Science:Math Exam Resources/Courses/MATH104/December 2013/Question 05/Solution 1

To use the elasticity of demand, we evaluate $\epsilon$ at $p=4$. If $\epsilon <-1$, then demand is elastic and the price should be lowered to increase revenue. Otherwise, if $\epsilon >-1$, then demand is inelastic and the price should be raised.

A quick calculation using the given price-demand relationship confirms that if $q=1800$, then $p=4$.

To evaluate ${\frac {dq}{dp}}$, we opt to use the chain rule on the price-demand relationship. Differentiating it with respect to $p$, gives

${\begin{aligned}{\frac {d}{dp}}p&={\frac {d}{dp}}\left({\frac {q-3000}{600}}\right)^{2}\\1&=2\left({\frac {q-3000}{600}}\right){\frac {1}{600}}{\frac {dq}{dp}}\\1&=\left({\frac {q-3000}{600}}\right){\frac {1}{300}}{\frac {dq}{dp}}\end{aligned}}$

Solving for ${\frac {dq}{dp}}$ we obtain

${\frac {dq}{dp}}=300\left({\frac {600}{q-3000}}\right)$

Putting together all the pieces to evaluate $\epsilon$ at $p=4$, we get

${\begin{aligned}\epsilon &={\frac {p}{q}}{\frac {dq}{dp}}\\&={\frac {1}{q}}\left({\frac {q-3000}{600}}\right)^{2}300\left({\frac {600}{q-3000}}\right)\\&={\frac {300}{q}}\left({\frac {q-3000}{600}}\right)\end{aligned}}$

Hence, when $q=1800$, then $\epsilon =-6/18$.

Thus the price should be raised in order to increase revenue.

Note: If we weren't told that the current number of passengers is 1800, we would have to consider the alternate solution $p=4,q=4200$. This gives us a positive price elasticity of $3/21$, which again tells us that increasing the price will increase revenue.

Note that while it may happen that a good has positive elasticity, and thus it does not follow the law of demand (e.g. Do expensive handbags become more or less desirable to wealthy shoppers if the price drops?), this is unlikely to be the case in our scenario. Clearly the demand equation we are using cannot hold for all $q$ and $p$ (otherwise it would suggest that the price should be increased to infinity!), reminding us that mathematical models are only valid within certain bounds.