 36 hours, 50 hours and 60 hours respectively. Profits are Rs. 20 per chair and Rs.30 per table. Formulate the above as a linear programming problem to maximize the profit. **(3)**

Answer.

Machine A_{1} | Machine A_{2} | Machine A_{3} | Profit | |

Chair | 3hrs | 5hrs | 2hrs | Rs. 20 |

Table | 3hrs | 2hrs | 6hrs | Rs.30 |

36hrs | 50hrs | 60hrs |

Let x chairs and y tables are manufactured.

Then LPP is

To maximize P = 20x + 30y

Subject to constraints, x ≥0, y ≥ 0

3x +3y ≤36, 5x + 2y ≤50,2x + 6y≤ 60

Question 5.

Solve the following problem graphically: Minimize, Z = 3x+ 2y

Subject to constraints: x+ y≥8 **(3)**

3x + 5y≤15

x ≥0, y≥0

Answer.

Plotting the inequations x + y≥8,3x + 5y≤15 and x ≥ 0, y≥0, we notice there is no common shaded portion.

Hence no feasible solution; so no minimum Z.

Question 6.

a. The region other than the feasible region is called…………………**(1)**

b. If the feasible region of a linear programming is bounded, then it is always a ………… **(1)**

c. Who invented Linear programming **(1)**

Answer.

a. Infeasible region

b. Convex polygon

c. Leonid

Question 7.

The feasible region for the constraints is

(a. Area GFH b. Line segment FH c. Area ABH d. Line segment AH)

b. A person deals only two items, cycles and scooters. He has Rs. 1,20,000 to invest and a space to store at most 38 pieces. One scooter costs him Rs. 12,000 and a cycle costs him Rs. 800. He can sell S a scooter at a profit of Rs. 1,500 arid a cycle at a profit of Rs.200, Assuming that he can sell all the items he buys, how should he invest his money in order that he may maximize his profit. Formulate the problem mathematically. **(2)**

Answer.

a. b. Line segment FH .

b. Suppose the person buys x cycles,

y scooters x≥0, y ≥0

By the given conditions, x + y ≤ 38

800x+ 12000Y≤ 120000

i..e.. x + 15y ≤ 150

Let Z be the total profit

Z = 200x + 1500y

Then the required LPP is :

Maximise Z = 200x + 1500y

Subject to:

x+y ≤38, x+15y≤150 and x,y≥0.

Question 8.

i. Find the linear constraints for the shaded area from the figure. **(1)**

ii.

Find the maximum value of 5x+ y subject to the above constraints. **(2)**

Answer.

i. x+2y≤8, 2x+y≥ 2, x-y≤1 and

x ≥ 0, y ≥ 0

ii. The comer points are A (0,4 ), B (0, 2), C(1, 0) and The value of z are respectively 4,2,5 and 19. Hence the maximum value of z =5x + y subject to the above constraints is 19,at

Question 9.

Maximise the corner points of the feasible region determined by the following system of linear inequations: 2x + y<10, x + 3y ≤15 , x,y ≥ 0 are (0,0),(5,0),(3,4) and (0,5). Let , Condition on p and q so that the maximum of Z occurs at both (3,4) and (0,5). **(3)**

Answer.

The maximum value of Z is unique.

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-q

Question 10.

A firm manufactures 3 products A,B and C. The profits are Rs.3, Rs.2 and Rs.4 respectively. The firm has 2 machines and below is the required processing time in minutes for each mahine on each product:

Machine M_{1 }and M_{2} have 2000 and 2500 machine minutes respectively. The must manufacture 100 A’s, 200 B’s and 50 C’s but not more than 150A’s. Set up a LPP to maximize the profit. **(3)**

Answer.

Max. Z = 3x+ 2y + 4z

Subject to

4x+ 3y + 5z ≤ 2000

2x+ 2y + 4z ≤ 2500

100≤ x ≤ 150

y ≥ 200

z ≥ 50

x≥ 0, y ≥ 0, z ≥ 0

**Long Answer Type Questions**

**(Score 4)**

Question 1.

Determine graphically the minimum value of the objective function Z = -50x+20y. Subject to constraints:

2x – y ≥ -5, 3x+ y ≥ 3

2x – 3y ≤ 12, x ≥ 0 , y ≥ 0 **(4)**

Answer.

