Solve the problem.
-If the price charged for a candy bar is $\mathrm { p } ( \mathrm { x } )$ cents, then $\mathrm { x }$ thousand candy bars will be sold in a certain city, where $\mathrm { p } ( \mathrm { x } ) = 24 - \frac { \mathrm { x } } { 22 }$. How many candy bars must be sold to maximize revenue?
A) 264 thousand candy bars
B) 528 candy bars
C) 264 candy bars
D) 528 thousand candy bars

Question

Quizplus · Accepted Answer

To maximize revenue, differentiate $p(x) = 24 - \frac{x}{22}$ with respect to $x$ and set the derivative equal to zero to find the critical point. The derivative of $p(x)$ with respect to $x$ is $-\frac{1}{22}$. However, since this is a linear function, its slope is constant, and there's no maximum point in the traditional sense. Instead, we maximize revenue $R(x) = x \cdot p(x)$. Setting up the revenue function: $R(x) = x(24 - \frac{x}{22})$. Differentiating $R(x)$ with respect to $x$ and setting it to zero gives the quantity $x$ that maximizes revenue. The derivative of $R(x)$ is $R'(x) = 24 - \frac{2x}{22}$. Setting $R'(x) = 0$ gives $x = 264$. Therefore, 264 thousand candy bars must be sold to maximize revenue.

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