1 / 29

Application of derivatives

Application of derivatives. Presented by; Jihad Khaled Becetti Kariman Mahmoud Malak Abbara Fatma Hussein Amna Al-Sayed Wadha Al mohannadi. The content. The definition of derivatives. The history of derivatives. The demand function. The cost function. The revenue function.

shad-gonzales
Download Presentation

Application of derivatives

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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Application of derivatives

  2. Presented by; Jihad Khaled Becetti Kariman Mahmoud Malak Abbara Fatma Hussein Amna Al-Sayed Wadha Al mohannadi

  3. The content • The definition of derivatives. • The history of derivatives. • The demand function. • The cost function. • The revenue function. • The profit function.

  4. The derivatives • In calculus, the derivative is a measurement of how a function changes when the values of its inputs change. • In finance, the derivative is a financial instrument that is derived from an underlying asset's value.

  5. The history of derivatives • The ancient period introduced some of the ideas of integral calculus. • In the medieval period, the Indian mathematician Aryabhata used the notion of infinitesimals and expressed an astronomical problem in the form of a basic differential equation. This equation eventually led Bhāskara II in the 12th century to develop an early derivative.

  6. The demand function Definition; • A demand function is a fundamental relationship between a dependent variable (i.e., quantity demanded) and various independent variables (i.e., factors which are supposed to influence quantity demanded) www.classwork.busadm.mu.edu

  7. The MOLL Cinema The Moll cinema obtains 750 viewers at 30QR in the regular days, and obtains 500 Viewers at price 38QR in special occasions.

  8. Find 1- The demand function

  9. REMEMBER • Law of demand: ( The quantity of a good demanded in a given time period increases as it’s prices falls, and visa versa)

  10. A- The two points (750,30) (500,38) We could conclude that ; B- By finding the slope; M= 38-30\500-750= -0.032 C- The equation of the line; P(X)-30= -0.032(x-750) P(X)=-0.032 .(X-750)+30

  11. So, The demand function is P (x)= -0.032x+24+30 P (x)= -0.032x+54

  12. The Cost Function The cost function is a function of input prices and output quantity. Its value is the cost of making that output given those input prices. C(x)= p(x) • x

  13. Example The cost of producing 100 units of good in is 500,000 QR, what is the total cost to produce this amount of output?

  14. Solution C(x)= p(x) • x = ( x • p ) • x = ( 100 • 500,000 ) • 100 = 50000000 QR

  15. TheRevenue Function • Revenue in economics means: • Amount received or to be received from customers for sales of products or services. R(x)=x.p(x)

  16. If R(x) is the revenue received from the sale of x units as some commodity then the derivative R is called the managerial revenue. • Economists use this to measure the rate of increase in revenue per unit increase in sale.

  17. Casio company • The demand equation of CASIO Company is: • P(x) = 5- 1/3 x Find the revenue: R(x)=x.p(x) = x (5 – 1/3 x) R(x) = 5x – 1/3x 2

  18. The Profit Function Is the difference between the revenue function R(x) and the total cost function C(x) • P(x) = R(x) – C(x) = x p(x)– C(x)

  19. Chocolate company • If we Know that the production cost of a chocolate company is = 2x2 + 8000 + 1200000 ,and the price = -2x + 16000. Find The maximum profit, and number of units that should be produced for the factory to obtain maximum profit. Then The price of each unit.

  20. The solution • P(x)=R(x)-C(x) • = xp(x) – C(x) • = x(-2x + 16000) – (2x2 + 8000 + 1200000) • = -4x2 + 16000x – 1208000

  21. Maximizing Profit If x0 is a number at which P′(x)= 0 , while P′′ (x) is negative, then x0 is a point of local maximum. To check whether this is a point of absolute maximum, we have to consider the other values of the function over its given domain.

  22. Maximizing Profit • If P' (x) =0 ,and P' (x) < 0 We will get the maximum profit. • P(x) = = -4x2 + 16000x – 1208000 • → P'(x)= -8x + 16000 • P'(x)= -8 (x-2000) • P'(x)= 0 if x = 2000

  23. Letting P′(x) = 0 , we get: x = 2000 • Thus x = 2000 is a critical point We also have: P′′ (x) = - 8 → P′′ (2000) = - 8 < 0 Thus x=2000 is a point of local maximum

  24. The profit P(x) = -4x2 + 16000x – 1208000 At x=2000, we have: The profit: P(2000) = - (2000)2 + 16000 (2000) – 1208000 = 26792000

  25. The Price • p(x) = -2x + 16000 At x=2000, we have: The price: p(2000) = - 2(2000) + 16000 = - 4000 + 16000 = 12000

  26. Graphing P(x) P(x) = -4x2 + 16000x – 1208000 P(x) intersects the x-axis at X = 76.98 and x = 3923.018

  27. Thank you for your attention

More Related