The feasible region of the system of inequations given in constraints is shown in figure. We observe that the feasible region is unbounded.

The values of the objective function Z at the corner points are given in the following table.

Comer point (x,y) | Value of the objective functionZ = -50x +20y |

(0,5) | 100 |

(0,3) | 60 |

(1,0) | -50 |

(6,0) | -300 |

Clearly -300 is the smallest valuce of Z at the corner point (6,0). Since the feasible region is unbounded. Therefore, to check whether -300 is the minimum value of Z, we draw the line -300 = -50x + 20y and check whether the open half plane – 50x + 20y < – 300 has points in common with the feasible region or not. From Figure, we find that the open half plane represented by -5 Ox + 20y <¬300 has points in common with the feasible region. Therefore, Z = -50x + 20y has no minimum value subject to the given constraints.

Question 2.

Represents the feasible region of the given LPP graphically. Z = 3x+5y

Subject to 3x – 4y + 12 ≥ 0

2x – y + 2 ≥ 0

2x + 3y – 12 ≥o

0≤ x≤ 4,

y ≥2 **(4)**

Answer.

Question 3.

A company produces two articals X and Y. There are two different departments through which the articles are processed namely assembly and finishing. The potential capacity of the assembly department is 60 hours, a week and that of finishing department is 48 hours a week. Production of one unit of X requires 4 hours of assembly and 2 hours of finishing. Each of the unit Y requires 2 hours in assembly and 4 hours in finishing. If profit is Rs 8 for each unit of X and Rs 6 for each unit of Y,

i. Formulate an LPP to maximize the profit

ii. Find the number of units of X and Y to be produced each week to get maximum profit. **(4)**

Answer.

i. Let x units of X and y units of Y be produced.

Total profit Z = 8x + 6y .

The following are the constraints 4x+ 2y ≤ 60 or 2x+y ≤ 30 (Assembly Dept.)

2x+ 4y ≤ 48 or x+ 2y ≤ 24 (Finishing – Dept.)

Hence the LPP is Max. Z = 8x + 6y,

subject to the constraints 2x + y ≤ 30,

x + 2y ≤ 24 and x ,y≥ 0

ii. The feasible region determined by the system of constraints and non-negativity restrictions is given in the following graph.

We observe that the feasible region OABC is bounded. So,we now use comer point method to determine the maximum value of Z.

The co-ordinates of the comer points: O, A, B and C are (0,0), (15,0), (12,6) and (0, 12) respectively

Now we evaluate Z at each comer points

Comer Point | Corresponding value of Z 8x + 6y |

O (0,0) | 0 |

A (15,0) | 120 |

B(12,6) | 132 Max |

C(0,12) | 72 |

Hence maximum value of Z is 132 at the point (12, 6). Thus x = 12, y = 6.

Question 4.

A cooperative society of farmers has 50 hectares of land to grow two crops10500 and Rs.9000 respectively. The A and B. The profits from crops A and control weeds, a liquid herbicide has

to be used for crops A and B at the rates of 20 litres and 10 litres per hect are, respectively. Further not more than 800 litres of herbicide should be used in order to protect fish and wild life using a pond which collects drainage from his land. Keeping in mind that the protection of fish and other wild life is more important than earning profit, how much land should be allocated to each crop so as to maximize the total profit? From an LPP B per hectare are estimated as Rs. from the above and solve it graphically. Do you agree with the message that the protection of wild life is utmost necessary to preserve the balance in environment ? **(4)**

Answer.

Let x hectares and y hectares of land is used to grow crop A and crop B respectively.

LPP is to maximize Z = 10500x + 9000y

Subject to constraints x≥0,y≥0

x + y≤50

20x + 10y≤800 ⇒ 2x + y≤80

Plotting the above inequations, we notice,

we have shaded portion as feasible solution. Possible points for maximum

Point | Z= 10500 x 9000 y | Value (in Rs) |

A (40,0) | 420000+0 | 420000 |

B(30,20) | 315000+180000 | 495000 |

C(0,50) | 0+450000 | 450000 |

i.e., x = 30, y = 20

Hence crop A should be grown at 30 hectares of land and crop B at 20 hectares of land for a maximum profit of Rs. 495000.

