# Price Elasticiyt of Demand & (AR - MR)

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.

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

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 $$\frac{MR}{p}$$

Questions:

2. How does their answer relate to the distance between MR & AR?

• Does your $\epsilon$ include the negative sign or is it the absolute value of the elasticity? May 6, 2023 at 17:55
• 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! May 6, 2023 at 18:07

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 $$|\epsilon|$$.

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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, $$MR = MC$$.

Therefore,

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

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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.

• 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! May 20, 2023 at 21:48