Question

A mill owner buys two types of machines $A$ and $B$ for his mill. Machine $A$ occupies $1,000$ sqm of area and requires $12$ men to operate it; while machine $B$ occupies $1,200$ sqm of area and requires $8$ men to operate it. The owner has $7,600$ sqm of area available and $72$ men to operate the machines. If machine $A$ produces $50$ units and machine $B$ produces $40$ units daily, how many machines of each type should he buy to maximize the daily output? Use Linear Programming to find the solution.

Answer 1

The data given in the problem are as under:

	Machine A	Machine B	Maximum available
Area needed	$1,000$ sq. m	$1,200$ sq.m	$7,600$ sq.m
Labour force	$12$	$8$	$72$
Daily output	$50$ units	$40$ units	-

Let $x$ and $y$ be the number of machines A and B respectively
$\begin{array}{c}1,000x+1,200y\le 7,600\\ 10x+12y\le 76\\ 5x+6y\le 38\\ 12x+8y\le 72\\ 3x+2y\le 18\\ x\ge 0,y\ge 0\end{array}\}constraints$

Total output $Z=50x+40y$

The problem is to maximise
$Z=50x+40y$ subject to constraints
$5x+6y\le 38$
$3x+2y\le 18$
$x\ge 0,y\ge 0$
$3x+2y=18$

$x$	$6$	$0$	$4$
$y$	$0$	$9$	$3$

$5x+6y=38$

$x$	$\frac{38}{5}$	$0$	$4$
$y$	$0$	$\frac{19}{3}$	$3$

The vertices of the feasible region ORPD are $O(0,0),R(6,0),P(4,3)$ and $D(0,\frac{19}{3})$

Point	Value of $Z=50x+40y$
At $O(0,0)$	$Z=0$
At $R(6,0)$	$Z=50\times 6+0=300$
At $P(4,3)$	$Z=50\times 4+40\times 3=320$
At $D(0,\frac{19}{3})$	$Z=50\times 0+40\times \frac{19}{3}$ $=\frac{760}{3}=\mathrm{2.53.33}$

Thus we see that $Z$ is maximum at $(4,3)$.
$\therefore$ Number of machine $A=4$
Number of machine $B=3$
The graph of these in equation is as shown