Yes, protection of wild life is utmost necessary to preserve the balance in environment.

Question 5.

A company produces two types of goods A and B that require gold and silver. Each unit of type A requires 1 gram of silver and 2grams of gold. Type B requires (each unit) 2 grams of silver and 1 gram of gold. The company has only 100 grams of silver and 80 grams of gold. Each unit of type A brings a profit of Rs.500 and each unit of type B brings a profit of Rs. 400. Find the number of units of each type the company should produces to maximise the profit. **(4)**

Answer.

Let x units of type A and y units of type B are produced.

LPP is to maximize Z = 500x + 400y

Subject to constraints,

x ≥ 0,y ≥0

x + 2y≤100,2x + y≤80

Plotting the inequations on graph we notice shaded portion is feasible solution. Possible points for maximum Z are A(40,0), B(20,40) and C (0, 50).

Point | Z =500x + 400y | Value |

A (40,0) | 20000+0 | 20000 |

B(20,40) | 10000+16000 | 26000 |

C(0,50) | 0+20000 | 20000 |

Z is maximum for B(20,40) i.e., x = 20, y = 40

Hence 20 units of type A and 40 units of type B should be produced for maximum profit of Rs. 26000.

**Very Long Answer Type Questions (Score 6)**

Question 1.

A dietitian wishes to mix two types of food in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 unit/ kg of vitamin A and

1 unit/kg of vitamin C while food II contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs.50 per kg to produce Food II and Rs. 70 per kg to produce Food 11. Find the minimum cost of such a mixture. Formulate the above LPP mathematically and then solve it. **(6)**

Answer.

Let x kg of Food I and y kg of Food II are mixed. LPP is to minimize Z = 50x + 70y. Subject to constraints On plotting the in equations on graph we notice shaded portion is optimum solution. Possible points for minimum Z are A(10,0), B(2,4), C(0,8).

Point | Z = 50x+70y | Value |

A(10,0) | 500+0 | 500 |

B(2,4) | 100+280 | 380 |

C(0,8) | 0+560 | 560 |

We notice Z is minimum for B(2,4). i.e., x = 2, y = 4.

Hence 2 kg of Food I and 4 kg of Food II must be mixed for a minimum cost of Rs.380.

Question 2.

Solve the following problem graphically.

Minimize and Maximize Z = 3x+9y

Subject to constraints: x+3y ≤ 60

x + y ≥ 10, x≤ y, x ≥ 0,y ≥0 **(6)**

Answer.

Plotting the inequations x + 3y ≤ 60,

x + y ≥ 10, x≤ y, x ≥ 0,y ≥0

We notice common shaded portion is the feasible solution. Possible points for maximum and minimum Z are A (5,5), B(15, 15), C(0,20),D(0,10).

Point | Z = 3x + 9y | Value |

A(5,5) | 15 + 45 | 60 |

B(15, 15) | 45 + 135 | 180 |

C(0,20) | 0+180 | 180 |

D(0,10) | 0+90 | 90 |

Minimum Z is at A (5,5). i.e., x = 5, y = 5, Minimum Z = 60.

Maximum Z is at B (15, 15). i.e., x = 15, y = 15 and

C(0, 20) i.e., x = 0 , y = 20, Maximum Z = 180.

Question 3.

An aeroplane can carry a maximum of 200 passengers. A profit of Rs. 500 is made on each executive class ticket out of which 20% will go to the welfare fund of the employees. Similarly a profit of Rs. 400 is made on each economy ticket out of which 25% will go for the improvement of facilities provided to economy class passengers. In both cases, the remaining profit goes to the airline’s fund. The airline reserves at least 20 seats for executive class. However at least four 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 net profit of the airline? Make the above as an LPP and solve graphically. Do you think, more passengers would prefer to travel by such an airline than by others? **(6)**

Answer.

Let x tickets of executive class and y tickets of economy class are sold.

LPP maximize

i.e., Z = 400x + 300y

Subject to constraints

x≥0,y≥0,x + y≤200,x ≥20,y≥4x

Plotting the inequations on graph we notice shaded portion is optimum solution. Possible points for maximum profits are A(20,80),B(40,160),C(20,180).

