• Problem
• Detailed Solution
• Summary Solution

### Revenue as a function of price

Let the price p (in mu/qu) of a good be given by

$p(x)=150-0.4x$,

with x being the product quantity in qu.

Determine the price level p for which a price increase of 0.1 mu/qu would make revenue go down by about 0.5 mu.

### Preliminary remarks

To solve this problem, you need the following competences:

• The definition of a differential
• Differentiation rules in general (see learning sequence 2, block II, of our script)

### Sketching the problem structure

Let us first sketch a diagram to better understand the problem and to get an idea of how to solve it. This might give us a starting point for a formal approach. Since we need to know how the price works on revenue, it makes sense to map revenue as a function of price to start with: R = R(p).

In exercise 10 of series 5, we already did this:

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

Thus, revenue is a quadratic function of price. Since the quadratic term, i.e. $2.5p^2$, has a negative sign, we know that the graph of R must be concave down. This means that the arms of that graph point downwards, and the vertex must be the maximum of R. In the exercise it says that an increased price must make revenue decrease (the amount of this reaction is of secondary importance, here). Thus we must be in the right hand section of the graph.

We now try to be more precise: if in point P 1, a price increase of 0.1 (called a differential) makes revenue go down by 0.5 (equally called a differential), the quotient of reaction by cause, that is dR/dp, must be -0.5/0.1, which is minus 5 (see diagram).

Our problem consists of finding the price p for which this is the case. The differentials dp and dR help us to find a precise solution by a formal approach.

### Formal approach

There are two ways of writing down a differential quotient:

$R'(p) \;$ (this is Leibniz' version)    or    $\large{\frac{dR}{dp}} \;$ (this is Newton's version).

Since both terms express the same thing, we may equate them: $\quad R'(p)=\large{\frac{dR}{dp}}$.

Since R(p) is a polynomial function, it can be differentiated easily: $R'(p) =375-5p$.

Newton's differential quotient shows us that in our problem, this term must be just the same as -.5/0.1 = -5, the quotient of differentials we found above.

We now have one equation with one unknown:

$375-5p=-5$

The solution can be obtained in just two steps:

$380=5p$

$p = 76$ mu/qu

$\large{\frac{dR}{dp}}$ $=\large{\frac{-0.5}{0.1}}$ $=-5$

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

$\large{\frac{dR}{dp}}$ $= R'(p) \quad \Rightarrow \quad 375-5p=-5$

$p = 76$ mu/qu