+ > for all x> 1, then f ( x) < 0 for the. (1 x) 0. = f between x= 0 and x= 0.5.

Transcription

1 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 Problem Le f ( ) = + ln( + ) a Show ha he graph of f is always concave down b Find he linear approimaion for f ( ) = + ln( + ) near = 0 c Use your linear approimaion from Par b o esimae f (05) d Is your answer o Par c an overesimae or underesimae? Eplain e Find a bound on he error for your approimaion f Use your linear approimaion from Par b o approimae where f ( ) = 0 g Is your answer o Par f an overesimae or underesimae? Eplain Soluion a f ( ) = + so f ( ) = Since + ( + ) enire domain of f + > for all >, hen f ( ) < 0 for he ( ) 0 b f (0) = 0 and f (0) = so he angen line passes hrough (0,0) wih slope Thus he equaion of he angen line is y= c Using f ( ), we ge f (05) d The esimae in Par c is an overesimae because he graph of f is concave down, which means he angen line lies above he graph e Using he error bound formula for a angen line approimaion, Since f ( ) = ( + ) so M = Thus M E (05) < 05 where M ma = f beween = 0 and = 05, is maimum on he relevan inerval is when is smalles ( = 0 ) E (05) < 05 = 8 f We can approimae where f ( ) = 0 by finding where he angen line y= = 0 is zero Specifically, we solve for o obain = 0 g The approimaion = 0 for where f ( ) = 0 is an underesimae Since f ( ) < 0, he graph of f is concave down, so he angen line will be above he graph of f and hus will reach he heigh y= 0 before he acual funcion does Problem Le U and V be posiive consans and F( ) = Ue + Ve a Find all criical poins of F

2 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 b Use he firs derivaive o deermine if your criical poins resul in local minima, local maima, or neiher Soluion a F ( ) = Ue Ve so he criical poins are when Ue Ve = 0 Solving for we ge V e = or = U ln V U b We need o know wheher F ( ) = Ue Ve is posiive or negaive o he lef and righ of he criical poin = ln V We can deermine his by noicing ha as, he U eponenial in he firs erm goes o 0 while he eponenial in he second erm goes o Thus ( ) F Ue Ve = < 0 when < ln V Similarly as, he eponenial in he firs U erm goes o while he eponenial in he second erm goes o 0 Thus F ( ) Ue Ve = > 0 when > ln V This means F is decreasing o he lef of he criical U poin and increasing o he righ, so F has a local minima a = ln V U Problem 3 A warehouse selling cemen has o decide how ofen and in wha quaniies o reorder I is cheaper, on average, o place a smaller number of larger orders On he oher hand, larger orders mean higher sorage coss The warehouse always reorders cemen in he same quaniy, q The oal weekly cos, C, of ordering and sorage is given by where a and b are posiive consans a C= + bq q a Which of he erms a q or bq, represens he ordering cos and which represens he sorage coss? Eplain b Wha value of q gives he minimum oal cos? Soluion a a q represens he weekly cos of ordering since i is a decreasing funcion of q and placing larger orders (larger q) reduces he weekly ordering cos On he oher hand bq is an increasing funcion of q and placing larger orders increases he weekly sorage cos dc a b Firs noe ha he pracical domain of C as a funcion of q is 0< q< = + b which dq q is undefined when q= 0 and zero when a q= Since q= 0 is no in he domain of C, we b

3 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 dc only have one criical poin For values of q near 0, < 0 and for large values of q, dq dc dq a > 0 Thus C has a global minimum a q= b Problem 4 For which posiive number is Hin: Take he log of boh sides implici differeniaion y / larges? Jusify your answer / = so you can move he / ou of he eponen Then use Soluion Using he hin, we firs simplify differeniaion, ln = = So y d dy ln y / = so / ln ln ln y= = Using implici dy ln ln = = This is undefined a d = 0, bu ha is no in he domain of he original funcion So he only criical poin is when he derivaive is zero This happens when ln = 0 which is a = e Problem 5 A recangular swimming pool is o be buil wih an area of 800 square fee The owner wans 5-foo wide decks along eiher side and 0-foo wide decks a he wo ends Find he dimensions of he smalles piece of propery on which he pool can be buil saisfying hese condiions Soluion We skech a generic picure of he configuraion of he pool and decks: Deck 5 f 0 f Pool (Area = 800 f ) y The area of he pools is y= 800 f and we wan o minimize he enire deck+pool area which is A= ( + 0)( y+ 0) So we can rewrie his as a funcion of only as ( ) A( ) = ( + 0) = The pracical domain of his funcion is 0< < Taking he derivaive, we ge A ( ) = which does no eis a = 0 and is zero when = 60 Since = 0 is no in he domain and here are no endpoins, he local erema occurs a = 60 Plug =60 ino =800 f o ge =30 We can verify i is a

