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 $z=11x+9y$
$\therefore$ 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 $C=11\times \frac{25}{3}+9\times 0=\frac{275}{3}=91.67$
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 = $₹100$.