Linear Programming Test

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Linear Programming Test - 7

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Weekly Quiz Competition

Question 1

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There are two types of fertilizers 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 fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

120 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1200

110 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1100

130 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1300

100 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1000

Solution

Let number of kgs. of fertilizer F1 = x
And number of kgs. of fertilizer F2 = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 6x +5y , subject to the constraints : 10/100 x + 5/100y ≥ 14 and 6/100x + 10/100y ≥ 14, i.e. 2 x + y ≥ 280 and 3x + 5y ≥ 700, x,y ≥ 0.,

Corner points	Z =6x +5 y
A ( 0 , 280 )	1400
D(700/3,0 )	1400
B(100,80)	1000………….(Min.)

Corner points Z =6x +5 y A ( 0 , 280 ) 1400 D(700/3,0 ) 1400 B(100,80) 1000………….(Min.) Here Z = 1000 is minimum.
i.e. 100 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1000.

Question 2
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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

Solution

Here Z = px +qy , subject to constraints :
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0
As it is given that Z is maximum at ( 3 ,4 ) and ( 0, 5 ).
Therefore , 3p + 4q = 0p + 5q , which gives 3p = q .

Question 3

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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?

5 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs2250

4 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 2150

3 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 1950

6 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs2350

Solution

Let number of bags of cattle feed of brand P = x
And number of bags of cattle feed of brand Q = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 250x +200y , subject to the constraints : 3 x + 1.5y ≥ 80, 2.5x + 11.25y ≥ 45, 2x + 3y ≥ 24 , x,y ≥ 0.,

Corner points	Z =250x +200 y
C( 0,12 )	2400
B (18,0)	4500
D(3,6 )	1950…………………(Min.)
A(9,2)	2650

Here Z = 1950 is minimum.
i.e. 3 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 1950 .

Question 4

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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:

Food	Vitamin A	Vitamin B	Vitamin C
X	1	2	3
Y	2	2	1

One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?

Least cost of the mixture is Rs 122 (2 kg of Food X and 4 kg of food Y).

Least cost of the mixture is Rs 112 (2 kg of Food X and 4 kg of food Y).

Least cost of the mixture is Rs 132 (3 kg of Food X and 4 kg of food Y).

Least cost of the mixture is Rs 142 (2 kg of Food X and 5 kg of food Y).

Solution

Let number of kgs of food of brand X = x
And number of kgs of food of brand Y = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 16x +20y , subject to the constraints : x + 2y ≥ 10,2x + 2y ≥12,3x + y ≥8,x,y ≥ 0.,

Corner points	Z =16x +20 y
C(10 , 0 )	160
B (0,8)	160
D(1,5 )	116
A(2,4)	112……………..(Min.)

Here Z = 112 is minimum.
i.e. Least cost of the mixture is Rs 112 (2 kg of Food X and 4 kg of food Y).

Question 5
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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 ?

A
44 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 146000

B
45 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 156000

C
40 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 136000

D
42 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 139000

Solution

Let number of executive class tickets = x
And number of economy class tickets = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 1000x +600y , subject to the constraints : x + y ≤ 200, 4x - y ≤ 0, x ≥ 20,x,y ≥ 0.,

Corner points

Z =1000x +600 y

C(20 ,80 )

68000

B (40,160)

136000

D(20,180 )

12800

Here Z = 136000 is maximum.
i.e. 40 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 136000 .

Question 6

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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:

Transportation cost per quintal (in Rs)
From/To	A	B
D	6	4
E	3	2
F	2.5	3

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

From A : 10,50, 40 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 510

From A : 12,52, 40 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 530

From A : 10,53, 44 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 570

From A : 10,52, 42 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 550

Solution

Let the number of units of grain transported from godown A to D = x And the number of units of grain transported from godown A to E = y Therefore , the number of units of grain transported from godown A to F = 100 – (x+y) Therefore , the number of units of grain transported from godown B to D = 60 – x The number of units of grain transported from godown B to E = 50 – y The number of units of grain transported from godown B to F = x + y – 60 . Here , the objective function is :Minimise Z = 2.5x + 1.5y + 410 . , subject to constraints : 60 - x ≥ 0,50 - y ≥ 0 ,100 – (x + y) ≥ 0 , (x + y) - 60 ≥ 0 , x,y ≥ 0.,

Corner points	Z = 2.5x +1.5 y + 410
C(60 , 0 )	560
B (60,40)	620
D(50,50 )	610
A(10,50)	510……………..(Min.)

Here Z = 510 is minimum.i.e. From A : 10,50, 40 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 510 .

Question 7

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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:

Distance in (km)
From/To	A	B
D	7	3
E	6	4
F	3	2

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

From A: 520, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4600

From A: 540, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4700

From A: 500, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4400

From A: 550, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4800

Solution

Here objective function is : Z = 3x + y + 39500 , subject to constraints : : x + y ≤ 7000, x ≤ 4500, x + y ≥ 3500, , y ≤ 3000x , x,y ≥ 0

Corner points	Z = 3x + y + 39500
C(3500 , 0 )	50000
B (4500,0)	53000
D(4500,2500 )	55500
A(4000,3000)	54500
E(500 , 3000)	44000……………….(Min.)

Here Z = 4400 is minimum.i.e. . From A: 500, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4400 .

Question 8

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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? should the delivery be scheduled in order that the transportation cost is minimum?

