A farmer’s crop will receive a grade of A, B, or C. if the crop is graded A, the
ID: 3253268 • Letter: A
Question
A farmer’s crop will receive a grade of A, B, or C. if the crop is graded A, the farmer will receive $5.20 per bushel. For grades B and C, the prices per bushel will be $4.00 and $3.20, respectively. The probabilities for the grade of the harvest when done at the normal time are P(A) = .35, P (B) =.55, and P(C) = .10. The yield of the harvest when done at the normal time is expected to be 66,000 bushels. If the harvest is early, the probabilities are P(A) = .50, P(B) = .50, and P(C) = 0, but the yield is reduced to 58,000 bushels. Should the farmer harvest early or at the normal time if he wishes to maximize expected revenue?
Explanation / Answer
In the normal harvest time 66,000 bushels are produced:
P(A) = 0.35. Therefore Bushels of A = 0.35*66000 = 23,100
P(B) = 0.55. Therefore Bushels of B = 0.55*66000 = 36,300
P(C) = 0.10. Therefore Bushels of C = 0.10*66000 = 6,600
Price per Bushel for A = $5.2. Therefore Revenue from A = 5.2 * 23100 = $120,120
Price per Bushel for B = $4.0. Therefore Revenue from B = 4.0 * 36300 = $145,200
Price per Bushel for C = $3.2. Therefore Revenue from C = 3.2 * 6600 = $21,120
Therefore Total Revenue from the normal harvest = 120,120 + 145,200 + 21,120 = $286,440
In the early harvest time 58,000 bushels are produced:
P(A) = 0.5. Therefore Bushels of A = 0.5*58000 = 29,000
P(B) = 0.5. Therefore Bushels of B = 0.5*58000 = 29,000
P(C) = 0.0. Therefore Bushels of C = 0.10*66000 = 0
Price per Bushel for A = $5.2. Therefore Revenue from A = 5.2 * 29000 = $150,800
Price per Bushel for B = $4.0. Therefore Revenue from B = 4.0 * 29000 = $116,000
Price per Bushel for C = $3.2. Therefore Revenue from C = 3.2 * 0 = $0
Therefore Total Revenue from the normal harvest = 150,800 + 116,000 + 0 = $266,800
The revenue while harvesting in the normal time ($286,440) is better than when harvesting early ($266,800).
Therefore the farmer should harvest during the normal time.
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