﻿
6.1 Optimization
107
localmaximummustbetheglobalmaximumandthesmallestlocalminimummustbethe
globalminimum. It t isusuallyeasier,however,tocompute thevalueoff f ateverypoint
atwhichtheglobalmaximumorminimummightoccur;thelargestoftheseistheglobal
maximum,thesmallestistheglobalminimum.
Sowecomputef(−2)=4,f(0)=0,f(1)=1. Theglobalmaximumis4atx=−2
andtheglobalminimumis0atx=0.
Itispossible thatthereisno globalmaximumor minimum. . It t isdiﬃcult, , and not
particularlyuseful, to o expressasetprocedure for determining whether thisisthe case.
Generally,thebestapproachistogainenoughunderstandingoftheshapeofthegraphto
decide.Fortunately,onlyaroughideaoftheshapeisusuallyneeded.
Therearesomeparticularlynicecasesthatareeasy.Ifyouhaveacontinuousfunction
onaclosedinterval[a,b],thereisalwaysbothaglobalmaximumandaglobalminimum,
soexaminingthecriticalvaluesandtheendpointsisenough:
THEOREM6.2
Extreme valuetheorem
Iff iscontinuousonaclosedinterval
[a,b],thenithasbothaminimumandamaximumpoint.Thatis,therearerealnumbers
canddin[a,b]sothatforeveryxin[a,b],f(x)≤f(c)andf(x)≥f(d).
Anothereasycase: Ifafunctioniscontinuousandhasasinglecriticalvalue,thenif
thereisalocalmaximumatthecriticalvalueitisaglobalmaximum,andifitisalocal
minimumitisaglobalminimum. Theremayalsobeaglobalminimumintheﬁrstcase,
or a globalmaximuminthe second case, , but t that willgenerallyrequire more eﬀortto
determine.
EXAMPLE 6.3
Letf(x)=−x
2
+4x−3. Findthe e maximumvalueof f(x) onthe
interval[0,4]. Firstnotethatf
(x)=−2x+4=0whenx=2,andf(2)=1.Nextobserve
thatf(x)isdeﬁnedforallx,sotherearenoothercriticalvalues. Finally,f(0)=−3and
f(4)=−3.Thelargestvalueoff(x)ontheinterval[0,4]isf(2)=1.
EXAMPLE 6.4
Letf(x)=−x
2
+4x−3. Findthe e maximumvalueof f(x) onthe
interval[−1,1].
Firstnotethatf
(x)=−2x+4=0whenx=2. Butx=2isnotintheinterval,so
wedon’t useit. . Thustheonlytwopointstobecheckedaretheendpoints;f(−1)=−8
andf(1)=0. Sothelargestvalueoff(x)on[−1,1]isf(1)=0.
EXAMPLE 6.5
Findthemaximumandminimumvaluesofthefunctionf(x)=7+
|x−2|for xbetween1and4inclusive. . Thederivativef
(x)isnever zero,but f
(x)is
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108
Chapter6 Applications s oftheDerivative
undeﬁnedatx=2,sowecomputef(2)=7.Checkingtheendpointswegetf(1)=8and
f(4) =9. . Thesmallestofthesenumbersisf(2)=7,whichis,therefore,the e minimum
valueoff(x)ontheinterval1≤x≤4,andthemaximumisf(4)=9.
−2
−1
2
1
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Figure6.2
f(x)=x
3
−x
EXAMPLE 6.6
Find all l local maxima and d minima a for r f(x) = = x
3
−x, and d deter-
mine whether there isa globalmaximumor minimumon theopeninterval(−2,2). . In
example 5.2 we found a local l maximum at t (−
3/3,2
3/9) and a local l minimum at
(
3/3,−2
3/9). Sincetheendpointsarenotintheinterval(−2,2)theycannotbecon-
sidered.Isthelonelocalmaximumaglobalmaximum?Herewemustlookmorecloselyat
thegraph.Weknowthatontheclosedinterval[−
3/3,
3/3]thereisaglobalmaximum
atx=−
3/3andaglobalminimumatx=
3/3. Sothequestionbecomes: whathap-
pensbetween−2and−
3/3,andbetween
3/3and2? Sincethereisalocalminimum
atx =
3/3,the graphmust continue up tothe right,sincetherearenomorecritical
values. Thismeansnovalueoff f willbelessthan−2
3/9between
3/3and2,butit
3/9.