4 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 minimum by noing ha if < 60 hen A ( ) < 0 bu if > 60 hen A ( ) > 0 Thus he smalles piece of propery for he pool and deck is 80 f by 40 f Problem 6 The markeing deparmen of compuer gian, Pear, deermines ha he revenue (in billions of dollars) generaed by selling q million unis of heir popular ppods is given by ( ) = 08ln(+ 5 q ) R q while he cos (also in billions of dollars) is deermined by C( q) = q Find he number of ppods ha Pear should produce o maimize heir profi Wha is heir profi in his case? Soluion The profi is The derivaive is P( q) = R( q) C( q) = 08ln(+ 5 q ) (+ 05 q) wih domain 0 q< 08(5 q) (5q 4q+ 05) P ( q) = 05= + 5q + 5q which is zero when 4± 4 4(5)(5) q= = 6± 470 We can find he global maimum by checking he (5) endpoin, q= 0, and he wo criical poins, q= 03 and q= 307 : P (0) = 07, P (03) = 073, and P (307) = 0956 Thus Pear should sell 3,070,000 ppods which will generae a profi of $956,000,000 Problem 7 Gasoline is pouring ino a cylindrical ank of radius 3 fee When he deph of he gasoline is 4 fee, he deph is increasing a 0 f/sec How fas is he volume of gasoline changing a ha insan? Soluion The volume and deph of gasoline in he ank are relaed by he equaion = dv dh h=9 h Treaing V and h as funcions of ime, we differeniae o ge = 9π d d dh dv Since = 0 f/s, we can solve for = 9π 0= 5655 f 3 /s d d Problem 8 Suppose a recangular beam is cu from a cylindrical log of radius 5 cm The srengh of a beam is proporional o he produc of he widh and he square of he heigh Find he widh and heigh of he beam wih maimum srengh ha can be cu from he log Fully jusify your answer Soluion Firs we draw a diagram

5 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 w Since he radius of he log is 5 cm, he diagonal of a beam cu from i will be 30 cm Labeling he widh as w, we can use he Pyhagorean heorem o find he heigh of he beam Then he srengh of he beam is ( ) S( w) kw 900 w 900kw kw 3 = =, wih domain 0< w< 30 The derivaive of his funcion is So S has criical poins when ( ) = S w k kw 3kw = 900 w = 300 k w= 0 3 As w 0 or as w 30, eiher he widh or he heigh of he beam goes o 0, hus S( w) 0 Since S( w ) > 0 for all w in he domain, he criical poin mus correspond o a maimum Thus he widh and heigh of he sronges beam ha can be cu from he log are cm and cm, respecively Problem 9 Represenaive values of he derivaive of a funcion f ( ) are shown in he able below Assume f ( ) is a coninuous funcion and ha he values in he able are represenaive of he behavior of f ( ) f ( ) a Esimae he -value of he global maimum and minimum of f ( ) on he closed inerval [0,3] Jusify your answers based on he daa in he able b Can you ell from hese daa if f ( ) has any inflecion poins? If so, esimae he -value of any inflecion poins and indicae how you know heir locaions If no, eplain why no Soluion

6 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 a We noe ha f ( ) > 0 for < and f ( ) < 0 for > Thus f ( ) has a local maimum a = Furher, because here is only one change of sign in he derivaive, we know ha his is he global maimum The global minimum will occur a one of he endpoins I is no easy o ell a which endpoin his occurs, bu because he negaive slopes are of smaller magniude (for > ) han he posiive slopes (for < ), we epec ha he global minimum occurs a = 0 b We know ha an inflecion poin occurs when f ( ) goes from increasing o decreasing or vice versa We can see from hese daa ha f ( ) is decreasing unil someime beween = and = 5, and increasing hereafer Thus here is an inflecion poin a an somewhere in (, 5) Problem 0 Recall ha arccosine is he inverse of cosine, so ha cos(arccos( )) = d a Differeniae boh sides of his equaion hen solve for arccos Epress your final d answer in a form ha does no depend on any rigonomeric funcions b Use he local linearizaion of arccos near 0 Hin: You will need o use he value of arccos( 0 ) arccos = o approimae ( ) c Is your answer o par b an overesimae or underesimae? Eplain your answer graphically Hin: You will need o describe he shape of he graph of y arccos( ) = Soluion a Differeniaing boh sides of he equaion cos(arccos( )) and solving for he derivaive, we ge = yields d sin(arccos( )) (arccos( )) =, d d (arccos( )) d = sin(arccos( )) We can simplify his by applying he Pyhagorean heorem o a righ-riangle wih angle arccos( ) :

