Question

A dealer wishes to purchase a number of fans and sewing machines. He has only $?\, 5760$ to invest and has space for at most $20$ items. A fan and sewing machine cost $?\, 360$ and $? \,240$ respectively. He can sell a fan at a profit of $?\,22$ and sewing machine at a profit of $? \,18$. Assuming that he can sell whatever he buys, how many fans and sewing machines respectively he sell in order to maximize his profit?

• A. $8, 12$
• B. $6, 10$
• C. $10, 6$
• D. $12, 8$

Answer 1

The Correct Option is A

Let the dealer purchase $x$ fans and $y$ sewing machines. Then, the profit function $z$ is given by $z = 22x + 18y$ The two variables $x$ and $y$ satisfy following constraints : $360x + 240y \le 5760$ or $3x + 2y\le 48$ $ x + y \le 20, x \ge 0, y \ge 0$ Hence mathematical formulation of the given $LPP$ is : Maximize $z = 22x + 18y$ subject to constraints : $3x + 2y \le 48 $ $x + y \le 20$ $x \le 0,7 \ge 0$

Now we draw the lines $ l_1 : 3x + 2y = 48$ $l_2 : x + y = 20 $ $l_3: x = 0, l_4 : y = 0$ Lines $l_1$ and $l_2$ intersect at $P( 8,12)$. The shaded region $OAPD$ represents the feasible region, which is bounded. Vertices of the feasible region are $O(0, 0), A(16,0), P(8, 12)$ and $D(0, 20)$ Since, maximize $z = 22x + 18y$ $\therefore$ At $O(0, 0), z = 0$ At $A(16, 0), z = 22 \times 16+ 18 \times 0 = 352$ At $P(8,12), z = 22\times 8 + 18 \times 12 = 392$ At $D(0, 20), z = 22 \times 0 + 18 \times 20 = 360$ Thus profit is maximum at $P(8, 12)$. Hence, profit is maximum when $8$ fans and $12$ sewing machines are purchased and sold and the maximum profit is $?\, 392$.