CALCULUS FOR TECHNOLOGY (BETU 1023)

Transcription

1 CALCULUS FOR TECHNOLOGY (BETU 103) WEEK 7 APPLICATIONS OF DIFFERENTIATION 1 KHAIRUM BIN HAMZAH, IRIANTO, 3 ABDUL LATIFF BIN MD AHOOD, 4 MOHD FARIDUDDIN BIN MUKHTAR 1 khaium@utem.edu.my, iianto@utem.edu.my, 3 latiff@utem.edu.my, 4 faiduddin@utem.edu.my

2 OPTIMIZATION PROBLEMS TABLE OF CONTENTS

3 LEARNING OUTCOMES At the end of this topic, students should be able to: Undestand the definition of optimization Distinguish between maimum o minimum poblem Solve application poblems using absolute value method Solve application poblems using second deivative test

4 OPTIMIZATION?? The pocess of finding maimum o minimum values is called optimization. In optimization poblems we ae looking fo the lagest (maimum) value o smallest (minimum) value. In ode to detemine the function is maimum o minimum we can use Absolute Value o Second Deivative Test.

5 STEPS IN SOLVING OPTIMIZATION PROBLEMS Undestand the poblem Daw a diagam (if needed) Intoduce notation Epess the equation with one unknown (vaiable) only which is the equation to be maimized o minimized Used the suitable method to find maimum o minimum value Method of absolute value Second deivative test

6 Method of Absolute Value?? Find the values of equation at the endpoints of the inteval (only applied fo closed inteval). Find the values of equation at the citical numbes of equation whee: Set up the fist deivative of equation is equal to zeo Maimum value when the value in step 1 and above give the lagest value. Minimum value when the value in step 1 and step above give the smallest value.

7 Second Deivative Test?? Find the values of equation at the citical numbes of equation whee: Set up the fist deivative of equation is equal to zeo Maimum value when the second deivative fo the citical numbes is negative value. Minimum value when the second deivative fo the citical numbes is positive value.

8 EXAMPLE 1 We need to enclose a field with a fence. We have 500 feet of fencing mateial and a building is on one side of the field and so won t need any fencing. Detemine the dimensions of the field that will enclose the lagest aea?

9 SOLUTION 1 To find the dimensions (length and width) of the field that give maimum aea. Let = length of the field y = width of the field

10 SOLUTION 1 Epess the equation of aea since we want to find the dimensions of maimum aea. Aea = length width A 500 A y 500 y y y 500 y

11 SOLUTION 1 Solve using the method of absolute value Note that y 0 and y 50, so the function we wish to maimize is A y y500 y, 0 y 50 The deivative is A' y 500 4y A'( y) 0 y 15 METHOD OF ABSOLUTE VALUE

12 SOLUTION 1 The maimum aea is A A A , maimum Theefoe the dimensions of the field that give maimum aea is width, length, y 15 ft 50 ft

13 SOLUTION 1 Solve using the second deivative test The second deivative is da dy da dy d A dy 500 4y 0 y ma Theefoe the dimensions of the field that give maimum aea is width, length, y ft ft SECOND DERIVATIVE TEST

14 EXAMPLE We want to constuct a bo with a squae base and we only have 10 m of mateial to use in constuction of the bo. Assuming that all mateial is used in the constuction pocess, detemine the maimum volume that the bo can have.

15 SOLUTION To find the maimum volume of the bo. Let = length of the bo with squae base y = height of the bo

16 SOLUTION Epess the equation of volume since we want to find the maimum volume. Volume = aea of the base height V 10 A 10 y 4y 5 y 5 V

17 SOLUTION Solve using the method of absolute value Note that 0, so the function we wish to maimize is V 5 The deivative is dv d dv d , 3 0 METHOD OF ABSOLUTE VALUE

18 SOLUTION Theefoe the maimum volume is V V V y m 3

19 SOLUTION Solve using the second deivative test The second deivative is dv d d V d ma y Theefoe the maimum volume is V V y.1517 m 3 SECOND DERIVATIVE TEST

20 EXAMPLE 3 A manufactuing needs to make an open top cylindical can that will hold 1.5 lite of liquid. Detemine the dimensions of the can that will minimize the amount of mateial used in its constuction.

