application of derivatives to business and economics n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Application of derivatives to Business and economics PowerPoint Presentation
Download Presentation
Application of derivatives to Business and economics

Loading in 2 Seconds...

play fullscreen
1 / 19

Application of derivatives to Business and economics - PowerPoint PPT Presentation


  • 231 Views
  • Uploaded on

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

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Application of derivatives to Business and economics' - prentice


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
application of derivatives to business and economics

Application of derivatives to Business and economics

Presented by:

Amal Al-Kuwari

Bashayer

Noof

Hind Nader

presentation content
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

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

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 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
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

slide7
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

slide8
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

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
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

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)

slide12
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

example1
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
slide14
Find:
  • The number of units that should be produced for the factory to obtain maximum profit.
  • The price of the unit.
slide15
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

slide16
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

slide18
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