Point | Z = 500x + 400y | Value |

A (20,80) B(40,160) C(20,180) | 8000+ 24000 16000+48000 8000 + 54000 | 32,000 64000 6200 |

Profit is maximum at B(40,160). i.e., x = 40, y = 160

Hence 40 tickets of executive class and 160 tickets of economy class must be sold to get a’maximum profit of Rs. 64000.Yes, as they care for employees and safety and facilities for passengers.

Question 4.

A factory makes two type of A and B, made of plywood. One piece of item A requires5 minutes for 5 minutes for Cutting and 10 minutes for assembling. One piece of item B requires 8 minutes for cuttin and 8 minutes for assembling. There are 3 hours and 20 minutes available for Cutting and 4 hours for assembling. The profit on one piece of item A is Rs. 5 and that on item B is Rs.6. How many pieces of each type should the factory make so as to maximise profit? Make it as an LPP and solve it graphically.**(6)**

Answer.

Cutting< 3hr 20 min | Assembling<4hr | Profit | |

Item A (x) | 5min | 10 min | Rs.5 |

Item B (y) | 8 min | 8 min | Rs.6 |

Let x pieces of item A and y pieces of iteam B be produced

∴ LPP is to maximize Z = 5x+6y

Subject to constraints,

x ≥ 0, y ≥ 0

5x + 8y≤ 200

10x + 8y≤ 240

and On plotting the inequalities, we notice shaded portion represents the optimum solution. Some points for maximum Z are A(24,0), B(8,20), C(0,25).

We notice that profit is maximum at B(8, 20). i.e., x = 8, y = 20. Hence 8 pieces of item A and 20 pieces of item B be produced to get a maximin profit of Rs. 160.

Question 5.

A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has 30 and 17 units of workers (male and female) and capital respectively, which he uses to produce two types of goods A and B. To produce one unit of A, 2 workers and 3 units of capital are required while 3 workers and 1 unit of capital is required to produce one unit of B. If A and B are priced at Rs.100 and Rs.120 per unit respectively, how should he use his resources to maximise the total revenue? Form the above as an LPP and solve graphically. **(6)**

Answer.

Let the objective function is

z_{max} = 100x + 120y

Subject to 2x + 3y ≤30

3x + y ≤ 17

x, y ≥ 0

Consider 2x + 3y = 17 and x = 0

y = 0

Shaded region is the feasible region.

The comer points are

O (0, 0), A, c (3, 8) and D (0, 10)at O (0, 0), z = 0

at A,z=

at C (3, 8), z = 1260 at D (0, 10), z= 1200

∴ Maximum revenue = Rs. 1260.

**Edumate Questions & Answers**

Question 1.

i. Choose the correct answer from the bracket. If an LPP is consistent, then its feasible region is always

{(a) Bounded (b) Unbounded (c) Convex region (d) Concave region}

ii. Maximize Z=2x+3y subject to the constraints

x + y≤ 4, x≥0,y ≥0

Answer.

i. c. convex region

Comer points of the feasible region are as follows

Corner points | Z=2x+3y |

0(0,0) | 0 |

A(0,4) | 12 →Maximum |

B(4,0) | 8 |

the maximun value of Z is 12 attained at (0,4)

Question 2.

A manufacture makes tea cups A and B. Three machines are needed for the manufacturing and the time in minutes required for each cups in the machine is given below .

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 cup A is 75 paise and that on each cup B is 50 paise . How many cups of types A and type B manufactured in a day to get maximum profit?

Answer.

Let x and y be the number of tea cups of types A and B

Then the LPP is , Maximize Z=75x+50y

subject to the constraints

12x + 6y ≤ 360

18x + 0y≤360

6x + 9y≤ 360

Comer point | Z=75x+50y |

0(0,0) A(0,40) B(15,30) C(20,20) D(20,0) | 0 2000 2625→ maximum 2500 1500 |

15 tea cups A and 30 tea cups B required to get maximum profit

Question 3.

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 progaramming problem . Find the miminum costs for diet that consists of mixture of these two foods and also meets the minimal nutritional requirments

Answer.

Let x units of food F_{1 }and y units of food

F_{2} be in the diet.

