Application Of Extremum to
Economics
The Demand Function:
The demand function is the function relating
the price p(x) of a unit of a product to the
number of units x produced
The Cost Function:
The cost function is the function relating the
total cost C(x) of producing an x number
of units of a product to that number x
The Revenue Function
The revenue R(x) function is the product of
the number of units produced x and the
price p(x) of a unit.
Thus:
R(x) = x p(x)
The Gain (Profit) Function
The Profit P(x) function is the difference between the
revenue function R(x) and the total cost function C(x)
Thus
P(x) = R(x) – C(x)
= x p(x) – C(x)
Maximizing the Profit
To find the number of units x of the product that
should be produced to attain the greatest profit,
We inestigate the local extremun for the profit
function P(x), by finding the critical point(points) for
that function This will be the point(points) at
which the first derivative P′(x) of P(x) either zero or
nonexistent.
Checking for Maximum
If x0 is a number at which P′(x) = 0 , while
P′′ (x) is negative, then x0 is a point of local
maximum.
To check whether this is a point of absolute
maximum, we have to consider the the other
values of the function over its given domain.
Examples
A factory director discovered that the total
cost C(x) of producing x units of extremely
sophisticated PC’s is:
C(x) = 0.5x2 + 4000x + 48000000
And that the price p(x) of the unit is subject
to the demand equation:
p(x) = -0.5x + 18000
Questions
Find
The number of units that should be produced for
the factory to obtain maximum profit.
The maximum profit obtained.
The price of the unit.
Solutions
We have:
P(x) = R(x) – C(x)
= xp(x) – C(x)
= x (-0.5x + 18000) – (0.5x2 + 4000x +
48000000)
= - x2 + 14000x - 48000000
The maximum Profit
P(x) = - x2 + 14000x – 48000000
→P′(x) = - 2x + 14000
= - 2 ( x – 7000)
Letting P′(x) = 0 , we get: x = 7000
Thus x = 7000 is a critical point
We also have:
P′′ (x) = - 2
→ P′′ (7000) = - 2 < 0
Thus x=7000 is a point of local maximum
P(x) = - x2 + 14000x – 48000000
At x=7000, we have:
The profit:
P(7000) = - (7000)2 + 14000(7000) – 48000000
= ( - 49 + 98 - 48). 106
= ( 98 – 97 ). 106
= 106
The Price
p(x) = -0.5x + 18000
At x=7000, we have:
The price:
p(7000) = - 0.5(7000) + 18000
= - 3500 + 18000
= 14500
Graphing P(x)
P(x) = - x2 + 14000x – 48000000
P(x) intersects the x-axis at
X = 6000 and x = 8000
Why?
P(x) =0
iff - x2 + 14000x – 48000000 = 0
Iff x2 - 14000x + 48000000 = 0
Iff ( x – 6000) ( x – 8000 ) = 0
Iff x = 6000 0r x = 8000
The Graph of P(x)
• Graph of P(x)
• Graph of P′(x)
• Graph of P′′ (x)
Graph of P
P(x) = - x2 + 14000x – 48000000
= - ( x – 6000) ( x – 8000
P
106
6000
- 48
.106
7000
8000
x
Graph of the derivative P′ of P
• P′(x) = - 2x + 14000
•
= - 2 ( x – 7000)
p′
14000
x
7000
Graph of the Second Derivative P′′ of P
P′′ (x) = - 2
P′′
x
-2
Comparing P , P′ and P′′