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 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 is the function that relates the total cost C(x) of producing an x number of units of a product to that number x
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.
m = (20-30)/(35,000 - 30,000)
P(x) - 20 =(-1/500).(x - 35,000)
P(x)=-1/500.(x - 35,000)+20
So the demand function is :
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 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)
The profit function is nothing but the revenue function minus the cost function
So the formula is :
A maximum profit is reached when:
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
=-0.01x2 + 50x-10000
P(x) =-0.01x2 + 50x-10000
So the P'(X)=0 When x=2500
- Now we will find the second derivative:
So x=2500 is at a local maximum
= 100-25= $75