Q. A factory owner wants to purchase two types of machines, $A$ and $B$, for his factory. The machine $A$ requires an area of $1000 m^2$ 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 $7600 m^2$ 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?
Linear Programming
Solution:
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) $12 x + 87 \le 72 \Rightarrow 3x + 27 \le 18$
$ 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.
| $A(x)$ | $B(y)$ | Maximum available capacity |
|
|---|---|---|---|
| Area $(m^2)$ | 1000 | 1200 | 7600 |
| Man power | 12 | 8 | 72 |
| Output | 50 | 40 |