Solve the problem.
-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 $\mathrm{p}(\mathrm{x})=106-\frac{\mathrm{x}}{28}$. How many candy bars must be sold to maximize revenue?
A) 2968 thousand candy bars
B) 1484 thousand candy bars
C) 1484 candy bars
D) 2968 candy bars

Question

Quizplus · Accepted Answer

Revenue is maximized when the derivative of the revenue function $R(x) = x \cdot p(x)$ is zero. Solving $R(x) = x(106 - \frac{x}{28})$ gives the maximum at $x = 1484$ thousand candy bars.

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