Question 1139948: A chain saw requires 3 hours of assembly and a wood chipper 5 hours. A maximum of 45 hours of assembly time is available. The profit is $ 160 on a chain saw and $ 240 on a chipper. How many of each should be assembled for maximum profit?
To attain the maximum profit, assemble
nothing chain saws and
nothing wood chippers.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i solved this graphically as shown below:
the area of the graph that is not shaded is the region of feasibility.
to find that region, i graphed the opposite of the inequalities formed by the constraints of the problem.
the constraints of the problem were:
3x + 5y <= 45
x >= 0
y >= 0
i graphed:
3x + 5y >= 45
x <= 0
y <= 0
the corner points of the feasible region are where the maximum / minimum solutions lie.
the objective function, which was profit = 160x + 240y, was evaluated at each of these corner points.
at (0,9), the profit was 240 * 9 = 2160
at (15,0), the profit was 160 * 15 = 2400.
maximum profit was 2400 when 15 chains saws and zero chippers were assembled.
unfortunately, there was no happy middle ground for maximum profit.
for example, when x = 5, y = 6 and the profit was 160 * 5 + 240 * 6 = 2240.
that was actually better than one of the corner points but not better than the other.
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