Suppose x denotes the number of units a company plan to produce or sell, usaually, a revenue function R(x) is set up as follows: R(x)=( price per unit) (number of units produced or sold). Sometimes the price per unit is a function x, say, p(x). It is often called a demand function too because when a company produce (or sell) more, it means there is more demand for the prouct, and the price per unit should come down. Let's see the following
Example: A fast-food restaurant has determind that the monthly demand for their hamburgers is given by p(x) = (60,000-x)/20,000 . (a) Find the number of hamburgers this restaurant should sell in order that the revenue is maximized. (b) Find the maximum revenue. (c) When will the restaurant make no revenue at all?
We first set up the revenue function R(x) = x[(60,000-x)/20,000] = -x 2 /20,000 + 3x
Notice that y=R(x) is a parabola opening downward, it has a maximum at x = 30,000 In other words, when the restaurant sells 30,000 hamburgers. The maximum revenue is R(30,000)=$45,000. And the restaurant makes no revenue when R(x)=0, which means -x 2/20,000 + 3x =0. Using factoring technique, we get x=0 or x =60,000 . This means that the restaurant makes no revenue when they do not sell any hamburger or when they sell 60,000 hamburgers.