Try the Free Math Solver or Scroll down to Tutorials!












Please use this form if you would like
to have this math solver on your website,
free of charge.

Math 165 Linear Price - Demand Model and Parabolas

A typical problem about maximizing revenue

A manufacturer determines that when x hundred units of a particular commodity are
produced, they can all be sold for a unit price given by the demand function p = 80 - x
dollars. What is the maximum revenue (in dollars)?

The demand function is linear - the quantity (demand) x and the price p are related by
a linear relation. The problem is to set the price p (or demand x) so that the revenue R,

R = p ยท x hundred dollars

is maximized.

Notice that

R = (80 - x)x

is a quadratic function of x; moreover, the coefficient of x2 is negative, so the graph of
R(x) is a parabola opening downward. We even have that the parabola is presented in a
factored form with roots at x = 0 and x = 80.

If we graph the parabola, it appears that R is maximized at the vertex of the parabola
which occurs when x = 40, halfway between the roots!

We are already familiar with the quadratic formula for the roots of the equation

Ax2 + Bx + C = 0

occur at

with the usual remarks about the case B2 - 4AC ≤ 0.

The quadratic formula is very much related to the process called completing the square -

It is apparent that the vertex of the parabola is located at

If A > 0, the graph opens upward and the quadratic function has a minimum value at
the vertex.

If A < 0, the graph opens downward and the quadratic function has a maximum value
at the vertex.

In our example A = -1, B = 80, C = 0. The revenue is maximized when

x = -B/2A = 40(hundred units);
p = (80 - 40) dollars;