Question

A demand equation is an equation that describes the relationship between the price of an item and the number of items that would be sold at that price. For example, the demand equation q = 2140 − p/100 suggests that 0 items would be sold if the price were p = 214, 000 and 2140 items would be sold if the price were p = 0.

The revenue function, R(q), is the amount of money that would be generated as a function of the quantity sold. It can be calculated from the demand equation.

Suppose that the demand equation for manufacturing switchboards is q = 2140 − p/100, where p is measured in dollars.

1. We would like to find out what price would maximize the total revenue generated.

(a) Explain why R(q) = −100q² + 241, 000q.

(b) Use calculus to find the local maximum of R(q) on the interval 0 < q ≤ 2140 (there is exactly one). What is the total revenue at that value of q, and what price p should you choose in order to sell that many switchboards?

2. It is much more important to find out what price would maximize the profit, rather than the revenue, of course. If C(x) is the cost to manufacture the xth switchboard, then the total cost to manufacture q switchboards would be T(q) = C(1) + C(2) + C(3) + · · · + C(q − 1) + C(q).

In Part IV of this project, we will find an upper bound estimate on this total cost under the learning curve model we have been using: T(q) ≤ 319, 575q ^0.848 − 48, 575.

(a) Calculate the profit generated when revenue is maximized, assuming that T(q) ≈ 319, 575q ^0.848 − 48, 575.

(b) It turns out that under these assumptions the profit will be maximized at p = 158, 916, which is not the same price that maximizes revenue. Does that make sense, or does it indicate a flaw in our model? Explain.

Answer 1