Science:Math Exam Resources/Courses/MATH104/December 2016/Question 06 (b)/Solution 1

From demand curve, we have the relationship between $p$ and $q$ as $q=\frac{168-p}{10}$.

Following the hint, the revenue is $R=p\cdot q$ and hence

$\begin{array}{rl}R& =p\cdot \frac{168-p}{10}\\ \end{array}$

If we want to find the price that makes most revenue, then that point should be at the maximum of function $R$, where the derivative of function is zero.

Therefore setting $\frac{dR}{dp}=1\cdot \frac{168-p}{10}+p\cdot \frac{-1}{10}=\frac{1}{10}\cdot (168-p-p)=0$ gives us $2p=168\to p=84$

Since $\frac{d^{2}R}{d{p}^2}=-\frac{2}{10}<0$, the critical point $p=84$ is indeed the price which produces the most revenue.

Answer: $84$