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 hundreddollars.
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 ±
√
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
⎛ ⎝
⎛ ⎝
x +
B
2A
⎞ ⎠
2
+
⎛ ⎝
C
A
−
B2
4A2
⎞ ⎠
⎞ ⎠
.
It is apparent that the vertex of the parabola is located at
x
= −
B
2A
,
y
= A
⎛ ⎝
C
A
−
B2
4A2
⎞ ⎠
=
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 ( hundredunits),
p
= (80 − 40) dollars,
R
= −
802
−4
= 1600 ( hundreddollars) = $160,000.
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