Q. An oil company required $12000,20000$ 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-grade, medium-grade and low-grade oil, respectively, while refinery $B$ produces $200,400$ and $100$ barrels per day of high-grade, medium-grade and lowgrade oil, respectively. If refinery $A$ costs $₹ 400$ per day and refinery $B$ costs $₹ 300$ per day to operate, then the days should each be run to minimize costs while satisfying requirements are
Linear Programming
Solution:
The given data may be put in the following tabular form
Refinery
High grade
Medium grade
Low grade
Cost per day
A
100
300
200
$₹ 400$
B
200
400
100
$₹ 300$
Minimum Requirement
12000
20000
15000
Suppose refineries $A$ and $B$ should run for $x$ and $y$ days respectively to minimize the total cost. The mathematical form of the above is Minimize $Z=400 x+300 y$ Subject to
$100 x+200 y \geq 12000$
$300 x+400 y \geq 20000$
$200 x+100 y \geq 15000$
and $x, y \geq 0$
The feasible region of the above LPP is represented by the shaded region in the given figure.
The corncr points of the feasible region are $A_{2}(120,0)$, $P(60,30)$ and $B_{3}(0,150)$
The value of the objective function at these points are given in the following table
Point $(x, y)$
Value of the objective function $Z=400 x+300 y$
$A_{2}(120,0)$
$Z=400 \times 120+300 \times 0=48000$
$P(60,30)$
$Z=400 \times 60+300 \times 30=33000$
$B_{3}(0,150)$
$Z=400 \times 0+300 \times 150=45000$
Clearly, $Z$ is minimum when $x=60,\, y=30$.
Hence, the machine $A$ should run for 60 days and the machine $B$ should run for 30 days to minimize the cost while satisfying the constraints.
| Refinery | High grade | Medium grade | Low grade | Cost per day |
| A | 100 | 300 | 200 | $₹ 400$ |
| B | 200 | 400 | 100 | $₹ 300$ |
| Minimum Requirement | 12000 | 20000 | 15000 |
| Point $(x, y)$ | Value of the objective function $Z=400 x+300 y$ | |
|---|---|---|
| $A_{2}(120,0)$ | $Z=400 \times 120+300 \times 0=48000$ | |
| $P(60,30)$ | $Z=400 \times 60+300 \times 30=33000$ | |
| $B_{3}(0,150)$ | $Z=400 \times 0+300 \times 150=45000$ | |