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

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

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 $2p=168\rightarrow p=84$

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

Answer: $\color {blue}84$