# Show that the revenue decreases with price on a elastic demand curve

Let us define revenue (R) as price (P) times quantity (Q).

Furthermore let us suppose that the price depends linearly on the quantity.

Then by the product rule we have that

$$\frac{\%\Delta R}{\%\Delta P} = Q + P \frac{\%\Delta Q}{\%\Delta P}$$

Furthermore by the linearity we have:

$$\frac{\%\Delta R}{\%\Delta P} = Q_0 + 2 P \frac{\%\Delta Q}{\%\Delta P}$$

According to the course on economics that I am taking, the revenue should decrease with the price in a linear demand curve if the elasticity $$\frac{\%\Delta Q}{\%\Delta P}$$, which is always negative, has module greater than one.

But that does not seem obvious from the equation above, what is going on?

To show that revenue decreases with increasing price $$P$$, you only need to show that

• $$\frac{\Delta R}{\Delta P} <0$$ (In words: increasing price gives decreasing revenue.)

The following is given:

• demand is $$\color{blue}{\mbox{elastic}}$$ (wrt. price) which means $$\epsilon(P) = \frac{\frac{\Delta Q}{Q}}{\frac{\Delta P}{P}} = \frac{\Delta Q}{\Delta P}\cdot \frac{P}{Q} \color{blue}{<- 1}$$
• You do not need further assumptions about the form of the demand function here.

Now, you get

$$\frac{\Delta R}{\Delta P} = Q + P\underbrace{\frac{\Delta Q}{\Delta P}}_{= \epsilon(P)\cdot \frac{Q}{P}} = Q + \epsilon(P)Q= (1+\underbrace{\epsilon(P)}_{<-1})Q <0$$

Let’s define the price-elasticity of some quantity $$X$$ to be $$\frac{dX}{dP}\frac PX.$$

Hence, the price-elasticity of revenue is $$1 + \frac{dQ}{dP}\frac PQ,$$ where the second term is the price-elasticity of demand.

Thus, revenue is decreasing (increasing) in price when the price-elasticity of demand is less (greater) than $$-1$$. Intuitively, when a $$1 \%$$ price increase leads to a fall in demand greater than $$1 \%$$, then revenue must fall when prices increase.

Note that I rely only on the differentiability of the demand function. In particular, we do not require a linear relationship between price and quantity for this to hold.