7 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 arccos( ) From his diagram, we can read off he sin(arccos( )) d (arccos( )) = d = Thus b A = 0, he derivaive of arccos( ) is 0 = π Since arccos(0) =, he angen line π passes hrough 0, wih slope : π arccos( ) Evaluaing his a =, π arccos( ) 0708 c Observe arcos = / afer simplificaion Observe ha for 0< < he second derivaive is negaive which means ha he graph of y= arccos( ), is concave down for 0< <, which implies he angen line will lie above he graph in his region Thus our approimaion is an overesimae Problem Consider he funcion f ( ) = e Find and classify all criical poins of his funcion as minima, maima or neiher, and use his informaion o skech he graph of f ( ) Where is f ( ) increasing? Decreasing? On your graph, mark he approimae locaion of he inflecion poins (you do no need o find hem algebraically) How many inflecion poins are here? Does his funcion have an absolue minimum? Absolue maimum? Find he absolue minimum and absolue maimum of f ( ) on he inerval [,3] Where are hey aained? Soluion The domain of f is all real numbers, < <, so any place where he derivaive is zero or undefined is a criical poin The derivaive is

8 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 f ( ) = e e = e ( ) which is coninuous for all Since e is never zero, he criical poins are when = 0, ha is, =± Evaluaing he derivaive a =, = 0, and = will hen give us complee informaion on where f is posiive and where i is negaive (since f is coninuous and can only swich sign a is zeroes): f ( ) = e < 0, f (0) = > 0, and f () = e < 0 Thus f is decreasing on (, ) ( ), and increasing on ( ), a and a local maimum a and has a local minimum Also noe ha since f ( ) 0 as ±, hese are in fac he global minimum and maimum for f Specifically, f ( ) = 0489 and ( ) Using his informaion, we can plo he graph of f: f = 0489 There are hree inflecion poins (where he graph of f changes concaviy) marked on he graph Alhough he problem only asked us o esimae he locaions of he inflecions poins on he graph, we can locae hem precisely from ( ) ( 3 ) = which is zero when = 0 f e 3 and =± Finally, since he inerval [,3] conains he global minimum and maimum for f, hese are also he absolue erema for he inerval Problem The hypoenuse of a righ riangle has one end a he origin and one end on he 3 Curve y= e, wih 0 One of he oher wo sides is on he -ais, he oher side is parallel o he y-ais Find he maimum area of such a riangle A wha -value does i occur?

9 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 Soluion We are looking for he maimum value of he area of a riangle wih base and heigh 3 y= e 3 3 Tha is, we wan o maimize A( ) = e for values of 0 The derivaive is A ( ) = e e = ( ) e, which is zero when = 0 and = The only facor in his epression ha can be negaive is he ( ) facor, so A ( ) is posiive jus o he lef of and negaive jus o he righ Thus A has a maimum a = The maimum area is hen 3 A() = e 0049 y y= e 3 Problem 3 A model for he amoun of an anihisamine in he bloodsream afer a paien akes k a( ) = A e e In his a dose of he drug gives he amoun, a, as a funcion of ime,, o be ( ) equaion, A is a measure of he dose of anihisamine given o he paien, and k is a ransfer rae beween he gasroinesinal rac and he bloodsream A and k are posiive consans, and for pharmaceuicals like anihisamine, k> a Find he locaion = Tm of he non-zero criical poin of a( ) b Eplain why = Tm is a global maimum of a( ) by referring o he epression for a( ) or a ( ) c The funcion a( ) has a single inflecion poin Find he locaion = TI of his inflecion poin You do no need o prove ha his is an inflecion poin d Using your epression for T m from Par a, find he rae a which T m changes as k changes Soluion a The maimum will occur a an endpoin or a a criical poin, when a ( ) = 0 The criical k poins are hus where a A( e ke ) e ( k ) = k, or = T = ln k m ( ) = + = 0 Solving, we have e k ke k =, so ha

