1. Application Of Extremum to Economics

2. 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

3. 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

4. The Revenue Function R(x) = x p(x)
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)

5. 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)

6. 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.

7. 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.

8. Examples
A factory director discovered that the total cost C(x) of producing x units of extremely sophisticated PC’s is:
C(x) = 0.5x x
And that the price p(x) of the unit is subject to the demand equation:
p(x) = -0.5x

9. 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.

10. Solutions We have: P(x) = R(x) – C(x) = xp(x) – C(x)
= x (-0.5x ) – (0.5x x )
= - x x

11. The maximum Profit P(x) = - x2 + 14000x – 48000000
Letting P′(x) = 0 , we get: x = 7000
Thus x = is a critical point
We also have:
P′′ (x) = - 2
→ P′′ (7000) = - 2 < 0
Thus x=7000 is a point of local maximum

12. P(x) = - x x –
At x=7000, we have:
The profit:
P(7000) = - (7000) (7000) –
= ( ). 106
= ( 98 – 97 ). 106
= 106

13. The Price p(x) = -0.5x + 18000 At x=7000, we have: The price:=
= 14500

14. 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 - x x – = 0
Iff x x = 0
Iff ( x – 6000) ( x – 8000 ) = 0
Iff x = r x = 8000

15. The Graph of P(x)
Graph of P(x)
Graph of P′(x)
Graph of P′′ (x)

16. Graph of P P(x) = - x2 + 14000x – 48000000 = - ( x – 6000) ( x – 8000
7000
8000
106 P x

17. Graph of the derivative P′ of P
P′(x) = - 2x
= - 2 ( x – 7000)
14000
7000 p′ x

18. Graph of the Second Derivative P′′ of P
P′′ (x) = - 2 x
P′′
- 2

19. Comparing P , P′ and P′′