Total cost Z = 4x

Then the LPP is

Minimize Z = 4x + 6y

Subject to the constraints

3x+6y≥80

4x+3y≥100

x,y > 0

As the feasible region is unbounded , 104 may not be the minimum value of Z. For this we draw a graph of the inequality 4x+6y<104 or 2x+3y<52 and check whether the resulting half plane has in common with the feasible region or not. It can be seen that the feasible region has no common points with 2x+3y<52. Therefore minimum cost of the mixture will be 104

Question 4.

The graph of a liner programming problem is given below . The shaded region is the feasible region. The objective function is Z=px+qy.

i. What are the co-ordinates of the corners of the feasible region.

ii. Write the constrains.

iii. If the Max .Z occurs at A and B, what is the relation between p and q?

iv. If q=1, write the objective function

v. Find the Max Z.

Answer.

.i. Corner points are (0,0),(5,0),(3,4),(0,5)

ii. Constraints are 2x+y≤ 10

x+3y≤ 15,x≥0,y≥0

At(3,4),Z = 3p + 4q

At(5,0),Z= 5p

⇒ 3p + 4g = 5p

⇒ P = 2q

iii. If q = 1, p = 2

Then the objective function is

Maximize Z = 2x + y

iv. At (3,4) Z = 2 x 3 + 4=10 is the maximum value.

**NCERT Questions & Answers**

Question 5.

A dietician wishes to mix two types of food in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units per kg of vitamin A and 1 unit per kg of vitamin C, while Food II contains 1 unit per kg of vitamin A and 2 units per kg of vitamin C. It costs Rs 5 per kg to purchase food I and Rs 7 per kg to purchase Food II.

i. Formulate the above problem as an LPP to minimize the cost of such a mixture.

ii. Determine the minimum cost of the mixture which will produce the diet.

Answer.

i. Let the dietician mix x kg of Food I and y kg of Food II.

Clearly x ^{3} 0, y ^{3} 0; Let Z denote the total cost. Then Z = 5x + 7y

By the given conditions, the constraints are 2x + y ≥ 8 (vitamin A)

x + 2y≥ 10 (vitamin C)

Hence the LPP is as follows:

Min. Z=5x +7y subject to the constraints 2x + y ≥ 8 . x + 2y ≥ 10 and x, y ≥ 0

ii. The shaded region in the following graph is the feasible region of the LPP

The co-ordinates of the corner points of this region are A(10,0), B(2, 4) and C (0.8).

The values of the objective function

Z = 5x + 1y at the corner points of the feasible region are given in the following table.

Corner Point | Corresponding value of Z |

A (10, 0) B (2, 4) C (0, 8) | Z = 5x 10 + 7×0 = 50 Z = 5×2 + 7×4 = 38Min Z = 5×0 + 7×8 = 56 |

Clearly Z is minimum at x = 2, y = 4. The minimum value of Z = 38.

Question 6.

There is a factory located at each of the two places P and Q. From these locations, a certain commodity is delivered to each of these 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:

To/From | Cost (in Rs) | ||

A | B | C | |

P | 16 | 10 | 15 |

Q | 10 | 12 | 10 |

How many units should be transported from each factory to each depot in order that the transportation cost is minimum? Formulate the above LPP mathematically and then solve it.

Answer.

Let the factory at P transports x units of commodity to depot at A and y units to depot at B. Since the factory at P has the capacity of 8 units of the commodity, the left out (8- x-y) units will be transported to depot at C. The requirements are nonnegative quantities.

Therefore, x ≥ 0, y ≥ 0 and 8 -x-y ≥ 0 ⇒ x ≥ 0, y ≥ 0 and x +y ≤ 8.

As the requirements at the depots at A. B and C are always non-negative.

5 – x ≥0, 5 – y ≥ 0 and x + y – 4 ≥ 0

⇒ x ≤ 5, y ≤ 5 and x + y ≥ 4.

Let Z be the total transportation cost.

Then Z = 16x + 10y + 15(8 -x-y) + 10(5 -x) +12(5 -y) + 10(x + y – 4)

⇒ x- 7y + 190

The following diagram will be helpful to write the rest of the constraints.

