12th NCERT Linear Programming Exercise 12.2 Questions 11

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Question (1)

Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P cost 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 Q contains 4units/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.Solution

Food | A | B | Cost | Quantity |

P | 3 | 5 | 60 | x |

Q | 4 | 2 | 80 | y |

-- -- | ≥ 8 | ≥ 11 | -- -- |

Let x unit of food P mixed with y units of food Q

minimise Z = 60x + 80y

Subject to constraint : 3x +4y ≥ 8 and 5x+2y ≥ 11

x ≥ 0, y ≥ 0

line l

x | 0 | 2 |

y | 2 | 0.5 |

line l

x | 0 | 2.2 |

y | 5.5 | 0 |

Z | ||

A | (0, 5.5) | 60(0) + 80(5.5) = 440 |

B | (2, 0.5) | 120 +40 = 160 |

C | ($\frac{8}{3}$ , 0) | 160 + 0 = 160 |

Minimum z = 160 at all the point of line segment joining (2, 0.5 ) and ($\frac{8}{3}$ , 0)

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 5 kg of flour and 1kg of fat assuming that there is no shortage of the other ingredients used in making the cakesSolution

type | flour | fat | -- -- |

A | 200 | 25 | x |

B | 100 | 50 | y |

≤ 5kg | ≤ 1 kg | ||

≤ 5000 | ≤ 1000 |

Let we prepare x cake of type A nad y cake of type B.

max. z = x + y

subject to 200x + 100y ≤ 5000

2x + y ≤ 50

25x + 50y ≤ 1000

x + 2y ≤ 40

x ≥ 0, y ≥ 0

2x + y = 50

x | 0 | 25 |

y | 50 | 0 |

x + 2y = 40

x | 0 | 40 |

y | 20 | 0 |

corner | point | z=x+y |

A | (0, 20) | 0+20=20 |

B | (20, 10) | 20 + 10 = 30 max |

C | (25,0) | 25+0=25 |

O | (0, 0) | 0+0=0 |

Maximum 30 cakes can be produced 20 of one type and 10 of other type

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 machin 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 racket and on a bat is Rs20 and RS 10 respectively, find the maximum profit of the factory when it works at full capacity

Solution

machin | craft | profit | ||

Tennis | 1.5 hr | 3 hr | x | 20 |

Cricket bats | 3 hr | 1 hr | y | 10 |

≤ 42 | ≤ 24 |

Let x tennis racketes and y cricket bats are produced

(i) max. z = x + y subject to constant

1.5x + 3y ≤ 42

x + 2y ≤ 28

and 3x + y ≤ 24

x ≥ 0 , y ≥ 0

line l

x | 0 | 28 |

y | 14 | 0 |

line l

x | 0 | 8 |

y | 24 | 0 |

corner | point | z=x+y |

A | (0, 14) | 0+14=14 |

B | (4, 12) | 4 +12 =16 max |

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

O | (0, 0) | 0 + 0= 0 |

max. number of bats produced = 16.

4 tennis rackets and 12 cricket bats

(ii) max. profit = 20x + 10y

max. profit = 20(4) + 10(12) = 200

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 Rs. 17.50 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 maximize his profit, if he operates his machines for at the most 12 hours a day?Solution

machin A | machin B | proft | quantity | |

nuts | 1 | 3 | 17.50 | x |

bolts | 3 | 1 | 7.00 | y |

<12 | <12 |

Let x bags of nut and y packets of bolts is prepared

max. z = 17.50 x + 7y

Subject to constraints :

x + 3y ≤ 12

3x + y ≤ 12

x ≥ 0, y ≥ 0

line l

x | 0 | 12 |

y | 4 | 0 |

line l

x | 0 | 4 |

y | 12 | 0 |

Corner | Point | Z=17.5x+7y |

A | (0, 4) | 28 |

B | (3, 3) | 73.5 max |

C | (4, 0) | 70 |

D | (0, 0) | 0 |

Maxmum profit = 73.5 at x = 3, y =3. He should produced 3 packs of nuts and 3 packs of bolts.

Question (5)

A factory manufacturers 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 screw A, while it takes 6 minutes on automatic and 3 minutes on 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 a day in order to maximize his profit? Determine the maximum profit.Solution

Type screw | Auto | hand operated | profit | quantity |

A | 4 min | 6 min | 7 | x |

B | 6 min | 3 min | 10 | y |

≤4hr = 240 min | ≤4hr = 240 min |

Max. z = 7x + 10y

Subject to

4x + 6 y ≤ 240 ⇒ 2x + 3y ≤ 120

6x + 3 y ≤ 240 ⇒ 2x + y ≤ 80

x ≥ 0, y ≥ 0

Line l

x | 0 | 60 |

y | 40 | 0 |

Line l

x | 0 | 40 |

y | 80 | 0 |

Corner | point | x=7x+10y |

A | (0, 40) | 7(0)+10(40) = 400 |

B | (30, 20) | 7(30)+10(20) = 410 max |

C | (40, 0) | 7(40)+10(0) = 280 |

O | (0, 0) | 0 + 0 =0 |

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 maximize his profit?Solution

grinding/cutting | sparay | profit | quantity | |

Pedestal lamp | 2 | 3 | 5 | x |

wooden shade | 1 | 2 | 3 | y |

≤ 12hr | ≤ 20hr |

Let x lamp and y sheds are produced

maximise Z = 5x + 3y

subject

2x + y ≤ 12, x ≥ 0

3x + 2y ≤ 20, y ≥ 0

Line l

x | 0 | 6 |

y | 12 | 0 |

Line l

x | 0 | 4 |

y | 10 | 4 |

Corner | point | Z = 5x + 3y |

A | (0, 10) | 0 + 30 = 30 |

B | (4, 4) | 20 + 12 = 32 (max) |

C | (6, 0) | 30 + 0 = 30 |

D | (0, 0) | 0 + 0 = 0 |

Maximum profit = Rs. 32, for it he should produced 20 lamps and 12 shades.

