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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.

[[[>50 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:

EQN:S=N-\alpha{p}

where EQN:\alpha is the "drop-off factor" and is greater than zero, and /p/ is the price charged.

Then profits are:

Then for a given price /p/ the profits are:

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

EQN:P=S(p-c)=(N-\alpha{p})(p-c)=Np-Nc-\alpha{p^2}+\alpha{c}

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 EQN: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 this by adjusting /p/ to get the largest /P,/ which is not necessarily

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:

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

We get a maximum (in this case - because the EQN:p^2 term is negative) when EQN:\frac{dP}{dp}=0,

and that happens when EQN: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./

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One of the enrichment activities on this site.