Question 1102106: Find the minimum and maximum values of z=9x+2y, if possible, for the following set of constraints.
4x+3y≥12
x+3y≥6
x≥0, y≥0
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Graph the equations of the boundary lines of the inequalities.
4x+3y=12 has intercepts (3,0) and (0,4).
x+3y=6 has intercepts (6,0) and (0,2).
The intersection of the two lines is at (2,4/3)
Since both inequalities have "y >= ...", the feasibility region is unbounded; the three vertices of the feasibility region are (0,4), (2,4/3), and (6,0).
Evaluate the objective function at each vertex of the feasibility region.
Note that there will be a minimum value of the objective function; but with the unbounded feasibility region, there will be no maximum value.
(0,4): 9(0)+2(4) = 0+8 = 8
(2,4/3): 9(2)+2(4/3) = 18+8/3 = 62/3
(6,0): 9(6)+2(0) = 54+0 = 54
The minimum value of the objective function subject to the given constraints is 8, at (0,4).
Note that it is not necessary to evaluate the objective function at every vertex of the feasibility region. In this fairly simple example, it was easy to do so. But in more complicated similar problems, you can get to the answer with much less work without evaluating the objective function at every vertex, as described below.
The objective function is z = 9x+2y. The slope of the graph of every function of this form is -9/2. By looking at even a rough sketch of the feasibility region, you can see that the only place where a line with slope -9/2 just touches the feasibility region, without passing through it, is at the vertex (0,4).
Furthermore, you can see that, since the feasibility region is unbounded, there is no place other than at (0,4) where a line with slope -9/2 will touch the feasibility region in only one place; that indicates that there is no maximum value to the objective function.
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