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Application of
derivatives
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Presented by;
Jihad Khaled BecettiKariman MahmoudMalak AbbaraFatma HusseinAmna Al-SayedWadha Al mohannadi
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The definition of derivatives.
The history of derivatives.
The demand function.
The cost function.
The revenue function.
The profit function.
The content
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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.
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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.
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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
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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
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Find1- The demand function
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REMEMBER• Law of demand:
( The quantity of a good demanded in a given time period increases as it’s prices falls, and visa versa)
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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
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So, The demand function is
P (x)= -0.032x+24+30
P (x)= -0.032x+54
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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
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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
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C(x)= p(x) • x
= ( x • p ) • x
= ( 100 • 500,000 ) • 100
= 50000000 QR
Solution
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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)
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• 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.
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• 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
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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
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• 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
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• P(x)=R(x)-C(x)
• = xp(x) – C(x)
• = x(-2x + 16000) – (2x2 + 8000 + 1200000)
• = -4x2 + 16000x – 1208000
The solution
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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.
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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
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• 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
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P(x) = -4x2 + 16000x – 1208000
At x=2000, we have:
The profit:
P(2000) = - (2000)2 + 16000 (2000) – 1208000
= 26792000
The profit
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The Price• p(x) = -2x + 16000
At x=2000, we have:
The price:
p(2000) = - 2(2000) + 16000
= - 4000 + 16000
= 12000
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Graphing P(x)P(x) = -4x2 + 16000x – 1208000
P(x) intersects the x-axis at
X = 76.98 and x = 3923.018
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