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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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