Question

A man has $₹\, 1500$ for purchase of rice and wheat. A bag of rice and a bag of wheat cost $₹\, 180$ and $₹\, 120$ respectively. He has a storage capacity of $10$ bags only. He earns a profit of $₹\, 11$ on each rice bag and $₹\, 9$ on each wheat bag. Find the maximum profit,

• A. $₹\,110 $
• B. $₹\,90 $
• C. $₹\,95 $
• D. $₹\,100 $

Answer 1

Answer: (D)

Let $x$ bags of rice and $y$ bags of wheat be purchased. Let $z$ be the total profit.
$\therefore z=11x + 9y$
According to question, $x$ and $y$ must satisfy the following conditions
$x + y\le 10$
$ 180x + 120y \le 1500$
i.e., $3x + 2y\le 25$
$ x \ge 0, y \ge 0$
Mathematical formulation of the $LPP$
is Maximize $z=11x + 9y$
subject to the constraints :
$ x + y \le 10$
$ 3x + 2y\le 25$
$ x \ge 0,y \ge 0$
Now, draw the lines

$l_1 : x + y= 10$
$ l_2 : 3x+2y = 25$
$l_3 : x = 0$ and $l_4 : y = 0$
Lines $l_1$ and $l_2$ meet at $E(5, 5)$
The shaded bounded region $OCEB$ is the feasible region of the given $LPP$.
Vertices of the feasible region are :
$O(0, 0),C\left(\frac{25}{3}, 0\right), E(5, 5)$ and $B(0,10)$
Maximize $z = 1 1x + 9y$
$\therefore $ The value of $z$ at $O = 0$
The value of $z$ at $B = 11 \times 0 + 9\times 10 = 90$
The value of $z$ at $C=11\times\frac{25}{3} +9\times0 = \frac{275}{3} = 91.67$
The value of $z$ at $E = 11 \times 5 + 9\times 5 = 100$
$\therefore $ For earning maximum profit, $5$ bags of rice and $5$ bags of wheat should be purchased and sold. Maximum profit = $₹\, 100$.