Question 1133644: A farm co-op wants to grow corn and soybeans. Each acre of corn requires 7 gallons of fertilizer and 0.50 hours to harvest. Each acre of soybeans requires 3 gallons of fertilizer and 1 hour to harvest. The co-op has available at most 40,500 gallons of fertilizer and 5,250 hours of labor for harvesting. If the profits per acre are $50 for corn and $45 for soybeans, how many acres of each should the co-op plant to maximize their profit? What is the maximum profit?
So far I have: 7x + 3y < 40,500
(0.5)x + y
P(x,y) = $50x + $45y
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
I'm not sure why one of your inequalities is "<" and the other is "<=". But it is irrelevant to solving the problem.
And of course x>=0 and y>=0 are also constraints.
Graph the non-zero constraint boundary lines and find their point of intersection.
The corners of the feasibility region will be (0,0), (0,5250), (4500,3000), and (approximately) (5786,0).
Evaluate the objective function 50x+45y at each corner of the feasibility region to find the solution to the problem.
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NOTE:
Nearly all resources will say you have to evaluate the objective function at every corner of the feasibility region (except (0,0)) in order to solve the problem. But in fact you don't.
You can tell which corner of the feasibility region is going to give the maximum value of the objective function by comparing the slope of the objective function to the slopes of the constraint lines.
In this problem, the slopes of the constraint lines are -7/3 and -1/2; the slope of the objective function is -10/9. The slope of the objective function is between the slopes of the two constraint lines; that tells you that the maximum value of the objective function will be at the intersection of the two constraint lines.
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