If the price charged for a candy bar is $p(x)$ cents, then $x$ thousand candy bars will be sold in a certain city, where $\mathrm{p}(\mathrm{x})=49-\frac{\mathrm{x}}{20}$. How many candy bars must be sold to maximize revenue?
A) 490 thousand candy bars
B) 980 thousand candy bars
C) 980 candy bars
D) 490 candy bars

Question

Quizplus · Accepted Answer

Revenue is maximized when the derivative of the revenue function $R(x) = x \cdot p(x)$ is zero. Here, $R(x) = x \cdot (49 - \frac{x}{20})$. Solving $\frac{dR}{dx} = 0$ gives $x = 490$ thousand candy bars.

If the Price Charged for a Candy Bar Is p(x)p(x)p(x) Cents, Then

If the Price Charged for a Candy Bar Is $p(x)$ Cents, Then