Question

An oil company requires $13000, 20000$ and $15000$ barrels of high grade, medium grade and low grade oil respectively. Refinery $A$ produces $100, 300$ and $200$ barrels per day of high, medium and low grade oil respectively, whereas the Refinery $B$ produces $200, 400$ and $100$ barrels per day respectively. If $A$ costs $₹\, 400$ per day and $$ costs $₹\, 300$ per day to operate, how many days refinery $A $ and $B$ should be run respectively to minimize the cost of requirement?

• A. $\frac{110}{3}, \frac{170}{3}$
• B. $\frac{140}{3}, \frac{160}{3}$
• C. $\frac{160}{3}, \frac{140}{3}$
• D. $\frac{170}{3}, \frac{110}{3}$

Answer 1

Answer: (D)

Let refineries $A$ and $B$ run for $x$ and $y$ days respectively.
Let $z$ be the cost.
According to question $x$ and $y$ must satisfy the following conditions:
$100x + 200y \ge 13000$ or $x + 2y \ge 130$
$300x + 400y \ge 20000$ or $3x + 4y \ge 200$
$200x + 100y \ge 15000$ or $2x + y \ge 150$
$x \ge 0, y \ge 0 $
Mathematical formulation of the $LPP$ is:
Minimize $z = 400x + 300y$
Subject to constraints:
$x + 2y \ge 130, 3x + 4y \ge 200,2x + y \ge 150,x \ge 0,y \ge 0$
Now, we draw the lines :
$ l_1 : x + 2y = 130, l_2 : 3x + 4y = 200, l_3 : 2x + y = 150$,
$l_4: x = 0$ and $l_5:y = 0$
Lines $l_1$ and $l_2$ meet at $P\left(\frac{110}{3}, \frac{170}{3}\right)$

The fesible region is shaded and it is unbounded region
with vertices $A, P$ and $F$.
Also minimize $z = 400x + 300y$
$\therefore $ At $A(130,0), z = 52000$
At $F(0,150), z = 45000$
At $P\left(\frac{170}{3}, \frac{110}{3}\right)$, $z\left(\frac{101000}{3}\right) = 33666.67$
$\therefore $ Cost is min. at $P\left(\frac{110}{3}, \frac{170}{3}\right)$ which is $₹\,33666.67$
Hence, refinery $A$ should run for $\frac{170}{3}$ days and refinery $B$ should run for $\frac{110}{3}$ days.