I have the following question:

Using this equation: $MR = P(1+\frac{1}{ε})$ and the attached graph. How does the vertical distance between the demand curve and MR curve at a given level of output depend on the PeD (ε) at that output level.

My Answer:

Vertical Distance $= AR - MR = \frac{p•q}{q} - p(1+\frac{1}{ε}) = -\frac{p}{ε}$

Book Answer:

The equation for ε implies that $\frac{MR}{p} = 1+\frac{1}{ε}$. Hence, the less elastic is the demand (assuming $ε < -1$), the smaller will be the ratio $\frac{MR}{p}$


  1. Is my answer correct?
  2. How does their answer relate to the distance between MR & AR?
  3. How does their answer relate to my answer?

enter image description here

  • 1
    $\begingroup$ Does your $\epsilon$ include the negative sign or is it the absolute value of the elasticity? $\endgroup$ May 6, 2023 at 17:55
  • $\begingroup$ I'm assuming that ε is negative, i.e.$-\frac{p}{ε} \ge 0$, thanks for checking!...And as a result of your comment, i have just fixed the typo. It now reads "(assuming $ε < -1$)" thanks! $\endgroup$
    – CormJack
    May 6, 2023 at 18:07

1 Answer 1


Notice that since $AR = P$, the ratio $\frac{MR}{P}$ is actually $\frac{MR}{AR}$.

Notice $\frac{MR}{AR} = 1 + \frac{1}{\epsilon}$ falls from $1$ to $0$ continuously as $\epsilon$ moves from $-\infty$ to $-1$.

This is what happens in the graph from $q = 0$ to $q = q_1$.

At $q = 0$,

$\epsilon = \frac{\partial q}{\partial P} \cdot \frac{P}{q} \to - \infty $ as $q \to 0$.

At $q = q_1$, we have

$MR = 0 \implies 0 = MR = \frac{\partial Pq}{\partial q} = P + q \frac{\partial P}{\partial q} \implies \frac{\partial P}{\partial q} = - \frac{P}{q}$

By the inverse function rule,

$\frac{\partial q}{\partial P} = - \frac{q}{P} \implies \epsilon = \frac{\partial q}{\partial P} \cdot \frac{P}{q} = -1 $

It is easy to check that $\frac{MR}{AR} = 1 + \frac{1}{\epsilon}$ for $\epsilon < -1$ is increasing in $\epsilon$, or equivalently, decreasing in $|\epsilon|$.


Your answer is mathematically correct, I think it’s more an issue of what the book’s author intended the exercise for.

From the textbook’s answer, we can express the markup formula of monopoly power, by noting that $AR = P$ and at the profit maximizing quantity, $MR = MC$.


$\frac{MC}{P} = 1 + \frac{1}{\epsilon} \implies \text{Markup} = \frac{P - MC}{P} = 1 - \frac{MC}{P} = - \frac{1}{\epsilon} = \frac{1}{|\epsilon|}$


Your answer also shows that for $\epsilon \to - \infty$, $AR - MR \to 0$, which is what happens as $q \to 0$.

The above is equivalent to the ratio $\frac{MR}{AR} \to 1$.

Notice your answer can be rewritten as $AR - MR = \frac{P}{|\epsilon|}$ which increases continuously as $\epsilon$ gets less negative (the demand is less elastic).

From your answer we can also see that when $\epsilon = -1$, $AR - MR = P$. This happens when $MR = 0$, since we always have $AR = P$.

The above is equivalent to the ratio $\frac{MR}{AR} = 0$.

Your answer is related to the textbook’s answer. The main difference is that yours is expressed in terms of arithmetic difference, while the textbook’s is expressed in terms of ratios.

  • $\begingroup$ Thanks a lot for the detail Nicholas, apologies you went to such great lengths and i didn't have chance to review it yet. I will do ASAP and let you know if i have any questions. Thank again! $\endgroup$
    – CormJack
    May 20, 2023 at 21:48

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