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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 for a given price /p/ the profits are:

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

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.