21 SOLUTION 3 To find the dimensions (adius and height) of the can that give minimum aea. Let = adius of the can h = height of the can

22 SOLUTION 3 Epess the equation of aea since we want to find the minimum aea (amount of mateial used elated with aea). Aea = aea of the side + aea of the base (open top) 1500 A h V h A 1500 ml h

23 SOLUTION 3 Solve using the method of absolute value Note that 0, so the function we wish to minimize is A 3000, The deivative is da 3000 d da 0 d METHOD OF ABSOLUTE VALUE

24 SOLUTION 3 The minimum aea is 3000 A 3000 A A Theefoe the dimensions of the can that give minimum aea is adius, height, h

25 SOLUTION 3 Solve using the second deivative test The second deivative is da 3000 d da 0 d d A d min SECOND DERIVATIVE TEST Theefoe the dimensions of the can that give minimum aea is adius, height, h

26 EXAMPLE 4 A window fame consists of a ectangle of height hmetes sumounted by a semicicle of adius metes as shown in the diagam below. If the peimete of the fame is constant at 10 metes, find the value of fo which the aea of the fame is a maimum. h h

27 Epess the equation of aea since we want to find the maimum aea Aea = aea of the ectangle + aea of the semicicle SOLUTION Peimete A A h A h h

28 SOLUTION 4 Solve using the method of absolute value Note that 0, so the function we wish to minimize is A 10 The deivative is A 10 da d da 0 d , 0 METHOD OF ABSOLUTE VALUE

29 SOLUTION 4 Solve using the second deivative test The second deivative is da d d A d ma SECOND DERIVATIVE TEST Theefoe the adius that give maimum aea is adius, 1.400

30 EXAMPLE 5 A sheet of cadboad measues 0 cm by 1 cm. Fou little squaes of side lengths cm ae cut out fom the cones of the cadboad and the emainde is tuned up to fom an open ectangula bo. Find the value of fo which the volume of the bo is a maimum.

31 SOLUTION 5 To find the value of that give maimum volume. ( 0 )cm ( 1 )cm

32 SOLUTION 5 Epess the equation of volume since we want to find the maimum volume Volume = aea of the base height V (0 )(1 ) ( )

33 SOLUTION 5 Solve using the method of absolute value Note that, so the function we wish to minimize is V 0 The deivative is dv d dv d ( 0 )(1 ), ,.43 METHOD OF ABSOLUTE VALUE Fo pactical measues.43 because the longest side value is 0 cm.

34 SOLUTION 5 Solve using the second deivative test The second deivative is dv d d V 18 4 d Fo.43 d V ma d Fo.77 d V d Theefoe the value of.43give maimum volume SECOND DERIVATIVE TEST

35 TRY IT YOURSELF 1 A window is being built and the bottom is a ectangle and the top is a semicicle. If the window has a peimete 1 metes, what must the dimensions of window so that the aea is maimize? Solution

36 TRY IT YOURSELF A pinte need to make a poste that will have a total aea of 00 cm and will have 1 cm magins on the sides, cm magin on the top and 1.5 cm magin on the bottom. What dimensions will give the lagest pinted aea? Solution

37 SUMMARY Undestand the definition of optimization Distinguish between maimum o minimum poblem Solve application poblems using absolute value method Solve application poblems using second deivative test

38 REFERENCES James, S. (01). Calculus (7 th ed.). Cengage Leaning. Bivens, I.C., Stephen, D., & Howad, A. (01). Calculus Ealy Tanscedentals (10 th ed.). John Willey & Sons Inc.