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An oil refinery refines types A and B of crude oil and can refine as much as 4000 barrels each week.
Type A crude has 2 kg of impurities per barrel, type B has 3 kg of impurities per barrel,
and the refinery can handle no more than 9000 kg of these impurities each week.
How much of each type should be refined in order to maximize profits, if the profit is R25/barrel
for type A and R30/barrel for type B?
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I will show here totally different method to solve,
very easy, simple and straightforward.
Let x be the amount of the oil type A to refine (in barrels).
Then the amount of oil type B to refine is 4000-x barrels.
We want to maximize profit
P(x) = 25x + 30*(4000-x) = 25x + 120000 - 35x = 120000 - 5x. (1)
under this restriction
2x + 3*(4000-x) =<= 9000, or 2x + 12000 - 3x <= 9000, or x >= 3000. (2)
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| According to (1), we want to subtract from 120000 |
| the minimal possible value of 5x. |
| |
| According to (2), x must be at least 3000. |
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So, it is logical to take for x the minimal possible value, which is 3000.
So, the ANSWER to the problem is this:
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| Under given restrictions, the optimal solution is |
| |
| x = 3000 barrels for oil type A |
| and 4000-x = 4000-3000 = 1000 barrels of oil type B. |
| |
| The profit is then P = 25*3000 + 30*1000 = 75000 + 30000 = 105000. |
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Solved - without using the Linear Programming method.