Howcanwetell?Sincethefunctionincreasestotherightof
3/3,weneedtoknowwhat
thefunctionvaluesdo “close to” 2. . Herethe e easiest test isto picka number anddoa
computationtoget someideaofwhat’sgoingon. . Since e f(1.9)=4.959>2
3/9,there
isnoglobalmaximumat−
3/3,andhencenoglobalmaximumatall. (Howcanwetell
that4.959>2
3/9? Wecanuseacalculatortoapproximatetherighthandside;ifitis
notevencloseto4.959wecantakethisasdecisive. Since2
3/9≈0.3849,there’sreally
noquestion.Funnythingscanhappenintheroundingdonebycomputersandcalculators,
however,sowemightbealittlemorecareful,especiallyifthevaluescomeoutquiteclose.
Inthiscase we can convert the relation 4.959 >2
3/9 into (9/2)4.959 >
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6.1 Optimization
109
whetherthisistrue. Sincetheleftsideisclearlylargerthan4·4whichisclearlylarger
than
3,thissettlesthequestion.)
Asimilaranalysisshowsthatthereisalsonoglobalminimum.Thegraphoff(x)on
(−2,2)isshowninﬁgure6.2.
EXAMPLE6.7
Ofallrectanglesofarea100,whichhasthesmallestperimeter?
Firstwemusttranslatethisintoapurelymathematicalprobleminwhichwewantto
ﬁndtheminimumvalueofafunction. Ifxdenotesoneofthesidesoftherectangle,then
tominimizeis
f(x)=2x+2
100
x
sincetheperimeteristwicethelengthplustwicethewidthoftherectangle.Notallvalues
ofxmakesenseinthisproblem:lengthsofsidesofrectanglesmustbepositive,sox>0.
Ifx>0thensois100/x,soweneednosecondconditiononx.
Wenextﬁndf
(x)andsetitequaltozero:0=f
(x)=2−200/x
2
. Solvingf
(x)=0
forx givesusx=±10. . Weareinterestedonlyin n x>0,so onlythevalue x=10 isof
interest.Sincef
(x)isdeﬁnedeverywhereontheinterval(0,∞),therearenomorecritical
values, and there e are noendpoints. . Istherea a localmaximum,minimum,or neitherat
x =10? ? The e second derivative is s f

(x)= 400/x
3
,and f

(10) >0, , so o there isa local
minimum. Sincethereisonlyone e criticalvalue,thisisalsotheglobalminimum,sothe
rectanglewithsmallestperimeteristhe10×10square.
EXAMPLE 6.8
Youwantto sellacertain number nof itemsinordertomaximize
yourproﬁt. Marketresearchtellsyouthatifyouset t thepriceat\$1.50,youwillbeable
tosell5000items,andforevery10centsyoulowerthepricebelow\$1.50youwillbeable
tosellanother 1000items. . Suppose e thatyourﬁxedcosts(“start-upcosts”) total\$2000,
andtheperitemcostofproduction(“marginalcost”)is\$0.50. Findthepricetosetper
itemandthe number ofitemssoldinorder to maximize proﬁt,and also determine the
maximumproﬁtyoucanget.
Theﬁrststepistoconverttheproblemintoafunctionmaximizationproblem. Since
we wanttomaximizeproﬁtbysettingthe priceperitem,weshouldlookfora function
P(x)representingtheproﬁtwhenthepriceperitemisx.Proﬁtisrevenueminuscosts,and
revenueisnumberofitemssoldtimesthepriceperitem,sowegetP =nx−2000−0.50n.
Thenumberofitemssoldisitselfafunctionofx,n=5000+1000(1.5−x)/0.10,because
(1.5−x)/0.10isthenumberofmultiplesof10centsthatthepriceisbelow\$1.50. Now
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110
Chapter6 Applications s oftheDerivative
wesubstitutefornintheproﬁtfunction:
P(x)=(5000+1000(1.5−x)/0.10)x−2000−0.5(5000+1000(1.5−x)/0.10)
=−10000x
2
+25000x−12000
We want to know the maximum value of this function n when n x x is s between 0 0 and d 1.5.
The derivative is P
(x) = −20000∗x+25000, which h is zero o when n x x = 1.25. . Since
P(x)=−20000<0,theremustbealocalmaximumat x=1.25,andsincethisisthe
onlycriticalvalueitmustbeaglobalmaximumaswell. (Alternately,wecouldcompute
P(0)=−12000,P(1.25)=3625,andP(1.5)=3000andnotethatP(1.25)isthemaximum
ofthese.) Thusthemaximumproﬁtis\$3625,attainedwhenwesetthepriceat\$1.25and
sell7500items.
