Question

A manufacturer makes two types of toys $A$ and $B$. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below

Type of toys	Machine $I$	Machine $II$	Machine $III$
$A$	$12$	$18$	$6$
$B$	$6$	$0$	$9$

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

Answer 1

Let $x$ and $y$ toys of type $A$ and $B$ respectively be manufactured in a day.
The given problem can be formulated as follows:
Maximise $z=7.5x+5y......\left(1\right)$
subject to the constraints
$2x+y\le \mathrm{60........}\left(2\right)$
$x\le 20........\left(3\right)$
$2x+3y\le \mathrm{120.........}\left(4\right)$
$x,y\ge \mathrm{0..........}\left(5\right)$
The feasible region determined by the constraints is as shown
The corner points of the feasible region are $A(20,0),B(20,20),C(15,30)$ and $D(0,40)$
The value of $z$ at these corner points are as follows.

Corner point	$Z=7.5x+5y$
$A(20,0)$	$150$
$B(20,20)$	$150$
$C(15,30)$	$262.5$	$\to$ Maximum
$O(0,40)$	$200$

The maximum value of $z$ is $262.5$ at $(15,30)$
Thus, the manufacturer should manufacture $15$ toys of type $A$ and $30$ toys of type $B$ to maximize the profit.