SOLUTION: I need your help with this problem A study done for management shows that the revenue, R, in dollars, as a function of number, x, of units produced can be modeled by R(x) = 1

Question 45497: I need your help with this problem
A study done for management shows that the revenue, R, in dollars, as a function of number, x, of units produced can be modeled by
R(x) = 10x - 0.001x^2
while the cost, C, in dollars, can be modeled by
C(x) = 5000 + 2x.
What will the maximum profit be, in dollars?
Answer by Fermat(136) (Show Source): You can put this solution on YOUR website!
Profit = Revenue - Cost
P(x) = R(x) - C(x)
P(x) = 10x - 0.001x� - (5000 + 2x)
P(x) = 8x - 0.001x� - 5000
To find the maximumn, first work out the turning point(s), which are where when dP/dx = 0.
Differentiating P(x),
dP/dx = 8 - 0.002x
When dP/dx = 0 then 8 - 0.002x = 0
8 - 0.002x = 0
8 = 0.002x
x = 8/0.002
x = 4000
========
There is thus a turning point at x = 4000.
To find out whether it is a maximum or a minimum, take the 2nd derivative.
This gives,
d�P/dx� = -0.002
The rule is:
d�P/dx� > 0 means the turning point is a minimum
d�P/dx� < 0 means the turning point is a maximum
Since our d�P/dx� is -0.002 which is < 0, then our turning point is a maximum.
Our maximum profit is at x = 4000 and is equal to,
P(4000) = 8(4000) - 0.001(4000)� - 5000
P = 32,000 - 16,000 - 5,000
P = 11,000
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