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y=a
(x,x
2
)
............. ............. ............. . ............. ............. . ............. ............. ............. ............. ............. ............. . ............. ............. . ............. ............. ............. ............. .............
Figure6.3
Rectangleinaparabola.
EXAMPLE 6.9
Findthe largest rectangle (that is,the rectangle with largest area)
thatﬁtsinsidethegraphoftheparabolay=x
2
belowtheliney=a(aisanunspeciﬁed
constantvalue).(Thetopsideoftherectangleshouldbeonthehorizontalliney=a;see
ﬁgure6.3.)
WewanttoﬁndthemaximumvalueofsomefunctionA(x)representingarea.Perhaps
the hardest part t of f thisproblemis s deciding g what x shouldrepresent. . The e lower right
corner ofthe rectangleisat(x,x2), , andoncethisischosenthe e rectangleiscompletely
determined. Sowe e canletthe xinA(x) bethex ofthe parabolaf(x)=x
2
. Thenthe
areaisA(x)=(2x)(a−x
2
)=−2x
3
+2ax. WewantthemaximumvalueofA(x)whenxis
in[0,
a].(Youmightobjecttoallowingx=0orx=
a,sincethenthe“rectangle”has
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6.1 Optimization
111
eithernowidthornoheight,soisnot“really”arectangle. Buttheproblemissomewhat
easierifwesimplyallowsuchrectangles,whichhavezeroarea.)
Setting0=A
(x)=6x
2
+2awegetx=
p
a/3astheonlycriticalvalue.Testingthis
andthe two endpoints,we have A(0)=A(
a) =0 and A(
p
a/3) =(4/9)
3a3/2. The
maximumareathusoccurswhentherectanglehasdimensions2
p
a/3×(2/3)a.
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(h−R,r)
Figure6.4
Coneinasphere.
EXAMPLE 6.10
Ifyouﬁtthelargestpossibleconeinsideasphere,whatfractionof
thevolumeofthesphereisoccupiedbythecone?(Hereby“cone”wemeanarightcircular
cone,i.e.,aconeforwhichthebaseisperpendiculartotheaxisofsymmetry,andforwhich
thecross-sectioncutperpendiculartotheaxisofsymmetryatanypointisacircle.)
theconeinsidethesphere.Whatwewanttomaximizeisthevolumeofthecone: πr
2
h/3.
HereR isaﬁxedvalue,butr andhcanvary. . Namely,wecouldchoose e rtobeaslarge
aspossible—equaltoR—bytaking theheight equalto R;or we couldmake thecone’s
heighthlargerattheexpenseofmakingralittlelessthanR. Seethesideviewdepicted
inﬁgure6.4. Wehavesituatedthe e pictureinaconvenientwayrelativetothe x and y
axes,namely,withthecenterofthesphereattheoriginandthevertexoftheconeatthe
farleftonthex-axis.
Noticethatthefunctionwewanttomaximize,πr2h/3,dependsontwovariables.This
isfrequentlythecase,butoftenthetwovariablesarerelatedinsomewaysothat“really”
thereisonlyonevariable. Soournextstepistoﬁndtherelationshipanduseittosolve
foroneofthevariablesintermsoftheother,soastohaveafunctionofonlyonevariable
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112
Chapter6 Applications s oftheDerivative
Thatis,
(h−R)
2
+r
2
=R
2
.
Wecansolvefor hin termsofr orfor r intermsofh. . Either r involvestaking asquare
root,butwenoticethatthevolumefunctioncontainsr
2
,notrbyitself,soitiseasiestto
solveforr
2
directly: r
2
=R
2
−(h−R)
2
.Thenwesubstitutetheresultintoπr
2
h/3:
V(h)=π(R
2
−(h−R)
2
)h/3
=−
π
3
h
3
+
2
3
πh
2
R
We want t to o maximize V(h) whenh h isbetween0 0 and 2R. . Now w we e solve 0 = = f
(h) =
−πh+(4/3)πhR, getting g h = = 0 0 or h = = 4R/3. . We e compute V(0) = = V(2R) =0 0 and
V(4R/3) =(32/81)πR
3
. Themaximumisthe e latter; ; since e thevolume ofthe sphere is
(4/3)πR
3
,thefractionofthesphereoccupiedbytheconeis
(32/81)πR
3
(4/3)πR3
=
8
27
≈30%.
EXAMPLE 6.11
You are making cylindricalcontainersto o contain n a given volume.
Suppose thatthe top andbottomare madeof a materialthat isN N timesasexpensive
(costperunitarea)asthematerialusedforthelateralsideofthecylinder.Find(interms
thecontainers.
