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 l1: 3x +4y = 8
 x 0 2 y 2 0.5

line l2 : 5x + 2y = 11
 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 cakes

Solution

 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 l1: x + 2y = 28
 x 0 28 y 14 0

line l2: 3x + y = 24

 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 l1: x + 3y = 12
 x 0 12 y 4 0

line l2: 3x + y = 12
 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 l1: 2x + 3y = 120
 x 0 60 y 40 0

Line l2: 2x + y = 80
 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
Maximum profit Rs410. For that he has to produce 30 packets of screw A and 20 packets of screws B is to be produced

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

Line l1 : 3x + 2y = 20
 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 l1: 5x + 8y = 200

 x 0 40 y 25 0

line l2: 5x + 4y = 120

 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 l1: x + y = 250

 x 0 250 y 250 0

line l2: 5x + 8y = 1400

 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> F1 4 3 4 x F2 6 6 3 y ≥ 80 ≥ 100

Let x unit of food F1 is mixed with y unit of food F2
minimise z = 4x + 6y
subject to constraint
3x + 6y ≥ 80
4x + 3y ≥ 100
x ≥ 0, y ≥ 0
Line l1: 3x + 6y = 80

 x 0 26.7 y 13.3 0

Line l1: 4x + 3y = 100

 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 F1 is mixes with $\frac{4}{3}$ unit of food F2.

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 F1 10% 6% 6/kg x F1 5% 10% 5/kg y ≥ 14 kg ≥ 14

Let x kg of F1 mix with y kg of F2.
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 l1: 10x+5y = 1400
 x 0 140 y 280 0

line l2: 6x + 10y = 1400
 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) $6\left( {\frac{{700}}{3}} \right) + 5\left( 0 \right) = 1400$
minimum cost = 1000
so 100kg of F1 should be mixed with 80 kg of F2.

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 + qy
subject 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
at (3, 4) and (0, 5) is max
∴ 3p + 4q = 5q
3p = q
So (D) is correct option