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Application of derivatives to Business and economics

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# Application of derivatives to Business and economics - PowerPoint PPT Presentation

Application of derivatives to Business and economics. Presented by: Amal Al-Kuwari Bashayer Noof Hind Nader. Presentation content:. Introduction to Application of derivatives and it’s importance in the Business field The demand function The cost function The revenue function

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### Application of derivatives to Business and economics

Presented by:

Amal Al-Kuwari

Bashayer

Noof

Hind Nader

Presentation content:
• Introduction to Application of derivatives and it’s importance in the Business field
• The demand function
• The cost function
• The revenue function
• The profit function
• Examples
The demand function

The demand function is the function that relates the price p(x) of a specific unit of a product to the number of units x produced.

p(x) is also called the demand function.

p(x)= p * x

The cost Function

The cost function is the function that relates the total cost C(x) of producing an x number of units of a product to that number x

• C(x) = p(x) * x
A general Example:

A football match takes place in a stadium that can take 60,000 people. At first the tickets cost QR30 , it was expected that around 30,000 people would come , then when the tickets value were lowered to QR20 the audience attendance was expected to rise to 5000.

• Find the demand function assuming that this is a linear Function?
Solution
• First we find the equation of the straight line through the following points:

(30000,30) ,(35000,20)

• Secondly we find the slope:

m = (20-30)/(35,000 - 30,000)

= -10/5000

= -1/500

So the equation of this line is:

P(x) - 20 =(-1/500).(x - 35,000)

P(x)=-1/500.(x - 35,000)+20

So the demand function is :

P(x)= -1/500x+70+20

= -1/500x+90

Total cost example:-

If BMW company wants to produce 10 cars and each car cost 200,000 QR, find total cost gain from selling this amount of cars:-

• The solution:-
• C(x) = P(x) * x
• C(x) = (10*200,000) * 200,000
• C(x) = 40,000,000,000 QR
The revenue function

The revenue function which represented in R(x) is simply the product of the number of units produced x by the price p(x) of the unit.

So the formula is:

R(x)=x . p(x)

Example
• The demand equation of a certain product is p=6-1/2x dollars.
• Find the revenue:

R(x)= x.p

=x(6-1/2x)

=6x-1/2x2

The Profit Function

The profit function is nothing but the revenue function minus the cost function

So the formula is :

P(x)= R(x)-C(x)

→x.p(x)-C(x)

Notice that :

A maximum profit is reached when:

• the first derivative of P(x) when P‘(x) is zero or doesn’t exist

And

2. The second derivative of P(x) is always negative P″(x)<0

*when P‘(x)=0 this is the critical point, and since P″(x)<0 the parabola will be concave downward

Another important notice is:

Since P(x)= R(x)- C(x)

Then a maximum profit can be reached when:

R'(x)-C'(x)=0 and R″ (x)-C″(x)<0

Example
• A factory that produces i-pods computed its demand, and costs and came to realize that the formulas are as follows:
• p=100-0.01x
• C(x)=50x+10000
Find:
• The number of units that should be produced for the factory to obtain maximum profit.
• The price of the unit.
R(x)=x.p

=x.(100-0.01)

=100x -0.01x2

Now:

P(x)=R(x)-C(x)

=100x-0.01x2 –(50x+10000)

=-0.01x2 + 50x-10000

Now the maximum point in the parabola will occur When P'(x)=0 ; P"(x)<0

P(x) =-0.01x2 + 50x-10000

P'(x)=-0.02x+50

→ -0.02x+50=0

→ -0.02x=-50

→ x=2500

P'(2500)= -50+50=0

So the P'(X)=0 When x=2500

- Now we will find the second derivative:

P″(x)= -0.02

P″(2500)= -0.02<0

So x=2500 is at a local maximum

Finally to find the price that is needed to be charged per unit we return to the demand function:

P=100-0.01(2500)

= 100-25= \$75