Optimal pricing is a difficult subject, with many unquantifiable issues to take into account. However, we can analyse a simple case, and build from there.

 This model is pretty simplistic, but the principle is sound. In particular, even if the price-sales function is more complex, each part of it is, when you zoom in, an approximation to a straight-line. The calculations still work too, but the function can't almost certainly can't be described by a simple function, so it must be done numerically. Additionally, the robustness of the solution should be considered. It may be that small changes to the price/sales response curve causes the solution to "leap" to another part of the curve, giving a completely different "optimal" solution. Finally, "optimal" is actually a relative term. Maximum profits might not be the objective of a business in a complex market. Preventing competition might be more important.

Suppose we have identical items to sell, each of which costs c to make. We have a market saturation at N, which is as many as we can dispose of even charging nothing, and that the sales drop off linearly as the price increases. That means our sales, S, are given by a formula:

$S=N-\alpha{p}$

where $\alpha$ is the "drop-off factor" and is greater than zero, and p is the price charged. Then for a given price p the profits are:

$\begin{eqnarray}P&=&S(p-c)\\&=&Np-Nc-\alpha{p^2}+\alpha{p}{c}\\&=&-\alpha{p^2}+(N+{\alpha}c)p-cN\end{eqnarray}$

This is a downwards facing parabola with a maximum for some value of p. As a quick sanity check, if p=0 then we lose cN because we dispose of N, items, each costing c to provide. Substituting $p=N/\alpha$ we get 0, which is right because we sell none, hence make none, hence no money changes hands anywhere.

We want to maximise profit by adjusting p to get the largest P, which is not necessarily obtained by getting the maximum sales!

We can use calculus to maximise P. We differentiate with respect to the thing we can change, p, and we get this:

$\frac{dP}{dp}=N-2{\alpha}p$

We get a maximum (in this case - because the $p^2$ term is negative) when $\frac{dP}{dp}=0,$ and that happens when $p=N/(2\alpha).$

Exercises:

• Compute the profit when p is this optimal value,
• Show that that varying the value of p slightly up or down makes P smaller
• ... thereby showing that the profit is a local maximum
• Show that the sales achieved are less than half the market saturation N.

One of the enrichment activities on this site.