Question

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?

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

Answer 1

Answer: (A)

$4,3$

Answer 2

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 : 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.