Question

A company that manufactures flash drives knows that the number of drives x it can sell each week is related to the price p , in dollars, of each drive by the equation x = 1000− 100p . a. Find the price p that will bring in the maximum revenue. Remember, revenue (R) is the product of price (p) and items sold (x) , in other words, R = xp .

Answer 1

To find the price p that will bring in the maximum revenue, we need to find the value of p that maximizes the function R = xp. Since x = 1000 - 100p, we can rewrite the function as R = p(1000 - 100p). To maximize this function, we can take the derivative of R with respect to p, set it equal to zero, and solve for p.

The derivative of R with respect to p is given by:

dR/dp = (1000 - 100p) + (-100 * p) = 1000 - 200p

To find the value of p that maximizes R, we need to set this derivative equal to zero and solve for p:

0 = 1000 - 200p

200p = 1000

p = 1000 / 200 = 5

Therefore, the price p = $5 will bring in the maximum revenue.

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