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

MATH104 December 2013

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Question 05

Currently 1800 people ride a commuter passenger ferry each day and pay $4 for a ticket. The number of people q willing to ride the ferry at price p is determined by the relationship $p=\left({\frac {q-3000}{600}}\right)^{2}.$ The company would like to increase its revenue. Use the price elasticity of demand $\epsilon$ to give advice to management on whether it should increase or decrease its price from $4 per passenger. Recall that $\epsilon ={\frac {p}{q}}{\frac {dq}{dp}}.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 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.

Hint 2
Try differentiating the price-demand relationship with respect to $p$ and using the chain rule to evaluate ${\frac {dq}{dp}}$.

Hint 3
Note that the price-demand relationship can give more than one value for q for any one value of p.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

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