Question

A factory owner wants to purchase two types of machines, $A$ and $B$, for his factory. The machine $A$ requires an area of $1000m^2$ and $12$ skilled men for running it and its daily output is $50$ units, whereas the machine $B$ required $1200m^2$ area and $8$ skilled men, and its daily output is $40$ units. If an area of $7600m^2$ and $72$ skilled men be available to operate the machine, how many machines $A$ and $B$ respectively should be purchased to maximize the daily output?

• A. $4,3$
• B. $2,6$
• C. $6,2$
• D. $3,4$

Answer 1

Answer: (A)

Let the number of machine $A$ be $x$
and number of machine $B$ be $y$.
Let $z$ be the daily output.
Now given information can be summarized as :

	$A(x)$	$B(y)$	Maximum available capacity
Area $(m^2)$	1000	1200	7600
Man power	12	8	72
Output	50	40

According to question, $x$ and $y$ must satisfy the following conditions:
(Area) $1000x+1200y\le 7600\Rightarrow 5x+67\le 38$
(Man power) $12x+87\le 72\Rightarrow 3x+27\le 18$
$x\ge 0,y\ge 0$
Mathematical formulation of the $LPP$ is
Maximize $z=50x+40y$
subject to constraints :
$5x+67\le 38,3x+27\le 18,x\ge 0,y\ge 0$
Now, we draw the lines
$l_1:5x+67=38$
$l_2:3x+2y=18$
$l_3:x=0$ and $l_4:7=0$
Lines $l_1$ and $l_2$ meet at $E(4,3)$.

The shaded region $OCEB$ is the feasible region which is bounded.
Vertices of the feasible region are
$0(0,0),C(6,0),E(4,3)$ and $B\left(0,\frac{19}{3}\right)$
Maximize $z=50x+40y$
At $0,z=50\times 0+40\times 0=0$
At $C,z=50\times 6+40\times 0=300$
At $E,z=50\times 4+40\times 3=320$
At $B,z=50\times 0+40\times \frac{19}{3}=253.33$
Clearly, the maximum output $=320$ is at $E(4,3)$, i.e.,
when $4$ machines $A$ and $3$ machines $B$ are purchased.