Hence the above LPP can be stated as follows:

Min. Z= x- 7y + 190 subject to the constraints.

x + y≤ 8, x + y ≥4, x ≤ 5 and x, y ≥0

The feasible region of the LPP is shaded in the following graph.

The corner points and corresponding values of the objective function are given in the following table:

Corner Point | Corresponding value of Z |

A (4, 0) | 194 |

B (5, 0) | 195 |

C (5, 3) | 174 |

D (3, 5) | 158 |

E (0, 5) | 155 Min. |

F (0, 4) | 162 |

The optimal transportation strategy will I be to deliver 0, 5 and 3 units from the factory at P and 5, 0 and 1 unit from the factory at Q to the depots at A, B and C respectively. The minimum transportation cost in this case is Rs 155.

Question 7.

A manufacturer has three machines II, II and III installed in his factory. 1 Machine I and II are capable of being operated for at the most 12 hours, where as machine III must be operated at least for 5 hours a day. He produces only two items, each require the use of the three machines. The number of hours required for producing 1 unit of each of the items A and B on the three machines are given below:

Item | Number of hours required on the machines |

I II III | |

A B | 1 2 1 2 1 5/4 |

He makes a profit of Rs. 60 on item A I and Rs. 40 on item B. Assuming that he can sell all that he produces, how many of each item should he produce so as to maximize his profit? Formulate the problem as a LPP.

Answer.

Let x be the number of items A and y be the number of items B produced.

Total profit on the production

Z. 60x+40y

Maximize z = 60x+ 40y.

The problem is subject to the constraints

x + 2y ≤ 12 , 2x + y ≤12,

x+ y>5 and x, y≥0

Corner point | Z = 60x + 40y |

(5, 0) | 300 |

(6, 0) | 360 |

(4, 4) | 400 ← |

(0. 6) | 240 |

(0. 4) | 160 |

We see that the point (4, 4) is giving the maximum value of Z. Hence, the manufacturer has to produce 4 units of each item to get the maximum profit of Rs. 400.

Question 8.

i. Draw the graph of x + 3y = 3, x+ y = 2, x, y = 0.

ii. Solve graphically, minimize Z = 3x + 5y subject to x + 3 y ≥3, x + y ≥ 2, x, y ≥ 0.

Answer.

ii. The objective function is Z = 3x+5y

The constraints are x+3y≥3, x+y≥2, x, y≥0.

The feasible region is shaded in the figure.

From the table the minimum value of Z is 7. Since the feasible region is unbounded, 7 may or may not be minimum value of Z. Consider the graph of the inequality 3x+5y<7. This half plane has no point in common with the feasible region. Hence the minimum value of Z is 7 at

Question 9.

A fruit grower can use two types offertilizer in his garden, brand P and brand I 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 atleast 240 kg of phosphoric acid, atleast 270 kg of potash and atmost 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 garder?

kg per bag | ||

Brand P | BrandQ | |

Nitrogen | 3 | 3.5 |

Phosphoric acid | 1 | 2 |

Potash | 3 | 1.5 |

Chlorine | 1.5 | 2 |

Answer.

Let x bags of brand P and y bags of brand Q be used in the garden.

We have the data as

Phosphoric acid constraint : x + 2y ≥ 240

Potash constraint : 3x +1.5y ≥ 270

Chlorine constraint : 1.5.x+2y≤310

Quantity of nitrogen : Z = 3x + 3.5 y

The L.P.P. is Minimise Z = 3x + 3.5y

subject to the constraints x+2y≥240,

3x + 1.5 y ≥ 270, 1 ,5.y + 2 y ≤ 310,x, y ≥0

The feasible region is shaded in the figure.

Comer points | Z = 3x + 3.5y |

A (140, 50) | Z = 3×140 + 3.5×50 =595 |

B (20, 140) | Z = 3×20 + 3.5×140 = 550 |

C (40, 100) | Z = 3×40 + 3.5×100 = 470 |

∴ Minimum value of Z is 470 at C (40, 100)

Hence 40 bags of brand P and 100 bags of brand Q are used.

We hope the given Plus Two Maths Chapter Wise Questions and Answers Chapter 12 Linear Programming will help you. If you have any query regarding Plus Two Maths Chapter Wise Questions and Answers Chapter 12 Linear Programming, drop a comment below and we will get back to you at the earliest.