SOLUTION: Maximize z = 3x + 5y subject to the constraints x &#8805; 0, y &#8805; 0, y <= 8, x + y &#8805; 2, 4x + y <= 12

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Question 251879: Maximize z = 3x + 5y subject to the constraints

x ≥ 0, y ≥ 0, y <= 8, x + y ≥ 2, 4x + y <= 12
Found 2 solutions by solver91311, jsmallt9:
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


Graph the five constraint inequalities. The area where the five solution sets overlap is the area of feasibility. For this problem, that area is an irregular pentagon. Each of the five vertices of the area of feasibility is a potential maximum, and the maximum can only exist at a vertex. Plot the five vertices and test the coordinates in the objective function. The coordinates of the vertex that provides the maximum value of the objective function are the solution set of your problem.


John
Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!
Start by graphing the constraints:

and we get a region bounded by the x-axis, the y-axis and the lines y = 8, x + y = 2 and 4x + y = 12. The maximum (and minimum) values are found at one of the vertices of this region. So we need to find the coordinates of each vertex and then figure out the z for each of these pairs of x and y values. One of the z's will be a maximum value and another will be the minimum value.

From the graph or by using Algebra we should be able to find that the vertices are: (0, 2), (0, 8), (1, 8), (3, 0), (2, 0). So we take each one of these and find its value for z. The z for one of these points will be the maximum. I'll get you started:
Vertex     z = 3x + 5y
(0, 2)     z = 3(0) + 5(2) = 0 + 10 = 10
(0, 8)     z = 3(0) + 5(8) = 0 + 40 = 40

I'll leave it up to you to finish the other three.
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