1. Application of derivatives to Business and economics Presented by: Amal Al-Kuwari Bashayer Noof Hind Nader

2. 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

3. 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

4. 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

5. 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?

6. 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

7. 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

8. 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

9. 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)

10. 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-1/2x 2

11. 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)

12. Notice that : A maximum profit is reached when: 1.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

13. 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

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

15. R(x)=x.p =x.(100-0.01) =100x -0.01x 2 Now: P(x)=R(x)-C(x) =100x-0.01x 2 –(50x+10000) =-0.01x 2 + 50x-10000

16. Now the maximum point in the parabola will occur When P'(x)=0 ; P"(x) < 0 P(x) =-0.01x 2 + 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

17. Graph

18. 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

19. ThankThank you you