• Problem
• Detailed Solution
• Summary Solution

Let a price function be given by $$\;p:x \to 150 - 0.4x \;$$ for demand $$x \ge 0$$.

Determine marginal revenue with respect to price for p = 120 mu/qu.

### Structure of the problem

What is needed here is the marginal revenue function R'(p). This function has to be evaluated for p = 120. To solve this problem, you need the following competences:

• Determine and interpret revenue and marginal revenue
• Differentiation rules (see block II learning sequence 2 of our script)

### Formal approach

To determine marginal revenue for p = 120, we have to take three consecutive steps:

(i) Find the revenue function R(p).

(ii) Determine the first order derivative R'(p) of that function.

(iii) Find the functional value R'(120).

Step (i): Find the revenue function R(p)

Revenue is price times quantity, so we have $$R=p \cdot x$$. Since p and x are related to each other, we have two options to express revenue as a function of either price or quantity.

Option 1: Revenue as a function of demand quantity x
(won't help us here, but might be helpful in a different setting; see, for example, exercise 14 of series 5)

Since $$p(x)=150-0.4x$$, we multiply this equation with x to obtain

$$R(x)=p(x) \cdot x=(150−0.4x) \cdot x=150x−0.4x^2$$.

Option 2: Revenue as a function of price p

We need x as a function of p to start with. This is the inverse function to p(x) given above:

$$p=150-0.4x \quad \Rightarrow \quad 0.4x+p=150 \quad \Rightarrow \quad 0.4x=150-p$$

$$\Rightarrow x=375-2.5p \quad \Rightarrow \quad x(p)=375-2.5p$$

All we need to do is multiply this expression by p to obtain

$$E(p)=p \cdot x(p)=p \cdot (375-2.5p)=375p-2.5p^2$$.

Please note that here, revenue depends on price only. Since our problem consisted of finding out how revenue reacts on different price levels, this is what we wanted to have.
In exercise 23 of series 5, we have a comparable situation, and the same approach must be chosen.

Graphically, the revenue function itself is (almost) enough to solve our entire problem: we might draw the tangent to the graph of R(p) at the point (120/R(120)) and then measure the slope of this tangent. The result must be the same number as R‘(120). However, drawing exercises are imprecise by their very nature, and so is measuring slopes. To compare the result of our calculus to the graphical approach please click here.

Step (ii): Determine the first order derivative R'(p) of that function

We differentiate $$R(p)=375p−2.5p 2$$ with respect to p.

Obviously, this leads to a marginal revenue function of $$R'(p)=375-5p$$.

This function can be evaluated for any x of the domain of R, which is all nonnegative numbers. The resulting number R'(p) is the slope of the tangent to the graph of R in the point (p, R(p)).

The economic meaning of that number is as follows: if, starting from a price of p mu/qu, the price increases by another mu/qu, the extra revenue will be approximately R'(p) mu.

Step (iii): Find marginal revenue for price p=120

This is the easiest part of the exercise: all you need to do is evaluate R' for p=120, that means determine R'(120).

$$R'(120)=375-5 \cdot 120=-225$$ mu/(mu/gu).

The corresponding diagram can be found here.

The economic meaning of this results is: if, starting from a price of 120 mu/qu, the price increases by another mu/qu, the revenue will decrease by about 225 mu (see diagram).

$$p=150-0.4x \quad \Rightarrow \quad x(p)=375-2.5p$$

$$E(p)=p \cdot x(p)=p \cdot (375-2.5p)=375p-2.5p^2$$

$$E'(p)=375p-5p$$

$$E'(120)=375-5 \cdot 120=-225$$ mu/(mu/qu)

If, starting from a price of 120 mu/qu, the price increases by another mu/qu, the revenue will decrease by about 225 mu.