Math 165: Maximizing Revenue

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Suppose the quantity (demand) q and the price p are related, e.g., by a relation of the form p = D(q).

The problem is find the quantity q so that the revenue, R = pq, is maximized. The revenue R can be expressed as a function of the quantity q by

R(q) = q * D(q).

A typical demand function looks like this:

Notice that the revenue, R(q) = p ·q = q ·D(q), is represented by the it area of the rectangle with opposite vertices at (0,0) and (q,D(q)).

Notice that when q is very small or near the right side, the area of the rectangle is small.

Now observe how the area (revenue) changes as q moves from 0 to 4.

It appears that for q small (high price and low demand) the revenue is small. For p small (low price, market is saturated), there is low revenue. For some reasonable price p, the revenue (area) is maximized.

The mathematical assumptions are:

·

Increasing p means decreasing demand.

·

There is maximum price D(0) the consumer will pay.

·

There is maximum consumer demand (In our example, q = 4).

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File translated from T_EX by T_TH, version 3.85.
On 23 Feb 2009, 21:45.