Instantaneous rate of change (IRC) at the point x Slope of tangent

Transcription

1 CHAPTER 2: Differentiation Do not study Sections 2.1 to Rates of change Rate of change (RC) = Two types Average rate of change (ARC) over the interval [, ] Slope of the line segment Instantaneous rate of change (IRC) at the point x Slope of tangent 1

2 Example of ARC The supply of a certain product (in 1000) is given by the following function ==6 + Calculate the ARC over the interval [5,10]. = = OR = = where 10 = and 5 = 2

3 2.5 The derivative of a function The derivative of a function can be found by differentiation. Example of IRC = then =2. Notation: 3

4 2.6 Rules of differentiation Rule 1 If =, where k is a constant, then =0. 1.=5 then = 2.=2ln 5 =3.219 then = 3.= = then = 4.= =1 then = 4

5 Rule 2 If =, where n is a real number and 0, then =. 1.= then = 2.= then = 3.= then = 4. = then = 5

6 5.= = then = Rule 3 If =, where k is a constant, then =. 1.=2 then = 2.= then = 3.h = 7 = 7 then h = 6

7 Rule 4 If =+h then = +h If = h then = h 1.=3+ then = 2. = Calculate and. First we calculate = In order to calculate we first rewrite r as: = 2 +2 then = 7

8 Example of Rules 1 to 4: The price of a product (in Rand) depends on the quantity of the product sold and is given by = Calculate the sales price if 40 items are sold. 2. Calculate the marginal income of the 40 items. Answers: 1.For =40 we find = = 2.The income function is == 8

9 The marginal income function: The marginal income for the 40 items is Rule 5 (Product Rule) If = h, then =h + h. 1.= 6 then = 2.= then = 9

10 3.h = + =5 + then h = Rule 6 (Quotient Rule) If =, then it follows that =. [] 1.= then = 2.= then = 10

11 3.Let =. We could calculate by rewriting r as = and we get = 3 =. This same question could also be answered using the quotient rule: = = = Rule 7 (Chain Rule) If = then =. 1.= then = 2. = then = 11

12 3. =. In this form we can find using the quotient rule. If we rewrite k we can find using the chain rule. Re-writing k we find, 1 = + +1 = + +1 = 4. = +5= +5 = 12

13 Example of the quotient rule The profit of tea produced is given by = with x the amount in 100 kg and the profit in R Calculate the profit if 1500 kg of tea is produced 2. Calculate the marginal profit function 13

14 3. Calculate the marginal profit if: a) 1500 kg of tea is produced b) kg of tea is produced 14

15 Example Calculate the equation of the tangent line of = at =3. Answer The equation of a line can be obtained using the equation = +. 15

16 2.7 Inverse functions and their derivatives Consider the following function =3 +2 The inverse function of = is: = = 2 = (Inverse function) == (Inverse function) 16

17 2.8 The derivatives of special functions A. Exponential functions Rules 9 and 10 If = then = If = then = 1. = then = 2.= then = 3.= then = 17

18 4.= then = Exponential function: Example 1 The demand of a product is given by == 980. where d is the demand in thousands and p is the price of the product in Rand 1. Calculate the demand if the price is R2 2. The marginal demand function is: 18

19 3. Calculate the marginal demand if the price is R3 Exponential function: Example 2 The demand (in 100) for a CD-player in terms of price (in R1000) is given by =. for p > 0. 1.Calculate the demand for a CD-player for a price of R The units in which income will be measured is:. 19

20 B. Logarithmic functions Rules 11 and 12 If =ln then = If =ln then = = 1. = 5 then = 2.= then = 3.= then = 4.=2 +1 then = 20

21 2.9 Higher derivatives Calculate the fourth order derivative of = = = =24 18 =24 =0 Calculate the second order derivative of = : = 6 = 21

22 2.10 Optimization problems Need to find max/min values Figure Absolute max: Relative max: Absolute min: Relative min: All max / min values are extreme values Critical values x values that might indicate extreme values Consider =0. How do we test if k will lead to a relative min or relative max 22

23 Calculate : <0 then k leads to a relative max value >0 then k leads to a relative min value =0 examine the values of the function at = inflection point 23

24 Figure : Rel max in the point k Figure Rel min in the point k 24

25 Example of optimization with one critical value Calculate the extreme and critical value(s) for: =16 = = Critical value(s): Set =0 Therefore, = is a critical value. Type of extreme value: 8 = 2<0 =8 leads to a relative maximum. Extreme value: 8 = 25

26 Example of optimization with two critical values Calculate the extreme and critical values for: =3 4 = = Critical values: Set =0 Therefore, = and = are the critical values. Type of extreme values: 1 = =12>0. =1 leads to a relative min. 0 = =0. =0 leads to an inflection point. 26

27 How do we know that this leads to an inflection point? We examine the values of the function at =0. x ) 3(0) 4( 0.1) 4(0) = =0 =3 4 3(0.1) 4(0.1) = Graphical representation of an inflection point: Extreme value: (1)= 27

28 Homework (work through this example on your own) The cost (in Rand) to manufacture x products: ( )= The income (in Rand) if x products are sold: ( )= How many products should be sold if we want to maximize the profit? ( )= ( ) ( ) ( )= ( ) = ( )= ( )= 0.1 Critical value(s): Set ( )= =0 Therefore, =500 is a critical value. 28

29 Type of extreme value: (500)= 0.1<0 =500 leads to a relative maximum. Hence, to earn the max profit we need to sell 500 products. Calculate the maximum profit. (500)= 0.05(500) +50(500) 1 500=

