<p>You want to show
<span class="math-container">$$
\frac{dMR}{dQ} &lt; 0.
$$</span>
As you point out in the comments
<span class="math-container">$$
\frac{dMR}{dQ}= Q\frac{d^2P(Q)}{dQ^2} + 2\frac{dP(Q)}{dQ}.
$$</span></p>
<br>
<p><strong>The linear case</strong></p>
<p>When <span class="math-container">$P(Q) = a - bQ$</span>, assuming <span class="math-container">$a,b&gt;0$</span>, you get
<span class="math-container">$$
\frac{dMR}{dQ}= Q \cdot 0 - 2b = - 2b&lt; 0.
$$</span></p>
<br>
<p><strong>The general case</strong></p>
<p><span class="math-container">$$
Q\frac{d^2P(Q)}{dQ^2} + 2\frac{dP(Q)}{dQ} &lt; 0
$$</span>
does not hold for all decreasing functions <span class="math-container">$P(Q)$</span>. A counterexample is <span class="math-container">$P(Q) = Q^{-2}$</span>. <br>In this case
<span class="math-container">$$
Q\frac{d^2P(Q)}{dQ^2} + 2\frac{dP(Q)}{dQ} = Q(-2)(-3)Q^{-4} + 2(-2)Q^{-3} = 2Q^{-3} &gt; 0.
$$</span></p>
<br>
<p><strong>The concave case</strong></p>
<p>A sufficient (but not necessary) condition for downward sloping <span class="math-container">$MR(Q)$</span> is to assume that <span class="math-container">$P(Q)$</span> is strictly decreasing and concave. <br> In this case <span class="math-container">$\frac{dP(Q)}{dQ}$</span> is negative and <span class="math-container">$\frac{d^2P(Q)}{dQ^2}$</span> is non-positive, so their (weighted) sum is negative, hence
<span class="math-container">$$
\frac{dMR}{dQ}= Q\frac{d^2P(Q)}{dQ^2} + 2\frac{dP(Q)}{dQ} &lt; 0.
$$</span></p>

CC BY-SA 4.0
2021-06-29T06:25:39.207Z
2021-06-29T06:59:19.743Z
2021-06-29T06:59:19.743Z
Giskard
<p>I want to show that the marginal revenue is negative for monopolists.</p>
<p>We assume <span class="math-container">$P(Q)$</span> is homogenous of degree 1, so it is linear (affine, strictly speaking): <span class="math-container">$P(Q)=a-bQ$</span>.</p>
<p>As we know, <span class="math-container">$\frac{dP(Q)}{dQ} \lt 0$</span> because the slope of the demand curve is negative.</p>
<p>So <span class="math-container">$\overbrace{Q\frac{dP(Q)}{dQ}+P(Q)}^{=MR} \lt P(Q)$</span> and <span class="math-container">$MR \lt P(Q)\ \forall \ Q \ge 0$</span>.</p>
<p>So I want to show that the marginal revenue is not constant (as it is on a competitive market) but also decreases.</p>
<p><span class="math-container">$$
\frac{dMR}{dQ}=\frac{d\bigl(Q\frac{dP(Q)}{dQ}+P(Q)\bigr)}{dQ}=\frac{d\bigl(Q\frac{dP(Q)}{dQ}\bigr)}{dQ}+\frac{dP(Q)}{dQ}=2\frac{dP(Q)}{dQ}.
$$</span></p>
<p>Is this approach correct and is it sufficient for showing that 1. the slope of MR will be negative for monopolists and 2. it will be exactly twice as great as the slope of the demand curve?
Is there a simpler/more logical explanation?</p>

CC BY-SA 4.0
2021-06-28T20:58:06.173Z
But if we have homothetic demand curves, the slope of the demand curve should be a constant shouldnt it? @Giskard
j3141592653589793238
CC BY-SA 4.0
2021-06-28T20:58:27.653Z
1. This assumption is not stated in your question.
Giskard
CC BY-SA 4.0
2021-06-28T20:58:57.373Z
2. What is a homothetic demand curve?
Giskard
CC BY-SA 4.0
2021-06-28T20:59:54.473Z
An affine transformation of an homogenous demand function of degree one? @Giskard
j3141592653589793238
CC BY-SA 4.0
2021-06-28T21:05:53.730Z
I think my assumption should be it's an affine demand curve $P(Q)=a-bQ$. @Giskard Sorry I forgot about that. So all of this would only hold for affine/linear demand functions?
j3141592653589793238
CC BY-SA 4.0
2021-06-28T21:06:46.223Z
Please [edit your question](https://economics.stackexchange.com/posts/45651/edit) accordingly.
Giskard
CC BY-SA 4.0
2021-06-28T21:12:04.487Z
In this (special) case it is indeed true that
$$
MR(Q) = \frac{d\bigl(Q\frac{dP(Q)}{dQ}\bigr)}{dQ}+\frac{dP(Q)}{dQ} = -2b < 0.
$$
Giskard
CC BY-SA 4.0
2021-06-28T21:14:41.507Z
Thanks. I reckon this will be an entirely different story for non-linear demand curves such as the one you proposed above? Or would there be a more general way of showing that $dMR/dQ$ is negative for monopolists? @Giskard Oh and I think you meant $dMR(Q)/dQ$, not $MR(Q)$?
j3141592653589793238
CC BY-SA 4.0
2021-06-28T21:21:07.630Z
Have you tried proceeding with the differentiation? Seems like using the product rule would yield results.
Giskard
CC BY-SA 4.0
2021-06-28T22:00:37.937Z
Hmm I got $\frac{d(Q\frac{dP(Q)}{dQ})}{dQ}+\frac{dP(Q)}{dQ}=Q\biggl(\frac{\frac{dP(Q)}{dQ}}{dQ}\biggr)+2\frac{dP(Q)}{dQ}$ but I can't really make much of the second derivative on the left side. The only thing I know is that $dP(Q)/dQ$ is negative. But that doesn't tell me anything about the second derivate. @Giskard
j3141592653589793238
CC BY-SA 4.0
2021-06-29T06:06:18.137Z
@HerrK. We moved on to non-linear curves in [this comment](https://economics.stackexchange.com/questions/45651/show-that-slope-of-mr-marginal-revenue-is-negative-for-monopolists#comment74293_45651).
Giskard
<microeconomics><mathematical-economics><monopoly>
Show that slope of MR (marginal revenue) is negative for monopolists
j3141592653589793238
# Show that slope of MR (marginal revenue) is negative for monopolists

