SOLUTION: Consider the following linear programming problem: Max. 3a + 3b s.t. 2a + 4b is less than or equal to 12 6a + 4b is less than or equal to 24 a.Find the optimal soluti

Question 717173: Consider the following linear programming problem:
Max. 3a + 3b
s.t.
2a + 4b is less than or equal to 12
6a + 4b is less than or equal to 24
a.Find the optimal solution using the graphical solution procedure
b. If the objective function is changed to 2a + 6b, what will the optimal solution be?
c. How many extreme points are there? What are the values of a and b at each extreme point?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Step 1: Graph the constraint inequalities. From the looks of the objective function, the values have to both be non-negative to achieve a maximum, so also graph

and

as additional constraints. Hint: When graphing constraints for a graphical solution to linear programming problems, it is often convenient to shade each inequality graph in the opposite sense than would ordinarily be done. That is to say shade the side of the line that is NOT the solution set. The result is that the feasibility area will be more visually obvious because it will be the only area of the graph with no shading at all.

Step 2: The feasibility area is where the solution sets overlap.

Step 3: Determine the critical points. These will be the vertices of the feasibility polygon (a quadrilateral in the case of the given problem).

Step 4: Test the objective function at the values of the coordinates of each of the critical points defined in Step 3. The optimum solution, if one exists, will be the set of vertex coordinates that make the objective function the maximum value. It is possible that two adjacent vertices give the same objective function result. In such case, there is no unique optimum. Rather, any point on the line segment that joins those two vertices gives an optimum result.

Laying out one of these problems is a significant amount of work; enough that I don't care to do it for free. Write back if you would care to negotiate a price for a complete solution to the posted problem.

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it

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