Science:Math Exam Resources/Courses/MATH104/December 2011/Question 05 (d)

MATH104 December 2011

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Question 05 (d)
Describe geometrically (in terms of the graph of the demand curve) how you would find the price that maximizes the revenue. It might be useful to note that the slope of a line connecting a point on the demand curve to the origin is q/p.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Revenue is maximized when elasticity is -1.

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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 maximize revenue, we need the elasticity $\epsilon ={\frac {p}{q}}{\frac {dq}{dp}}=-1$, which is equivalent to stating ${\frac {dp}{dq}}=-{\frac {q}{p}}.$ Consider the line joining the origin to the point $(p,q=f(p))$ on the demand vs price curve. As the hint suggests, the slope of such a line is q/p, which is the negative of what we want ${\frac {dp}{dq}}$ to be at the price where revenue is maximized. Thus, to maximize the revenue, we can take a series of lines joining the origin to the point $(p_{0},f(p_{0}))$ for different values of $p_{0}$ and determine the line such that if it is reflected across the line $p=p_{0}$ (so its slope is negated), it is the tangent line to the graph $q=f(p)$ at the point $(p_{0},q_{0})$. The slope of the tangent line is ${\frac {dq}{dp}}|_{p=p_{0},q=q_{0}}$. This has a nice geometric interpretation: the line segment joining the origin to the point $(p_{0},q_{0})$ has same length as the line segment of the tangent line joining $(p_{0},q_{0})$ to the $p-$axis. In fact, an isosceles triangle is formed. Note: it is tempting to say that the revenue is maximized at $p_{0}$ if the line joining the origin to $(p_{0},q_{0})$ meets the curve $q=f(p)$ at a right angle (recall two curves are orthogonal if the product of their slopes at their intersection point is $-1$). But the slope of the line is $q/p$ and not $p/q$, so this doesn't hold. We need ${\frac {p}{q}}{\frac {dq}{dp}}=-1$ and not ${\frac {q}{p}}{\frac {dq}{dp}}=-1$.