Linear Programming Question 121
Question: 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 low grade oil, respectively. If refinery A costs 400 per day and refinery B costs 300 per day to operate, then the days should each he run to minimize costs while satisfying requirements are
Options:
A) 30, 60
B) 60, 30
C) 40, 60
D) 60, 40
Show Answer
Answer:
Correct Answer: B
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 of for x and y days respectively to minimize the total cost. The mathematical form of the above is Minimize $ Z=400x+300y $ Subject to $ 100x+200y\ge 12000 $
$ 300x+400y\ge 20000 $
$ 200x+100y\ge 15000 $ And $ x,y\ge 0 $ The feasible region of the above LPP is represented by the shaded region in the given figure. The corner 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=400x+300y $ | |||||
| $ 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.
BETA
