Calculus question?

Question

The quantity demanded each month of the Walter Serkin recording of Beethoven's Moonlight Sonata, manufactured by Phonola Record Industries, is related to the price/compact disc. The equation
p= -.00042x + 7.6 (0 (those should be greater than or equal to)

where p denotes the unit price in dollars and x is the number of discs demanded, relates the demand to the price. The total monthly cost (in dollars) for pressing and packaging x copies of this classical recording is given by

C(x)= 600+2x-.00002x^2 (0 (Once again that should be greater than or equal to)

To maximize its profits, how many copies should Phonola produce each month?

Hint: The revenue is , and the profit is

R(x)=px and P(x)=R(x)-C(x)

x=_______ copies

Answer 1

You have been given the functions p(x) and C(x). If each disc is sold for price p(x) and there are x many discs, the total income from sales, or revenue, is R(x) = x*p(x), as described. The profit is equal to revenue minus cost, or P(x) = R(x) - C(x), again as described. It's a little irresponsible of the writer to use p and P as separate functions, but try to keep them straight.

You can write P(x) as a polynomial function of x, and you can simplify it by combining like terms. You should get a second-order polynomial. In order to find the maximum profit, you need to find the maximum point of the function, and this is done by setting the derivative equal to zero. You should know how to take the derivative of a polynomial function, and the derivative of a quadratic function will be linear. You should also be able to solve the linear equation (set equal to zero) for x, giving you the number of copies that should be produced each month.

If p(x) is only defined for 0 < x < 12,000 and C(x) is defined for 0 < x < 20,000, the combined function P(x) is defined for the more restricting domain only, 0 < x < 12,000. It's possible that the solution above will yield a value outside of this domain; it might be that the theoretical maximum profit comes at a production level greater than what is defined. In that case, the answer should instead be 12,000. Verify that P'(x) is positive at that point, to show that the profit is still increasing.

Answer 2

Price-Cost=Profit

-.00042x + 7.6 -600 -2x +.00002x^2 = Profit
.00002x^2 - 2.0042x -592.4 = P

dx/dp derivative and solve for 0

.0004X - 2.0042 = 0
.0004x = 2.0042
X = 5010