How are elasticity of demand, marginal revenue, and total revenue connected?

Question

I am completely new to economics, and am looking at some old exam papers. The question is as following:

"In a normal marked, where there is a negative correlation between quantity and price, which of the following are true for elasticity of demand, marginal revenue, and total revenue?"

The correct alternative was:

d) When the absolute value of elasticity is 1, the marginal revenue is 0, and total revenue is maximized

I was wondering:

Why is the total revenue maximized when marginal revenue is 0? I know that, usually, total revenue is maximized when marginal revenue = marginal cost. In this case, however, no information is given about the marginal cost, so it does not make sense to me that it should be 0.

Also, is the marginal revenue always 0, when elasticity is exactly equal to 1? How does that make sense?

Answer 1

Marginal cost is irrelevant to total revenue. It's relevant to profit, but not total revenue.

Total revenue is just price x quantity.

Marginal revenue is a monotonic decreasing function. That is, as quantity increases, marginal revenue decreases. So, let's take the quantity Q to be the point at which marginal revenue is zero.

For quantities below Q, marginal revenue is positive, so total revenue increases with quantity.

For quantities above Q, marginal revenue is negative, so total revenue decreases with quantity.

Hence Q is the turning point of the curve: it is the point of maximum total revenue.

Elasticity of 1 means that a tiny % change, $\Delta$ in price in one direction (equivalent to multiplying price by $1+\Delta$), will be met by a reciprocal proportional change in quantity in the other direction, equivalent to multipying it by $\frac{1}{1+\Delta}$: that's straight from the definition of elasticity.

So total revenue, the product of price and quantity, remains the same:

$$(1+\Delta) P \times \frac{1}{(1+\Delta)} Q = P \times Q$$

Hence marginal revenue always 0, when elasticity is exactly equal to 1.