<p>Notice that since <span class="math-container">$AR = P$</span>, the ratio <span class="math-container">$\frac{MR}{P}$</span> is actually <span class="math-container">$\frac{MR}{AR}$</span>.</p>
<p>Notice <span class="math-container">$\frac{MR}{AR} = 1 + \frac{1}{\epsilon}$</span> falls from <span class="math-container">$1$</span> to <span class="math-container">$0$</span> continuously as <span class="math-container">$\epsilon$</span> moves from <span class="math-container">$-\infty$</span> to <span class="math-container">$-1$</span>.</p>
<p>This is what happens in the graph from <span class="math-container">$q = 0$</span> to <span class="math-container">$q = q_1$</span>.</p>
<p>At <span class="math-container">$q = 0$</span>,</p>
<p><span class="math-container">$\epsilon = \frac{\partial q}{\partial P} \cdot \frac{P}{q} \to - \infty $</span> as <span class="math-container">$q \to 0$</span>.</p>
<p>At <span class="math-container">$q = q_1$</span>, we have</p>
<p><span class="math-container">$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}$</span></p>
<p>By the inverse function rule,</p>
<p><span class="math-container">$\frac{\partial q}{\partial P} = - \frac{q}{P} \implies \epsilon = \frac{\partial q}{\partial P} \cdot \frac{P}{q} = -1 $</span></p>
<p>It is easy to check that <span class="math-container">$\frac{MR}{AR} = 1 + \frac{1}{\epsilon}$</span> for <span class="math-container">$\epsilon &lt; -1$</span> is increasing in <span class="math-container">$\epsilon$</span>, or equivalently, decreasing in <span class="math-container">$|\epsilon|$</span>.</p>
<p>—————————-—————————-—</p>
<p>Your answer is mathematically correct, I think it’s more an issue of what the book’s author intended the exercise for.</p>
<p>From the textbook’s answer, we can express the markup formula of monopoly power, by noting that <span class="math-container">$AR = P$</span> and at the profit maximizing quantity, <span class="math-container">$MR = MC$</span>.</p>
<p>Therefore,</p>
<p><span class="math-container">$\frac{MC}{P} = 1 + \frac{1}{\epsilon} \implies \text{Markup} = \frac{P - MC}{P} = 1 - \frac{MC}{P} = - \frac{1}{\epsilon} = \frac{1}{|\epsilon|}$</span></p>
<p>—————————-—————————-——</p>
<p>Your answer also shows that for <span class="math-container">$\epsilon \to - \infty$</span>, <span class="math-container">$AR - MR \to 0$</span>, which is what happens as <span class="math-container">$q \to 0$</span>.</p>
<p>The above is equivalent to the ratio <span class="math-container">$\frac{MR}{AR} \to 1$</span>.</p>
<p>Notice your answer can be rewritten as <span class="math-container">$AR - MR = \frac{P}{|\epsilon|}$</span> which increases continuously as <span class="math-container">$\epsilon$</span> gets less negative (the demand is less elastic).</p>
<p>From your answer we can also see that when <span class="math-container">$\epsilon = -1$</span>, <span class="math-container">$AR - MR = P$</span>. This happens when <span class="math-container">$MR = 0$</span>, since we always have <span class="math-container">$AR = P$</span>.</p>
<p>The above is equivalent to the ratio <span class="math-container">$\frac{MR}{AR} = 0$</span>.</p>
<p>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.</p>

CC BY-SA 4.0
2023-05-20T21:48:50.577Z
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!
CormJack
CC BY-SA 4.0
2023-05-06T19:31:56.937Z
2023-05-06T19:37:19.303Z
2023-05-06T19:37:19.303Z
Nicolas Torres
<p>I have the following question:</p>
<blockquote>
<p>Using this equation: <span class="math-container">$MR = P(1+\frac{1}{ε})$</span> 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.</p>
</blockquote>
<p><strong>My Answer:</strong></p>
<p>Vertical Distance <span class="math-container">$= AR - MR = \frac{p•q}{q} - p(1+\frac{1}{ε}) = -\frac{p}{ε}$</span></p>
<p><strong>Book Answer:</strong></p>
<blockquote>
<p>The equation for ε implies that <span class="math-container">$\frac{MR}{p} = 1+\frac{1}{ε}$</span>. Hence, the less elastic is the demand (assuming <span class="math-container">$ε &lt; -1$</span>), the smaller will be the ratio <span class="math-container">$\frac{MR}{p}$</span></p>
</blockquote>
<p><strong>Questions:</strong></p>
<ol>
<li>Is my answer correct?</li>
<li>How does their answer relate to the distance between MR &amp; AR?</li>
<li>How does their answer relate to my answer?</li>
</ol>
<p><a href="https://i.sstatic.net/ScIWx.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ScIWx.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-05-06T17:55:52.427Z
Does your $\epsilon$ include the negative sign or is it the absolute value of the elasticity?
Nicolas Torres
CC BY-SA 4.0
2023-05-06T18:07:24.520Z
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!
CormJack
<mathematical-economics><elasticity><producer-theory>
Price Elasticiyt of Demand & (AR - MR)
CormJack
# Price Elasticiyt of Demand & (AR - MR)

(Asked by: username_0 on 2023-05-06)

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 $\frac{MR}{p}$

**Questions:**

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](https://i.sstatic.net/ScIWx.jpg)](https://i.sstatic.net/ScIWx.jpg)

**Tags:** mathematical-economics, elasticity, producer-theory

> **Comments:**
> **username_1**: Does your $\epsilon$ include the negative sign or is it the absolute value of the elasticity?  
> **username_0**: 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!  

---

## 1 answers

### Answer by username_1 on 2023-05-06. Score: 2 | Accepted answer

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, $MR = MC$.

Therefore,

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

—————————-—————————-——

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.

> **Comments:**
> **username_0**: 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!  

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