Question 1206902: Solve the following linear programming model geometrically, labeling all the corner points and shading the feasible or solution region
Maximize z = 200x+100y
Subject to: 3x+4y ≤ 24
X+y ≤ 16
X+3y≤30
X,y ≥0
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website! Maximize z = 200x+100y
Subject to: 3x+4y ≤ 24
X+y ≤ 16
X+3y≤30
X,y ≥0
Are you sure some of those first three ≤'s shouldn't have been ≥ ?.
The graph seems funny because the first two restraints are irrelevant,
So the feasible region is on or below all three lines, which means it's
just on or below the green shaded region below the bottom line.
That means there are only three corner points
(0,0), (0,6), and (8,0)
Substituting (x,y) = (0,0) in z = 200x+100y,
z = 200(0)+100(0) = 0 + 0 = 0
Substituting (x,y) = (0,6) in z = 200x+100y,
z = 200(0)+100(6) = 0 + 600 = 600
Substituting (x,y) = (8,0) in z = 200x+100y,
z = 200(8)+100(0) = 1600 + 0 = 1600
So the maximum value for z is 1600 when x = 8 and y = 0.
Edwin
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