12th NCERT Linear Programming Miscellineous Question 10
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Question (1)

Refer to Example 9.( Diet problem) A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30g) of food P contains 12 units of calcium, 4 units of iron, 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 equires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How may packets of each food should be used to minimise the amount of viamin A in the diet? What is the minimum amount of vitamin A?
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?

Solution

 Food calcium iron cholesterol Vit A P 12 4 6 6 x Q 3 20 4 3 y ≥ 240 ≥ 460 ≤ 300
Maximise X = 6x + 3y
Subject to
12x + 3y ≥ 240
4x + 20y ≥ 460
6x + 4y ≤ 300
x > 0 and y > 0
Line l1 : 12x + 3y = 240;
 x 0 20 y 80 0

Line l2 : 4x + 20y = 460;
 x 0 115 y 23 0

Line l3 : 6x + 4y = 300;
 x 0 50 y 75 0

 Corner point Z= 6x+3y A(2, 72) 6(2) +3 (72) = 12 + 216 = 228 B(15, 20) 6(15) + 3(20) = 90 + 60 = 150 C(40, 15) 6(40)+3(15)= 240 + 45 = 285 max.

max z = 285 at (40, 15)
40 units of P should be mixed with 15 units of Q

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?

Solution

 cost A B C quantity P 250 3 2.5 2 x Q 200 1.5 11.25 3 y ≥ 18 ≥ 45 ≥ 24

Line l1: 2x + y = 12
 x 0 6 y 12 0

Line l2: x + 4.5y = 18
 x 0 18 y 4 0

Line l3: 2x + 3y = 24
 x 0 12 y 8 0

 Corner points Z= 250x+200y A(0, 12) 250(0)+200(12)=2400 B(3, 6) 250(3) + 200(6) = 1950 minimum C(9, 2) 250(9)+200(2)=2650 D(18, 0) 250(18)+200(0)=4500
Maximum cost is Rs 1950 for it mix 3 bags of P with 6 bags of Q

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:
 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?

Solution

 Food A B C cost quantity x 1 2 3 16/kg x kg y 2 2 1 20/ kg y kg ≥ 10 ≥ 12 ≥ 8

minimise z = 16x + 20y
Subject to constraint
x + 2y ≥ 10
2x + 2y ≥ 12 ⇒ x + y ≥ 6
3x + y ≥ 8
x ≥ 0 and y ≥ 0
Line l1: x + 2y = 10
 x 0 10 y 5 0

Line l2: x + y = 6
 x 0 6 y 6 0

Line l2: 3x + y = 8
 x 0 2 y 8 2

 Corner Point z= 16x+20y A(0, 8) 16(0)+20(8)=160 B(1, 5) 16(1)+20(5)=116 C(2, 4) 16(2)+20(4)=112 minimum D(10, 0) 16(10)+20(0)=160

minimum cost =112 at x =1 and y = 5
1 kg of x should be mixed with 5kg of y

Question (4)

A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) requires for each toy on the machines is given below:
 Type of Toys Machines I II III A 12 18 16 B 6 0 9

Each machin is available for a maximum of 6 hours per day. If the profit on each 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.

Solution

 I II III profit A 12 18 6 7.5 x B 6 0 9 5.0 y ≤ 6hr = 360 min ≤ 6hr = 360 min ≤ 6hr = 360 min
maximise z = 7.5 x + 5y
Subject to constraints:
12 x + 6y ≤ 360 ⇒ 2x + y ≤ 60
18x + 0y ≤ 360 ⇒ x ≤ 20
6x + 9y ≤ 360 ⇒ 2x + 3y ≤ 120
x ≥ 0 and y ≥0
Line l1: 2x + y = 60
 x 0 30 y 60 0

Line l2: x = 20
Line l3: 2x + 3y = 120
 x 0 60 y 40 0

 Corner Point z= 7.5x+5y A(0, 40) 0+200=200 B(15, 30) 112.5+150=262.5 max C(20, 20) 150+100=250 D(20, 0) 150+0=150 D(0, 0) 0+0=0

Maximum profit is when he produce 15 toys of type A and 30 toys of type B, profit Rs 262.5

Question (5)

An aeroplane carries 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 maximize the profit for the airline. What is the maximum profit?

Solution

Let x tickets of excecutive and y tickets of economy class
 executive economy profit 1000 600 tickets x y ≥ 20 ≥ 80
Maxi. z = 1000x + 600 y
Subject to :
x + y ≤ 200
x ≥ 20
y ≥ 4x
4x-y ≤ 0
x > 0 and y > 0
Line l1: x + y = 200
 x 0 200 y 200 0

Line l2: x= 20
Line l3: 4x - y = 0
 x 0 200 y 0 50

 Corner point z=1000x+600y A(40, 160) 40,000 +96,000 = 1,36,000 max B(20, 80) 20,000+48,000 = 68,000 C (20, 180) 20,000+ 108000 = 128000
Max. profit = 1,36,000
40 passengers of executive and 160 of eco

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:
 Transporation cost per quintel ( in Rs) From/To A B D 6 4 E 3 2 F 2.5 3

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

Solution

Let x quintals is suppled from A to D and y quintals from A to E
cost = 6x+3y+2.5(100-x-y)+4(60-x)+2(50-y)+3(x+y-60)
cost = 6x+3y+250-2.5x-2.5y+240-4x+100-2y+3x+3y-180
minz =2.5x + 1.5y + 410
subject to constrains:
x > 0, y > 0, 100-x-y > 0 ⇒ x+y<100
60 - x > 0 ⇒ x < 60
50 - y > 0 ⇒ y < 50
x+y-60 > 0 ⇒ x + y > 60
Line l1: x + y = 100
 x 0 100 y 100 0

Line l2: x + y = 60
 x 60 0 y 0 60

Line l3: x=60
Line l4: y=50
 Corner point z=2.5x+1.5y+410 A(10, 50) 25+75+41=510 mini B(50, 50) 125+75+410=610 C (60, 40) 150+60+410=620 D (60, 0) 150+0+410=560
Minimum cost is Rs 510
10, 50, 40 units from A and 50, 0, 0 units from B Should be supplied so that transportation cost will be minimum.

