NCERT Solutions for Exercise 12.2 Class 12 Maths Chapter 12 - Linear Programming

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

Answer:

Let there be x cakes of first kind and y cakes of the second kind.Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in the table as :

Therefore,

$200x+100y\leq 5000$

$\Rightarrow \, \, \, \, 2x+y\leq 50$

. $\, \, 25x+50y\leq 10000$

$\Rightarrow \, \, x+2y\leq 400$

The total number of cakes, Z. Z=X+Y

Subject to constraint,

$\Rightarrow \, \, \, \, 2x+y\leq 50$

$\Rightarrow \, \, x+2y\leq 400$

$x\geq 0,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of the feasible region are $A(25,0),B(20,10),C(0,20),D(0,0)$

The value of Z at corner points is as shown :

The maximum cake can be made 30 (20 of the first kind and 10 of the second kind).

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 machine 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?

Answer:

Let number of rackets be x and number of bats be y.

the machine time availability is not more than 42 hours.

i.e. $1.5x+3y\leq 42$

craftsman’s time availability is 24 hours

i.e. $3x+y\leq 24$

The factory has to work at full capacity.

Hence, $1.5x+3y= 42...............1$

$3x+y= 24...............2$

Solving equation 1 and 2, we have

$x=4\, \, and\, \, \, y=12$

Thus, 4 rackets and 12 bats are to be made .

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 machine time and 24 hours of craftsman’s time.

(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.

Answer:

Let the number of rackets is x and the number of bats is y.

the machine time availability is not more than 42 hours.

craftsman’s time availability is 24 hours

The given information can be repreented in table as shown :

$1.5x+3y\leq 42$

$3x+y\leq 24$

$x,y\geq 0$

The profit on the bat is 10 and on the racket is 20.

$Z=20x+10y$

The mathematical formulation is :

maximise $Z=20x+10y$

subject to constraints,

$1.5x+3y\leq 42$

$3x+y\leq 24$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points are $A(8,0),B(4,12),C(0,14),D(0,0)$

The value of Z at corner points is as shown :

Thus, the maximum profit of the factory when it works at full capacity is 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 Rs17.50 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 maximise his profit, if he operates his machines for at the most 12 hours a day?

Answer:

Let packages of nuts be x and packages of bolts be y .Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in table as :

Profit on a package of nuts is Rs. 17.5 and on package of bolt is 7.

Therefore, constraint are

$x+3y\leq 12$

$3x+y\leq 12$

$x\geq 0,y\geq 0$

$Z= 17.5x+7y$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(4,0),B(3,3),C(0,4),D(0,0)$

The value of Z at corner points is as shown :

The maximum value of z is 73.5 at $B(3,3)$ .

Thus, 3 packages of nuts and 3 packages of bolts should be manufactured everyday to get maximum profit.

Question:5 A factory manufactures 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 screws A, while it takes 6 minutes on automatic and 3 minutes on the 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 in a day in order to maximise his profit? Determine the maximum profit.

Answer:

Let factory manufactures screws of type A and factory manufactures screws of type B. Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in the table as :

Profit on a package of screw A is Rs.7 and on the package of screw B is 10.

Therefore, the constraint is

$4x+6y\leq 240$

$6x+3y\leq 240$

$x\geq 0,y\geq 0$

$Z= 7x+10y$

The feasible region determined by constraints is as follows:

The corner points of the feasible region are $A(40,0),B(30,20),C(0,40),D(0,0)$

The value of Z at corner points is as shown :

The maximum value of z is 410 at $B(30,20)$ .

Thus, 30 packages of screw A and 20 packages of screw B should be manufactured every day to get maximum profit.

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 maximise his profit?

Answer:

Let the cottage industry manufactures x pedestal lamps and y wooden shades. Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in the table as :

Profit on a lamp is Rs. 5 and on the shade is 3.

Therefore, constraint is

$2x+y\leq 12$

$3x+2y\leq 20$

$x\geq 0,y\geq 0$

$Z= 5x+3y$

The feasible region determined by constraints is as follows:

The corner points of the feasible region are $A(6,0),B(4,4),C(0,10),D(0,0)$

The value of Z at corner points is as shown :

The maximum value of z is 32 at $B(4,4)$ .

Thus, 4 shades and 4 pedestals lamps should be manufactured every day to get the maximum profit.

Question:7 A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 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 maximise the profit?

Answer:

Let x be Souvenirs of type A and y be Souvenirs of type B .Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in table as :

Profit on type A Souvenirs is Rs. 5 and on type B Souvenirs is 6.