Question (7)

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A requires 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?Solution

cutting | assembly | profit | quantity | |

A | 5 min | 10 min | 5 | x |

B | 8 min | 8 min | 6 | y |

≤ 3h:20min = 200min | ≤ 4h=240 |

maximum z = 5x + 6y

Subject to 5x + 8y ≤ 200

10x + 8y ≤ 240 ⇒ 5x + 4y ≤ 120

line l

x | 0 | 40 |

y | 25 | 0 |

line l

x | 0 | 24 |

y | 30 | 0 |

corner | point | z = 5x + 6y |

A | (0, 25) | 0 + 150 = 150 |

B | (8, 20) | 40 + 120 = 160 (max) |

C | (24, 0) | 120 + 0 = 120 |

O | (0, 0) | 0 + 0 |

Maximum profit = 160. He has to produce 8 souvenirs of type A and 20 souvenirs of type B.

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.Solution

Type | cost | profit | quantity |

Desktop | 25,000 | 4,500 | x |

Portable | 40,000 | 5,000 | y |

≤ 70,00,000 | ≤250 |

Let he sells x desktop and y portable models.

Maximise Z = 4500 x + 5000y

subject to

x + y ≤ 250

25,000x + 40,000y ≤ 70,00,000

⇒ 5x + 8y ≤ 1400, x ≥ 0, y ≥ 0

line l

x | 0 | 250 |

y | 250 | 0 |

line l

x | 280 | 120 | 0 |

y | 0 | 100 | 175 |

corner | point | z = 4500x+5000y |

A | (0, 175) | 0+175(5000) = 8,75,000 |

B | (200, 50) | 4500(200)+5000(40) = 11,50,000 maximum |

C | (250, 0) | 4500(250) + 0=11,25,000 |

O | (0, 0) | 0+0=0 |

Maximum profit = 11,50,000 by selling 200 units of desktop and 50 units of protable model

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.Solution

Food | cost | vit A | minerals | quantity/td> |

F_{1} |
4 | 3 | 4 | x |

F_{2} |
6 | 6 | 3 | y |

≥ 80 | ≥ 100 |

Let x unit of food F

minimise z = 4x + 6y

subject to constraint

3x + 6y ≥ 80

4x + 3y ≥ 100

x ≥ 0, y ≥ 0

Line l

x | 0 | 26.7 |

y | 13.3 | 0 |

Line l

x | 17.5 | 25 |

y | 10 | 0 |

corner | point | z=4x+6y |

A | (0, $\frac{{100}}{3})$ | 4(0) + 6($\frac{{100}}{3}$) = 200 |

B | (24, $\frac{4}{3}$) | 4(24) + 6($\frac{4}{3}$)=104 min |

C | ($\frac{{80}}{3}$, 0) | 4($\frac{{80}}{3}$)+ 6(0) =106.67 |

Minimum cost is Rs104.

24 units of food F

Question (10)

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?Solution

Nitrogen | Phospharaus | cost | quantity | |

F_{1} |
10% | 6% | 6/kg | x |

F_{1} |
5% | 10% | 5/kg | y |

≥ 14 kg | ≥ 14 |

Let x kg of F

Minimise z = 6x +5y

subject to :

\[\frac{{10x}}{{100}} + \frac{{5y}}{{100}} \ge 14\] \[ \Rightarrow 10x + 5y \ge 1400\] \[\frac{{6x}}{{100}} + \frac{{10y}}{{100}} \ge 14\] \[ \Rightarrow 6x + 10y \ge 1400\] x ≥ 0, ≥ 0

line l

x | 0 | 140 |

y | 280 | 0 |

line l

x | 0 | $\frac{{700}}{3}$ |

y | 140 | 0 |

corner | point | z=6x+5y |

A | (0, 280) | 6(0)+5(280) =1400 |

B | (100, 80) | 6(100) +5(80) =1000 min |

C | ($\frac{{700}}{3}$, 0) | $6\left( {\frac{{700}}{3}} \right) + 5\left( 0 \right) = 1400$ |

so 100kg of F

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

Solution

max. z = px + qysubject to

2x + y ≤ 10

x +3y ≤ 15

x ≥ 0, y ≥ 0

Corner | point | z |

A | (0, 0) | 0 + 0 = 0 |

B | (5, 0) | 5p+0 = 5p |

C | (3, 4) | 3p + 4q (max) |

D | (0, 5) | 5q max |

∴ 3p + 4q = 5q

3p = q

So (D) is correct option