SOLUTION: 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 p

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Question 1181641: 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?
Found 3 solutions by CPhill, greenestamps, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Here's how to determine the optimal amount of each type of crude oil to refine:
**1. Define the variables:**
* Let 'x' be the number of barrels of type A crude oil.
* Let 'y' be the number of barrels of type B crude oil.
**2. Set up the constraints:**
* **Total barrels constraint:** x + y ≤ 4000 (The refinery can refine at most 4000 barrels per week)
* **Impurities constraint:** 2x + 3y ≤ 9000 (The refinery can handle no more than 9000 kg of impurities)
* **Non-negativity constraints:** x ≥ 0 and y ≥ 0 (We can't refine a negative number of barrels)
**3. Define the objective function (profit):**
Profit (P) = 25x + 30y (We want to maximize this)
**4. Solve the linear programming problem:**
This can be solved graphically or using linear programming techniques (like the simplex method or software tools). Here's the general approach:
* **Graphically:**
1. Plot the constraints on a graph.
2. Identify the feasible region (the area that satisfies all constraints).
3. Find the corner points of the feasible region.
4. Evaluate the objective function at each corner point.
5. The corner point that yields the highest profit is the optimal solution.
* **Linear Programming:** Specialized software or online calculators can solve this type of problem efficiently.
**Solution:**
The optimal solution is to refine:
* 3000 barrels of type A crude oil
* 1000 barrels of type B crude oil
This will result in a maximum profit of R105,000.

Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!


The AI solution from the other tutor is correct.

In that solution, the steps outlined say that you need to evaluate the objective function at each corner of the feasibility region. And that is what most references say.

However, in fact it is NOT necessary to evaluate the objective function at every corner of the feasibility region.

The corner of the feasibility region where the objective function is maximized can be determined from the slopes of the constraint boundary lines and of the objective function. The idea there is that the objective function will be maximized where a line with the slope of the objective function just touches the feasibility region.

In this problem, the constraint boundary line equations are x+y=4000 (slope -1) and 2x+3y=9000 (slope -2/3); the objective function is P=25x+30y (slope -5/6). Since -5/6 is between -1 and -2/3, the objective function is maximized at the intersection of the two constraint boundary lines.







ANSWER: Profit is maximized when 3000 barrels of type A and 1000 barrels of type B are refined each week.


Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
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?
~~~~~~~~~~~~~~~~~~~~~~~~~~~


       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)


    +------------------------------------------------------+
    |   According to (1), we want to subtract from 120000  |
    |         the minimal possible value of 5x.            |
    |                                                      |
    |     According to (2), x must be at least 3000.       |
    +------------------------------------------------------+



So, it is logical to take for x the minimal possible value, which is 3000.


So, the ANSWER to the problem is  this:


    +-----------------------------------------------------------------------+
    |   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. |
    +-----------------------------------------------------------------------+

Solved - without using the Linear Programming method.



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