30 Example of optimization with one critical value A manufacturer produces garden chairs at a cost of 20 a chair, and his overhead cost is a week. From previous experience he knows that he will sell chairs a week if he charge a chair. What must the price be, and how many chairs must he sell a week, to maximize his profit? Given: Cost per chair: R20 Overhead cost: R3000 Number of chairs: x Sales price: Rx 30

31 Answer: Profit per chair: x 20 Total profit ( )= ( )( 20) 3000 ( )= ( )= # of chairs Profit Overhead cost per chair Critical value(s): Set ( )=0 Therefore, = is a critical value. Type of extreme value: (35)= 80<0 =35 leads to a relative maximum. Number of chairs to be sold: 31

32 Example on page 97 (work through this example on your own). Know what the terms gross profit and nett profit mean. Questions 1 to 3 are based on the following information: An analysis of the financial statements of a coal mine indicates that when x tons of coal are extracted per day, the income and cost (in Rands) of the mine are, respectively: ( )= and ( )= The mine is taxed at a rate of 40% on its gross profit. 32

33 Question 1: Determine the value of x that maximises the income. Answer 1: ( )= ( )= ( )= 4 Critical value(s): Set ( )= =0 x = Therefore, x = is a critical value. Type of extreme value: (302.5 = 4<0 x = leads to a relative maximum. 33

34 Question 2: Calculate the gross profit and the value of x that maximises it: Answer 2: = = = = = 6 Critical value(s): Set = =0 =202 Therefore, =202 is a critical value. Type of extreme value: 202 = 6<0 =202 leads to a relative maximum. 34

35 Question 3: Calculate the nett profit and the value of x that maximises it: Answer 3: = 0.4 or =0.6 = ) = = =

36 Critical value(s): Set = =0 =202 Therefore, =202 is a critical value. Type of extreme value: 202 = 3.6<0 =202 leads to a relative maximum Partial differentiation Example of partial differentiation Suppose the manufacturer produce x units of product X and y units of product Y. The cost function in Rand =, =

37 Marginal cost function = Rate of change of z with respect to x for y constant = Rate of change of z with respect to y for x constant Interpretation of the marginal cost function at (100, 150) = =29.5 The cost increases with if an extra unit of product X is produced, when there are 150 units of product Y produced. 37

38 = =23 The cost increases with if an extra unit of product Y is produced, when there are 100 units of product X produced. More examples of partial differentiation 1. =3 + = = 38

39 2. =5 +7 = = Interpretation of partial derivatives The total production cost (in R100) of a factory per year, is a function of the number of spare parts produced (in 1000) and the number of workers employed, namely,= Where p is the number of spare parts produced (in 1000) and w is the number of laborers. 39

40 The interpretation of the marginal profit function ; = 1 is as follows: Cost decreases with R100 if spare parts increase from 1000 to 2000 and 5 workers are used. Higher partial derivatives Let =, and consider the second order partial derivatives 1. = 2. = 3. = Note: = 40

41 Example of higher partial derivatives Calculate the second order partial derivatives of = = = = = = = 41

42 Application to problems of optimization Steps in calculating extreme values of,: 1. Determine the critical point(s) of, A point, is a critical point when = and = is the solution when the two equations are solved simultaneous =0 and =0. 2. Determine the extreme values for the critical point, = = = = = 42

43 If >0 and <0, <0 then rel max >0, >0 then rel min If <0 then saddle point (no max or min) If =0 can t say anything Example of optimization using higher partial derivatives Calculate the critical points and extreme values of the following function,= Critical points: = = 43

44 Set =0 and = = =0 Typically, in order to use Cramer s Rule, we rewrite these equations so that the constants are on the right-hand side. 4 2 = = 10 Cramer s rule: = with = , = 16 and = 10. =, = and = = Therefore, the critical point is ( ). 44

45 Investigating the extreme values Recall that = and = = = = = = = = = = >0, >0, >0 and consequently (, ) reaches a at the point (3, -2) Extreme value: (3, 2)=2(3) 2(3)( 2)+( 2) 16(3)+ 10( 2)+12= 45

46 Example of optimization using higher partial derivatives Calculate the critical points and extreme values of the following function Critical points: = = (, )= Set =0 and = = =0 Note: You can t use Cramer s Rule! 46

47 2 +2 =0 simplifies to Substitute = into =0: Therefore, the critical points are (0, 0) and,. Investigating the extreme values: = =2 +2 = = = = = = = 47

48 = = For (0, 0) =96(0) 16= For, =96 16= =48 6= = Hence, is a relative Extreme value:, = = 48

49 Example of optimization using higher partial derivatives A manufacturer produces t-shirts from 2 types of material Type 1 If x shirts are produced the sales price is (80 3 ) rand per shirt Type 2 If y shirts are produced the sales price is (60 2 ) rand per shirt Given The cost function if x shirts of type 1 and y shirts of type 2 are produced is (, )= Calculate How many shirts should be produced that maximize the profit function Income function 49

50 (, )= Profit function (, )= (, ) (, ) Critical points: = = Set =0 and = = =52 Using Cramer s Rule we obtain the critical point: 50

51 Extreme values = = = = = = = = = = = >0, <0 and <0 Conclusion o Profit is maximized if 8 shirts of type 1 and 5 shirts of type 2 are produced o The max profit is then (8,5)= 402 o The sales price is Type 1: 80 3(8)= Rand per shirt Type 2: 60 2(5)= Rand per shirt 51