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File: OptimalPricing Optimal pricing is a difficult subject, with many unquantifiable issues to take into account. However, we can analyse a particularly simple case, and build from there. [[[>50 Of course, 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 price/sales response curve 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-ap/ where /a/ 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: P = S(p-c) = (N-ap)(p-c) = - ap^2 + (N+ac)p - cN 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/a 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: dP/dp = N-2ap+ac We get a maximum when dP/dp=0, and that happens when p=(N+ac)/(2a). Substitute that back into the formula for total sales and we get S=N-a(N+ac)/(2a) Simplifying, S=(N-ac)/2. Interestingly, this implies that, under this model, when your profit is maximised you have less than half the maximum possible market.