Optimal PricingYou are currentlybrowsing as guest. Click here to log in 

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.

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 "dropoff factor" and is greater than zero, and p is the price charged. Then for a given price p the profits are:
$\begin{eqnarray}P&=&S(pc)\\&=&NpNc\alpha{p^2}+\alpha{p}{c}\\&=&\alpha{p^2}+(N+{\alpha}c)pcN\end{eqnarray}$
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:
$\frac{dP}{dp}=N2{\alpha}p$
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:
Last change to this page Full Page history Links to this page 
Edit this page (with sufficient authority) Change password 
Recent changes All pages Search 