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.