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 4500 L, 3000 L and 3500 L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:
 Transporation cost per quintel ( in Rs) From/To A B D 7 3 E 6 4 F 3 2

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

Solution

Min. cost = $\frac{7}{{10}}x + \frac{6}{{10}}y + \frac{3}{{10}}\left( {7000 - x - y} \right) + \frac{3}{{10}}\left( {4500 - x} \right) + \frac{4}{{10}}\left( {3500 - y} \right) + \frac{2}{{10}}\left( {x + y - 3500} \right)$ $= \frac{{7x}}{{10}} + \frac{{6y}}{{10}} + 2100 - \frac{{3x}}{{10}} - \frac{{3y}}{{10}} + 1350 - \frac{{3x}}{{10}} + 1200 - \frac{{4y}}{{10}} + \frac{{2x}}{{10}} + \frac{{2y}}{{10}} - 700$ $\min \quad C = \frac{{3x}}{{10}} + \frac{y}{{10}} + 3950$ Subject to constrains:
x > 0 , y > 0
7000 -x - y > 0 ⇒ x + y < 7000
4500 -x > 0 ⇒ x < 4500
3000 - y > 0 ⇒ y < 3000
x + y -3500 > 0 ⇒ x + y > 3500
Line l1: x + y = 7000
 x 0 7000 y 7000 0

Line l2: x = 4500
Line l3: y = 3000
Line l4: x + y = 3500
 x 0 3500 y 3500 0

 Corner point $z = \frac{{3x}}{{10}} + \frac{y}{{10}} + 3950$ A(500, 3000) 150+3000+3950=4400 minimum B(4000, 3000) 1200+300+3950=5450 C (4500, 2500) 1350+250+3950=5550 D (4500, 0) 1350+0+3950=5300 E (3500, 0) 1050+0+3950=5000
Minimum cost = 4400 at A(500, 3000)
From depo A, 500L, 3000L, 3500L and from depo B, 4000l, 0L, 0L should be supplied to D, E and F

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 minimize 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?
 kg PER BAG Brand P Brand Q Nitrogen 3 3.5 Phosphoric acid 1 2 Potash 3 1.5 Chlorine 1.5 2

Solution

 nitrogen phosphate potash Chlorine P 3 1 3 1.5 x Q 3.5 2 1.5 2 y ≥240 ≥270 ≤ 310
Maxi z = 3x+3.5y
x+2y ≥ 240
3x+1.5y ≥ 270 ⇒ 2x+y ≥ 180
1.5x + 2y ≤ 310 ⇒ 3x +4y ≤ 620, x > 0, y > 0
Line l1: x + 2y = 240
 x 0 240 y 120 0

Line l2: 2x + y = 180
 x 0 90 y 180 0

Line l3: 3x + 4y = 620
 x 0 40 y 155 125

 Corner point z = 3x + 3.5 y A(140, 50) 420+175=595 B(20, 140) 60+490=550 C (40, 100) 120+350=470 (min)
Minimum cost = 470 at (40, 100)
∴ 40 bags of P should be mixed with 100 bags of Q to mnimise nitrogen

Question (9)

Refer to Question 8. If the grower wants to maximize 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?

Solution

 nitrogen phosphate potash Chlorine P 3 1 3 1.5 x Q 3.5 2 1.5 2 y ≥240 ≥270 ≤ 310
Maxi z = 3x+3.5y
x+2y ≥ 240
3x+1.5y ≥ 270 ⇒ 2x+y ≥ 180
1.5x + 2y ≤ 310 ⇒ 3x +4y ≤ 620, x > 0, y > 0
Line l1: x + 2y = 240
 x 0 240 y 120 0

Line l2: 2x + y = 180
 x 0 90 y 180 0

Line l3: 3x + 4y = 620
 x 0 40 y 155 125

 Corner point z = 3x + 3.5 y A(140, 50) 420+175=595 maximum B(20, 140) 60+490=550 C (40, 100) 120+350=470
Maximum value of Z is 595 at (140, 50) so
∴140 bags of P should mixed with 50 bags of Q for maximum 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 maximize the profit?

Solution

 Type quantity profit A x 12 B y 16 ≥ 1200
max Z = 12x + 16y
subject to :
x + y < 1200
y ≤ $\frac{x}{2}$ ⇒ x ≥ 2y ⇒ x - 2y ≥ 0
x - 3y ≤ 600
x > 0, y > 0
Line l1: x + y = 1200
 x 0 1200 y 1200 0

Line l2: x - 2y = 0
 x 0 400 y 0 200

Line l3: x - 3y = 600
 x 0 600 y -200 0

 Corner point z = 12x + 16 y A(800, 400) 9600+6400= 16,000 maximum B(1050, 150) 12600+2400 = 15000 C (600, 0) 7200+0=7200 D (0, 0) 0+0=0
maximum profit = 16000 at x = 800, y=400
A company should produce 800 dolls of type A and 400 dolls of type B