Math 165 Linear Price - Demand Model and Parabolas
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 ·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
A x2 + B x + C
= 0
occur at
x
=
- B ±
Ö
B2 - 4 A C
2 A
 ,
with the usual remarks about the case B2 - 4 A C £ 0.
The quadratic formula is very much related to the process called completing the square - write
A x2 + B x + C
= A (x2 + (B/A) x + (C/A))
= A æ
è 		æ
è 		x + B
2A
 ö
ø 		2

 
 + æ
è 		C
A
 - B2
4A2
 ö
ø 		ö
ø 		.
It is apparent that the vertex of the parabola is located at
x
= - B
2A
 ,
y
= A æ
è 		C
A
 - B2
4A2
 ö
ø
= 4AC - B2
4A
 .
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,
R
= - 802
-4
 = 1600 ( hundred dollars) = $160,000.


File translated from TEX by TTH, version 3.79.
On 08 Jan 2008, 14:39.