10 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 k = > and ha for large, a A( e ke ) b Noe ha a (0) A( k ) 0 ( ) = + < 0 since k> guaranees ha he second eponenial decays much faser han he firs Thus he criical poin mus be a maimum In addiion, because = Tm his is he only criical poin so we know i mus be he global maimum Alernaely, noe ha a (0) = 0 Because k> 0, a( ) 0 for all (he eponenial involving k will decay faser han e ) And for large, a( ) 0 Thus a = Tm, a( ) mus ake on a maimum value, and because i is he only criical poin his mus be he global maimum k c To find inflecion poins, we look for where a A( e k e ) have e = k, so ha ( k ) = T = ln k = ln k I k k ( ) = = 0 Solving for, we Alhough he problem did no ask, we could show ha his is an inflecion poin by a similar argumen o (b): because k> we know ha a (0) < 0 and as increases a ( ) mus evenually become posiive Thus his is an inflecion poin d We have T = ln k Thus m k dt m = ln k+ dk ( k ) k( k ) Problem 4 If C( q) = 00q q+ 8 is he oal cos o produce q lighsabers and R( q) = 48q is he revenue from selling q ligh sabers hen: a Find he average cos o produce q lighsabers b Find he average revenue from selling q lighsabers c Idenify he fied cos and he variable cos d Find he maimum profi e Find and inerpre C (8) in he cone of he problem Soluion a The average cos o produce q lighsabers is b The average revenue from selling q lighsabers is C( q) 8 = 00q + dollars per lighsaber q q R( q) = 48 dollars per lighsaber q c The fied cos is 8 dollars The variable cos is 00q q dollars d The profi is P( q) = R( q) C( q) = 00q + 49q 8 so P ( q) = 004q+ 49 which is zero when q= 30 Since P (000) = 9 and P (50) = 08, C is increasing o he lef of he criical poin q= 30 and decreasing o he righ Thus C (30) = $30,30 is he maimum profi

11 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 e C (8) = = 008 This means ha when producing 8 lighsabers, for some srange reason, i will acually cos abou 8 cens less o produce one more Problem 5 For he following funcion: f 3 ( ) = NO CREDIT will be given if here is no work shown a Find all criical value(s) b Classify all criical values from Par a as local maima, local minima, or neiher c Find all inflecion poins d Find he global maima and minima over he closed inerval [ 3,3] e Skech he graph labeling all of he poins found in he previous pars Soluion a ( ) = = 6( + )( ) so he criical poins are a f = and = b If <, boh facors in he derivaive will be negaive, so f ( ) > 0 If < <, he + facor will be posiive bu will be negaive, so f ( ) < 0 If >, boh facors in he derivaive will be posiive, so f ( ) > 0 This means ha f will have a local maimum a = and a local minimum a = c f ( ) = + 6, and if < hen f ( ) < 0, bu if > hen f ( ) > 0 Thus he concaviy of f swiches from concave down o concave up, so = is an inflecion poin d Evaluaing f a boh criical poins and boh endpoins, f ( 3) = 4, f ( ) = 5, f () = and f (3) = 40 Thus over he closed inerval [ 3,3], f has a global maimum of 40 a = 3 and a global minimum of a = Problem 6 Find he following is a b 0 + c (ln 5) d 0 e Soluion a " 0 " = so we can use L Hopial s Rule 0 ( ) = = 8 3 =

12 Mah 3 CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS) Fall 0 b c d + = " " so we will need o rewrie his by seing 0 + ln y= ln Now, ( ) + y= ln " 0 " ln y= = so we can now apply L Hopial s Rule: 0 + ( ) ln + ln y= = = So y= e 0 " 0 " = so we can use L Hopial s Rule: (ln 5) 0 " 0 " 0 5 (ln 5) 0 = (ln 5)5 (ln 5) = so we can use L Hopial s Rule again: (ln 5)5 (ln 5) + = (ln 5) 5 = (ln 5) 0 e hen = " " so we need o rewrie he funcion as a single fracion, e " 0 " = so we can now apply L Hopial s Rule: 0 e 0 ( ) ( ) e e " 0 " = = so we can use L Hopial s Rule again: 0 0 e e + e 0 e e = = e + e e + e + e 0 0