Question 203030: A fruit grower has 150 acres of land available to raise two crops, A and B. It takes 1
day to trim an acre of crop A and 2 days to trim an acre of crop B, and there are 240 days available each
year for trimming. It takes 0.3 days to pick an acre of crop A and 0.1 days to pick an acre of crop B, and
there are 30 days per year available for picking. The profit is $140 per acre for crop A and $235 per acre
for crop B. Determine the optimal acreage for each fruit and the optimal profit.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x = number of acres of crop A.
let y = number of acres of crop B.
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the equation we want to maximize is the profit equation.
the profit on crop A is 140 per acre so the total profit on crop A would be:
140 * x
the profit on crop B is 235 per acre so the total profit on crop B would be:
235 * y
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it takes 1 day to trim an acre of crop A and 2 days to trim an acre of crop B with a total of 240 days available for trimming.
1*x + 2*y <= 240
solving for y, we get:
y <= (240-x)/2
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it takes .3 days to pick an acre of crop A and .1 days to pick an acre of crop B with a total of 30 days available for picking.
.3*x + .1*y <= 30
solving for y we get:
y <= 300 - 3*x
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total number of acres has to be less than or equal to 150.
x + y <= 150
solving for y, we get
y <= 150
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the 3 constraints are shown in the following graph.
one of the points of intersection of all 3 lines should contain the maximum profit.
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since the points of intersection cannot be clearly seen, they can be calculated by taking sets of 2 equations and solving them simultaneously.
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equation 1 is: 1x + 2y <= 240_____ solving for y gets y <= (240-x)/2
equation 2 is: .3x + .1y <= 30____ solving for y gets y <= 300 - 3x
equation 3 is: x + y <= 150_______ solving for y gets y <= 150 - x
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for equation 1 and 2, the point of intersection = (72,84)
for equation 1 and 3, the point of ijntersection = (60,90)
for equation 2 and 3, the point of intersection = (75,75)
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since the maximum profit should lie at the intersection of these 3 lines, we can solve the profit equation at each of these points and pick the one that is maximum.
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the profit equation is again:
140x + 235y
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at point (72,84) the profit would be:
140*72 + 235*84 = $29,820
at point (60,90) the profit would be:
140*60 + 235*90 = $29,550
at point (75,75) the profit would be:
140*75 + 235*75 = $28,125
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It appears that maximum profit would be attained with 72 acres of crop A and 84 acres of crop B but this can't be because 72 + 84 > 150 which violates the maximum number of acres constraint (equation 3).
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The next choice would be 60 acres of crop A and 90 acres of crop B.
60 + 90 = 150 <= 150 so the constraint on maximum number of acres is not violated (equation 3).
60 + 2*90 = 60 + 180 = 240 <= 240 so the constraint on maximum number of days to trim is not violated (equation 1)
.3*60 + .1*90 = 18 + 9 = 27 <= 30 so the constraint on maximum number of days to pick is not violated (equation 2)
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It appears the answer is:
60 acres of crop A and 90 acres of crop B gives the maxiumum profit while meeting all the constraints.
the maximum profit is $29,550
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It appears to be logically correct since the profit on crop B acres is much greater than the profit on crop A acres so maximizing the acres for crop B makes sense.
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based on the constraints it appears the maximum acres of crop B are 90. I tried 91 acres of crop B and came up with a smaller profit because of lower total utilization on the acres that could be cropped (91 for B left only 58 for A because of the constraints). I suspect further analysis would yield the same results, i.e. the max profit is what is shown.
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