A used car dealership can hire both college graduates and non-college graduates.
ID: 2800249 • Letter: A
Question
A used car dealership can hire both college graduates and non-college graduates. The average number of cars it sells per hour is equal to q = 10 - 10*exp(-0.017c - 0.011n - 1.19), where q is the number of cars sold, c is the number of college graduates working during one hour, and n is the number of non-college graduates working during one hour. The dealership pays college graduates $35/hour and non-college graduates $11/hour. The dealership wishes to maximize its average profit during an hour. We assume that each car is sold for $29,000, and the variable cost (without considering the cost of labor) is $28,000 per car. If the dealership employs n = 8 non-college graduates each hour, how many college graduates should the dealership employ during one hour? Your answer can be a decimal to represent that an individual can work for a fraction of an hour.
Explanation / Answer
This question is based on concepts of calculus.
q = 10 - 10*exp(-0.017c - 0.011n - 1.19)
Revenue per hours = $29000* [10 - 10*exp(-0.017c - 0.011n - 1.19)]
Cost per hour = $28000 * [10 - 10*exp(-0.017c - 0.011n - 1.19)] + 35c + 11n
Profit per hour = Revenue per hour - Cost per hour = $1000 * [10 - 10*exp(-0.017c - 0.011n - 1.19)] -35c -11n
Given n= 8, find c such that profit is maximized
Profit per hour P = $1000 * [10 - 10*exp(-0.017c - 0.011*8 - 1.19)] -35c -11*8
= $1000 * [10 - 10*exp(-0.017c - 0.088 - 1.19)] -35c -88
= $10000-$88 - 10000 *exp(-0.017c - 1.278) -35c
= $9912 - 10000 *exp(-0.017c) * e(- 1.278)] -35c
= = $9912 - 2785.94 *exp(-0.017c) -35c
For maximum profit dP/dc = 0
differentiating above function, we get
dP/dc = 0- 2785.94 *(-0.017) * exp(-0.017c) - 35 = 0
47.36 * exp(-0.017c) = 35
exp(-0.017c) = 0.739005
taking ln on both sides (ln e^a = a
-0.017c = -0.30245
c = 17.79
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