Question 1104488: Maximize:Z=40x+88y
Subject to:2x+8y≤60
5x+2y≥60,x,y,>0
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52785)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The constraints are equivalent to
 and 
Here is a graph; the feasibility region is below the red line and above the green line. The vertices of the feasibility region are (12,0), (30,0), and (10,5).
The standard method for finding the maximum value of the objective function  is to evaluate that function at each vertex of the reasibility region:
(12,0): 40x+88y = 40(12) = 480
(30,0): 40x+88y = 40(30) = 1200
(10,5): 40x+88y = 40(10)+88(5) = 400+440 = 840
The maximum value is 1200 at (30,0).
In fact there is no need to evaluate the objective function at every vertex of the feasibility region. You can tell which vertex will produce the maximum value of the objective function by comparing the slope of the objective function with the slopes of the constraint lines.
The slope of the objective function is -40/88 = -5/11. So consider all the lines with slope -5/11:
A line with slope -5/11 passing through (12,0) will just touch a vertex;
a line with slope -5/11 passing through (5,10) will cut through the feasibility region;
a line with slope -5/11 passing through (30,0) will again touch the feasibility region just at a vertex.
We can conclude from that -- without evaluating the objective function at any of the three vertices -- that the minimum value will be at (12,0), the maximum value will be at (30,0), and an intermediate value will occur at (10,5).
Then, since the problem asks for the maximum value of the objective function, we only need to evaluate it at (30,0).
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