Question 1197606: A vendor makes and sells x strings of beads at her souvenir store every
month. The monthly cost and price - demand equations are, respectively,
C(x) = 0.00002x3 + 0.04x2 + 150x + 45000
x = 4000 - 200p
Given the information above
i. How many sets of beads should she manufacture each month to
maximize her profit?
ii. What is the maximum monthly profit?
iii. How much should the company charge for each set of beads to
achieve maximum profit?
Answer by Solver92311(821)   (Show Source):
You can put this solution on YOUR website!
First, find the domain of the demand function \ =\ 4000\,-\,200p\)) : If the price is zero, the demand would be 4000, but the revenue would be zero, so zero has to be a lower limit on the demand function (unless you are going to pay customers to take your stuff). On the other hand, if the price is 20, the demand is zero so you don't sell any.
You are given the Cost as a function of the number sold, namely:
)\ =\ 0.00002x^3\,+\,0.04x^2\,+\,150x\,+\,45000)
The Revenue as a function of the number sold is simple, i.e, the price times the number sold or:
)\ =\ px(p))
And the profit as a function of the number sold is just the Revenue minus the Cost:
)\ =\ R(x(p))\ -\ C(x(p)))
But to know the number of beads to sell to maximize the profit, you need to know the price that maximizes the profit. Hence, you have to maximize ) and to find that function, substitute the demand function for 
\ = \ P(x(p))\ =\ R(4000\,-\,200p)\ -\ C(4000\,-\,200p)\ =\ 4000p\ -\ 200p^2\ -\ \[0.00002(4000\,-\,200p)^3\,+\,0.04(4000\,-\,200p)^2\,+\,150(4000\,-\,200p)\,+\,45000\])
The next step is to simplify the Profit function to a cubic polynomial in  .
Then calculate the first derivative and solve:
}{dp}\ =\ 0) for a value of  in the domain for  you established at the beginning of the problem.
And then make sure the second derivative is negative in the interval of the domain of the demand function.
John
My calculator said it, I believe it, that settles it
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