# Show that slope of MR (marginal revenue) is negative for monopolists

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?

• But if we have homothetic demand curves, the slope of the demand curve should be a constant shouldnt it? @Giskard Jun 28 at 20:58
• 1. This assumption is not stated in your question. Jun 28 at 20:58
• 2. What is a homothetic demand curve? Jun 28 at 20:58
• An affine transformation of an homogenous demand function of degree one? @Giskard Jun 28 at 20:59
• In this (special) case it is indeed true that $$MR(Q) = \frac{d\bigl(Q\frac{dP(Q)}{dQ}\bigr)}{dQ}+\frac{dP(Q)}{dQ} = -2b < 0.$$ Jun 28 at 21:12

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 $$\frac{dMR}{dQ}= Q\frac{d^2P(Q)}{dQ^2} + 2\frac{dP(Q)}{dQ} < 0.$$