Question

A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires $9$ labour hours of fabricating and $1$ labour hour for finishing. Each type of B requires $12$ labour hours for fabricating and $3$ labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are $180$ and $30$ respectively. The company makes a profit of $80$ on each piece of type A and $120$ on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?

Answer 1

Let the pieces of type A manufactured per week be $x$

Let the pieces of type B manufactured per week be $y$

Therefore maximum profit $z=80x+120y$

Fabricating hours for $A$ is $9$ and finishing hour is $1$

Fabricating hours forours $B$ is $12$ and finishing hours is $3$

Maximum number of fabricating hours =$180$

$\therefore 9x+12y\le 180i.e^{3}x+4y\le 60$

Maximum number of finishing hours=$30$

$x+3y\le 30$

Now $z$ is$=80+120y$ is subject to

$3x+4y\le 60$

$x+3y\le 30$

$y\ge 0$

Now the area of th feasible region is as shown in above figure

$3x+4y=60$

$3x+3y=30$

$3x+4y=60$

$3x+9y=30$

$\therefore y=6$

and $x=12$

W can see that the post on the region are

$A(0,15),B(0,10),C(20,0)$ and $D(12,6)$

Th maximum profit is

Point(x,y)	$z=80x+120$
A(0,15)	$1800\to$ maximum
B(0, 10)	1200
C(20, 0)	1600
D(12,6)	1680

Hence the maximum profit is at $(0,15)$

$\therefore$ The teaching aid $A=0$ and teaching aid $B=15$ has to be made.