Application of

derivatives

Presented by;

Jihad Khaled BecettiKariman MahmoudMalak AbbaraFatma HusseinAmna Al-SayedWadha Al mohannadi

The definition of derivatives.

The history of derivatives.

The demand function.

The cost function.

The revenue function.

The profit function.

The content

The derivatives• In calculus, the derivative is a

measurement of how a function changes when the values of its inputs change.

• In finance, the derivative is a financial instrument that is derived from an underlying asset's value.

The history of derivatives

• The ancient period introduced some of the ideas of integral calculus.

• In the medieval period, the Indian mathematician Aryabhata used the notion of infinitesimals and expressed an astronomical problem in the form of a basic differential equation. This equation eventually led Bhāskara II in the 12th century to develop an early derivative.

The demand function

Definition;• A demand function is a fundamental relationship

between a dependent variable (i.e., quantity demanded) and various independent variables (i.e., factors which are supposed to influence quantity demanded)

www.classwork.busadm.mu.edu

The Moll cinema obtains 750 viewers at 30QR in the regular days, and obtains 500 Viewers at price 38QR in special occasions.

The MOLL Cinema

Find1- The demand function

REMEMBER• Law of demand:

( The quantity of a good demanded in a given time period increases as it’s prices falls, and visa versa)

We could conclude that ;

A- The two points (750,30) (500,38)

B- By finding the slope; M= 38-30\500-750= -0.032

C- The equation of the line;P(X)-30= -0.032(x-750)

P(X)=-0.032 .(X-750)+30

So, The demand function is

P (x)= -0.032x+24+30

P (x)= -0.032x+54

The Cost Function

The cost function is a function of input prices and output quantity. Its value is the cost of making that output given those input prices.

C(x)= p(x) • x

The cost of producing 100 units of good in is 500,000 QR, what is the total cost to produce this amount of output?

Example

C(x)= p(x) • x

= ( x • p ) • x

= ( 100 • 500,000 ) • 100

= 50000000 QR

Solution

The Revenue Function

• Revenue in economics means:

• Amount received or to be received from customers for sales of products or services.

R(x)=x.p(x)

• If R(x) is the revenue received from the sale of x units as some commodity then the derivative R is called the managerial revenue.

• Economists use this to measure the rate of increase in revenue per unit increase in sale.

• The demand equation of CASIO Company is:

• P(x) = 5- 1/3 x

Find the revenue:

R(x)=x.p(x)

= x (5 – 1/3 x)

R(x) = 5x – 1/3x 2

Casio company

Is the difference between the revenue function R(x) and the total cost function C(x)

• P(x) = R(x) – C(x)

= x p(x) – C(x)

The Profit Function

• If we Know that the production cost of a chocolate company is = 2x2 + 8000 + 1200000 ,and the price = -2x + 16000.

Find The maximum profit, and number of units that should be produced for the factory to obtain maximum profit. Then The price of each unit.

Chocolate company

• P(x)=R(x)-C(x)

• = xp(x) – C(x)

• = x(-2x + 16000) – (2x2 + 8000 + 1200000)

• = -4x2 + 16000x – 1208000

The solution

Maximizing ProfitIf 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 other

values of the function over its given domain.

Maximizing Profit• If P' (x) =0 ,and P' (x) < 0

We will get the maximum profit.

• P(x) = = -4x2 + 16000x – 1208000

• → P'(x)= -8x + 16000

• P'(x)= -8 (x-2000)

• P'(x)= 0 if x = 2000

• Letting P′(x) = 0 , we get: x = 2000

• Thus x = 2000 is a critical point

We also have:

P′′ (x) = - 8

→ P′′ (2000) = - 8 < 0

Thus x=2000 is a point of local maximum

P(x) = -4x2 + 16000x – 1208000

At x=2000, we have:

The profit:

P(2000) = - (2000)2 + 16000 (2000) – 1208000

= 26792000

The profit

The Price• p(x) = -2x + 16000

At x=2000, we have:

The price:

p(2000) = - 2(2000) + 16000

= - 4000 + 16000

= 12000

Graphing P(x)P(x) = -4x2 + 16000x – 1208000

P(x) intersects the x-axis at

X = 76.98 and x = 3923.018