Question 1162265: 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?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = number of acres of corn.
y = number of acres of soybeans.
make a table as shown below:
x y
corn soybeans
fertilizer gallons 7 3 <= 40,500
harvest hours .5 1 <= 5,250
profit 50 45 maximize
using the desmos.com calculator, graph the opposite of the constraints and then evaluate the objective function at the corner points of the feasible region.
the constraint functions are:
7x + 3y <= 40,500
.5x + y <= 5250
x,y >= 0
the objective function is:
profit = 50x + 45y
you are graphing the OPPOSITE of the constraint inequalities.
here's what the graph looks like:
maximum profit is at the point (4500,3000).
profit is 50*4500 + 45*3000 = 360,000
gallons of fertilizer used are 4500*7 + 3000*3 = 40,500 <= 40,500
hours of labor = 4500*.5 + 3000*1 = 5250 <= 5250
looks like all the available resources were used when the maximum profit was attained.
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