(Asked by: username_0 on 2021-06-28)

I want to show that the marginal revenue is negative for monopolists.

We assume $P(Q)$ is homogenous of degree 1, so it is linear (affine, strictly speaking): $P(Q)=a-bQ$.

As we know, $\frac{dP(Q)}{dQ} \lt 0$ because the slope of the demand curve is negative.

So $\overbrace{Q\frac{dP(Q)}{dQ}+P(Q)}^{=MR} \lt P(Q)$ and $MR \lt P(Q)\ \forall \ Q \ge 0$.

So I want to show that the marginal revenue is not constant (as it is on a competitive market) but also decreases.

$$
\frac{dMR}{dQ}=\frac{d\bigl(Q\frac{dP(Q)}{dQ}+P(Q)\bigr)}{dQ}=\frac{d\bigl(Q\frac{dP(Q)}{dQ}\bigr)}{dQ}+\frac{dP(Q)}{dQ}=2\frac{dP(Q)}{dQ}.
$$

Is this approach correct and is it sufficient for showing that 1. the slope of MR will be negative for monopolists and 2. it will be exactly twice as great as the slope of the demand curve?
Is there a simpler/more logical explanation?

**Tags:** microeconomics, mathematical-economics, monopoly
> **Comments:**
> **username_0**: But if we have homothetic demand curves, the slope of the demand curve should be a constant shouldnt it? @Giskard  
> **username_1**: 1. This assumption is not stated in your question.  
> **username_1**: 2. What is a homothetic demand curve?  
> **username_0**: An affine transformation of an homogenous demand function of degree one? @Giskard  
> **username_0**: I think my assumption should be it's an affine demand curve $P(Q)=a-bQ$. @Giskard Sorry I forgot about that. So all of this would only hold for affine/linear demand functions?  
> **username_1**: Please [edit your question](https://economics.stackexchange.com/posts/45651/edit) accordingly.  
> **username_1**: In this (special) case it is indeed true that $$ MR(Q) = \frac{d\bigl(Q\frac{dP(Q)}{dQ}\bigr)}{dQ}+\frac{dP(Q)}{dQ} = -2b < 0. $$  
> **username_0**: Thanks. I reckon this will be an entirely different story for non-linear demand curves such as the one you proposed above? Or would there be a more general way of showing that $dMR/dQ$ is negative for monopolists? @Giskard Oh and I think you meant $dMR(Q)/dQ$, not $MR(Q)$?  
> **username_1**: Have you tried proceeding with the differentiation? Seems like using the product rule would yield results.  
> **username_0**: Hmm I got $\frac{d(Q\frac{dP(Q)}{dQ})}{dQ}+\frac{dP(Q)}{dQ}=Q\biggl(\frac{\frac{dP(Q)}{dQ}}{dQ}\biggr)+2\frac{dP(Q)}{dQ}$ but I can't really make much of the second derivative on the left side. The only thing I know is that $dP(Q)/dQ$ is negative. But that doesn't tell me anything about the second derivate. @Giskard  
> **username_1**: @HerrK. We moved on to non-linear curves in [this comment](https://economics.stackexchange.com/questions/45651/show-that-slope-of-mr-marginal-revenue-is-negative-for-monopolists#comment74293_45651).  

---

## 1 answers

### Answer by username_1 on 2021-06-29. Score: 3 | Accepted answer

You want to show
$$
\frac{dMR}{dQ} < 0.
$$
As you point out in the comments
$$
\frac{dMR}{dQ}= Q\frac{d^2P(Q)}{dQ^2} + 2\frac{dP(Q)}{dQ}.
$$

  

**The linear case**

When $P(Q) = a - bQ$, assuming $a,b>0$, you get
$$
\frac{dMR}{dQ}= Q \cdot 0 - 2b = - 2b< 0.
$$

  

**The general case**

$$
Q\frac{d^2P(Q)}{dQ^2} + 2\frac{dP(Q)}{dQ} < 0
$$
does not hold for all decreasing functions $P(Q)$. A counterexample is $P(Q) = Q^{-2}$.   
In this case
$$
Q\frac{d^2P(Q)}{dQ^2} + 2\frac{dP(Q)}{dQ} = Q(-2)(-3)Q^{-4} + 2(-2)Q^{-3} = 2Q^{-3} > 0.
$$

  

**The concave case**

A sufficient (but not necessary) condition for downward sloping $MR(Q)$ is to assume that $P(Q)$ is strictly decreasing and concave.   
 In this case $\frac{dP(Q)}{dQ}$ is negative and $\frac{d^2P(Q)}{dQ^2}$ is non-positive, so their (weighted) sum is negative, hence
$$
\frac{dMR}{dQ}= Q\frac{d^2P(Q)}{dQ^2} + 2\frac{dP(Q)}{dQ} < 0.
$$


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