Kg per bag
	Brand P	Brand Q
Nitrogen Phosphoric acid Potash Chlorine	3 1 3 1.5	3.5 2 1.5 2

45 bags of brand P and 105 bags of brand Q; Minimum amount of nitrogen = 480 kg.

47 bags of brand P and 107 bags of brand Q; Minimum amount of nitrogen = 499 kg.

50 bags of brand P and 100 bags of brand Q; Minimum amount of nitrogen = 490 kg.

40 bags of brand P and 100 bags of brand Q; Minimum amount of nitrogen = 470 kg

Solution

Let the number of bags used for fertilizer of brand P = x And the number of bags used for fertilizer of brand Q = y . Here , Z = 3x + 3.5y subject to constraints : :1.5 x +2 y ≤ 310, x + 2y ≥ 240, 3x + 1.5y ≥ 270 , x,y ≥ 0

Corner points	Z =3x + 3.5 y
C(40 ,100 )	470……………….(Min.)
B (140,50)	595
D(20,140 )	550

Here Z = 470 is minimum i.e. 40 bags of brand P and 100 bags of brand Q; Minimum amount of nitrogen = 470 kg.

Question 9

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Kg per bag

Brand P

Brand Q

Nitrogen

Phosphoric acid

Potash

Chlorine

1.5

3.5

1.5

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?

140 bags of brand P and 50 bags of brand Q; Maximum amount of nitrogen = 595 kg

150 bags of brand P and 50 bags of brand Q; Maximum amount of nitrogen = 625 kg

160 bags of brand P and 52 bags of brand Q; Maximum amount of nitrogen = 635 kg

145 bags of brand P and 55 bags of brand Q; Maximum amount of nitrogen = 555 kg

Solution

Corner points	Z =3x + 3.5 y
C(40 ,100 )	470……………….(Min.)
B (140,50)	595……………………..(Max.)
D(20,140 )	550

Here Z = 595 is maximum i.e. 140 bags of brand P and 50 bags of brand Q; Maximum amount of nitrogen = 595 kg .

Question 10

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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?

800 dolls of type A and 400 dolls of type B; Maximum profit = Rs 16000

840 dolls of type A and 404 dolls of type B; Maximum profit = Rs 16500

820 dolls of type A and 420 dolls of type B; Maximum profit = Rs 16200

830 dolls of type A and 430 dolls of type B; Maximum profit = Rs 16300

Solution

Here , Maximise Z = 12x + 16y , subject to constraints : : x + y ≤ 1200, x - 2y ≥ 0, x - 3y ≤ 600 , x,y ≥ 0

Corner points	Z = 12x +16 y
C(0, 0 )	0
B (600,0)	7200
D(1050,150 )	15000
A(800,400)	16000……………..(Max.)

Here Z = 16000 is maximum.i.e. 800 dolls of type A and 400 dolls of type B; Maximum profit = Rs 16000 .

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Corner points	Z =1000x +600 y
C(20 ,80 )	68000
B (40,160)	136000
D(20,180 )	12800

Linear Programming Test - 7

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Linear Programming Test - 7

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Question 1

120 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1200

110 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1100

130 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1300

100 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1000

Solution

Question 2

p = q

p = 2q

p = 3q

q = 3p

Solution

Question 3

5 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs2250

4 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 2150

3 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 1950

6 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs2350

Solution

Question 4

Least cost of the mixture is Rs 122 (2 kg of Food X and 4 kg of food Y).

Least cost of the mixture is Rs 112 (2 kg of Food X and 4 kg of food Y).

Least cost of the mixture is Rs 132 (3 kg of Food X and 4 kg of food Y).

Least cost of the mixture is Rs 142 (2 kg of Food X and 5 kg of food Y).

Solution

Question 5

44 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 146000

45 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 156000

40 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 136000

42 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 139000

Solution

Question 6

From A : 10,50, 40 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 510

From A : 12,52, 40 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 530

From A : 10,53, 44 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 570

From A : 10,52, 42 units; From B: 50,0,0 units to D, E and F respectively and minimum cost = Rs 550

Solution

Question 7

From A: 520, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4600

From A: 540, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4700

From A: 500, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4400

From A: 550, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4800

Solution

Question 8

45 bags of brand P and 105 bags of brand Q; Minimum amount of nitrogen = 480 kg.

47 bags of brand P and 107 bags of brand Q; Minimum amount of nitrogen = 499 kg.

50 bags of brand P and 100 bags of brand Q; Minimum amount of nitrogen = 490 kg.

40 bags of brand P and 100 bags of brand Q; Minimum amount of nitrogen = 470 kg

Solution

Question 9

140 bags of brand P and 50 bags of brand Q; Maximum amount of nitrogen = 595 kg

150 bags of brand P and 50 bags of brand Q; Maximum amount of nitrogen = 625 kg

160 bags of brand P and 52 bags of brand Q; Maximum amount of nitrogen = 635 kg

145 bags of brand P and 55 bags of brand Q; Maximum amount of nitrogen = 555 kg

Solution

Question 10

800 dolls of type A and 400 dolls of type B; Maximum profit = Rs 16000

840 dolls of type A and 404 dolls of type B; Maximum profit = Rs 16500

820 dolls of type A and 420 dolls of type B; Maximum profit = Rs 16200

830 dolls of type A and 430 dolls of type B; Maximum profit = Rs 16300

Solution

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