Therefore, constraint are

$5x+8y\leq 200$

$10x+8y\leq 240$

$x\geq 0,y\geq 0$

$Z=5x+6y$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(24,0),B(8,20),C(0,25),D(0,0)$

The value of Z at corner points is as shown :

The maximum value of z is 160 at $B(8,20)$ .

Thus,8 Souvenirs of type A and 20 Souvenirs of type B should be manufactured everyday to get maximum profit.

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.

Answer:

Let merchant plans has personal computers x desktop model and y portable model

.Thus, $x\geq 0,y\geq 0$ .

The cost of desktop model is cost Rs 25000 and portable model is Rs 40000.

Merchant can invest Rs 70 lakhs maximum.

$25000x+40000y\leq 7000000$

$5x+8y\leq 1400$

the total monthly demand of computers will not exceed 250 units.

$x+y\leq 250$

profit on the desktop model is Rs 4500 and on portable model is Rs 5000.

Total profit = Z , $Z=4500x+5000y$

The mathematical formulation of given problem is :
$5x+8y\leq 1400$

$x+y\leq 250$

$x\geq 0,y\geq 0$

$Z=4500x+5000y$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(250,0),B(200,50),C(0,175),D(0,0)$

The value of Z at corner points is as shown :

The maximum value of z is 1150000 at $B(200,50)$ .

Thus, merchant should stock 200 desktop models and 50 portable models to get maximum profit.

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.

Answer:

Let diet contain x unit of food F1 and y unit of foof F2 .Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in table as :

Cost of food F1 is Rs 4 per unit and Cost of food F2 is Rs 6 per unit

Therefore, constraint are

$3x+4y\geq 4$

$6x+3y\geq 6$

$x\geq 0,y\geq 0$

$Z= 4x+6y$

The feasible region determined by constraints is as follows:

We can see feasible region is unbounded.

The corner points of feasible region are $A(\frac{80}{3},0),B(24,\frac{4}{3}),C(0,\frac{100}{3})$

The value of Z at corner points is as shown :

Feasible region is unbounded , therefore 104 may or may not be minimum value of Z .

For this we draw $4x+6y< 104$ or $2x+3y< 52$ and check whether resulting half plane has point in common with feasible region or not.

We can see feasible region has no common point with $2x+3y< 52$ .

Hence , Z has minimum value 104.

Question:10 There are two types of fertilisers 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 atleast 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 fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Answer:

Let farmer buy x kg of fertilizer F1 and y kg of F2 .Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in table as :

F1 contain 10% nitrogen and F2 contain 5% nitrogen .Farmer requires atleast 14 kg of nitrogen

$10\%x+5\%y\geq 14$

$\frac{x}{10}+\frac{y}{20}\geq 14$

$2x+y\geq 280$

F1 contain 6% phophoric acid and F2 contain 10% phosphoric acid .Farmer requires atleast 14 kg of nitrogen

$6\%x+10\%y\geq 14$

$\frac{6x}{100}+\frac{y}{20}\geq 14$

$3x+56y\geq 700$

Total cost is Z . $Z=6x+5y$

Subject to constraint ,

$2x+y\geq 280$

$3x+56y\geq 700$

$x\geq 0,y\geq 0$

$Z=6x+5y$

The feasible region determined by constraints is as follows:

It can be seen that feasible region is unbounded.

The corner points of feasible region are $A(\frac{700}{3},0),B(100,80),C(0,280)$

The value of Z at corner points is as shown :

Feasible region is unbounded , therefore 1000 may or may not be minimum value of Z .

For this we draw $6x+5y< 1000$ and check whether resulting half plane has point in common with feasible region or not.

We can see feasible region has no common point with $6x+5y< 1000$ .

Hence , Z has minimum value 1000 at point $,B(100,80)$

Question:11 The corner points of the feasible region determined by the following system of linear inequalities:

$2x+y \leq 10,x+3y \leq 15,x,y\geq 0$ 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

$(A) p=q$

$(B)p=2q$

$(C)p=3q$

$(D)q=3p$

Answer:

The maximum value of Z is unique.

It is given that maximum value of Z occurs at two points $(3,4)\, \, and\, \, \, (0,5)$ .

$\therefore$ Value of Z at $(3,4)$ =value of Z at $(0,5)$

$\Rightarrow \, \, \, p(3)+q(4)=p(0)+q(5)$

$\Rightarrow \, \, \, 3p+4q=5q$

$\Rightarrow \, \, \, q=3p$

Hence, D is correct option.