Letusﬁrstchooseletterstorepresentvariousthings: hfortheheight,rforthebase
ofthecylinder;V andcareconstants,handr r arevariables. . Nowwecanwritethecost
ofmaterials:
c(2πrh)+Nc(2πr
2
).
Again we have two variables; the relationship is provided by y the ﬁxed d volume of the
cylinder: V =πr
2
h. We e usethisrelationtoeliminate h(we couldeliminate r,butit’s
alittleeasierifweeliminateh,whichappearsinonlyoneplaceintheaboveformulafor
cost). Theresultis
f(r)=2cπr
V
πr2
+2Ncπr
2
=
2cV
r
+2Ncπr
2
.
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6.1 Optimization
113
We want toknowtheminimumvalueofthisfunctionwhen r isin (0,∞). . Wenowset
0 =f(r) =−2cV/r2+4Ncπr, , givingr r =
3
p
V/(2Nπ). Sincef(r) ) =4cV/r3+4Ncπ
ispositivewhenrispositive,thereisalocalminimumatthecriticalvalue,andhencea
globalminimumsincethereisonlyonecriticalvalue.
Finally,sinceh=V/(πr2),
h
r
=
V
πr3
=
V
π(V/(2Nπ))
=2N,
so theminimumcostoccurswhentheheight his2N N timesthe e radius. . If,forexample,
isequaltothediameter).
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x
a−x
A
D
B
C
b
Figure 6.5
Minimizingtraveltime.
EXAMPLE6.12
SupposeyouwanttoreachapointAthatislocatedacrossthesand
whichislessthanv,beyourspeedonthesand.RightnowyouareatthepointD,which
sandinordertominimizeyourtraveltimetoA?
LetxbethedistanceshortofCwhereyouturnoﬀ,i.e.,thedistancefromBtoC.We
wanttominimizethetotaltraveltime. Recallthatwhen n travelingat constantvelocity,
time=distance/velocity.
Youtravelthedistance
DBatspeedv,andthenthedistance
BAatspeedw. Since
DB=a−xand,bythePythagoreantheorem,
BA=
x2+b2,thetotaltimeforthetrip
is
f(x)=
a−x
v
+
x2+b2
w
.
114
Chapter6 Applications s oftheDerivative
We want to ﬁnd theminimumvalueof f f whenx x isbetween 0and a. . Asusualwe e set
f
(x)=0andsolveforx:
0=f
(x)=−
1
v
+
x
w
x2+b2
w
p
x2+b=vx
w
2
(x
2
+b
2
)=v
2
x
2
w
2
b
2
=(v
2
−w
2
)x
2
x=
wb
v2−w2
interestedonlyincriticalvaluesthatarein[0,a],andwb/
v2−wiseitherinthisinterval
ornot. Ifitis,wecanusethesecondderivativetotestit:
f

(x)=
b
2
(x2+b2)3/2w
.
Sincethisisalwayspositivethereisalocalminimumatthecriticalpoint,andsoitisa
globalminimumaswell.
Ifthecriticalvalueisnotin[0,a]itislargerthana. Inthiscasetheminimummust
occuratoneoftheendpoints. Wecancompute
f(0)=
a
v
+
b
w
f(a)=
a2+b2
w
but it isdiﬃcult to determine which ofthese issmaller bydirect comparison. . If,asis
likelyinpractice,weknowthevaluesofv,w,a,andb,thenitiseasytodeterminethis.
Withalittlecleverness,however,wecandeterminetheminimumingeneral.Wehaveseen
thatf(x)isalwayspositive,sothederivativef(x)isalwaysincreasing. Weknowthat
atwb/
v2−wthederivativeiszero,soforvaluesofxlessthanthatcriticalvalue,the
derivativeisnegative. Thismeansthatf(0)>f(a),sotheminimumoccurswhenx=a.
Sotheupshotisthis: IfyoustartfartherawayfromCthanwb/
v2−wthenyou
v2−wfrompointC.
IfyoustartcloserthanthistoC,youshouldcutdirectlyacrossthesand.
6.1 Optimization
115
Summary—Steps tosolveanoptimizationproblem.
1. Decide what the e variablesare e and d what t the e constants s are, , draw w a diagramif
appropriate,understandclearlywhatitisthatistobemaximizedorminimized.
2. Write e a formula for the e function n for r which you wish h to o ﬁnd the e maximum m or
minimum.
3. Expressthatformulaintermsofonlyonevariable,thatis,intheformf(x).
4. Setf
(x)=0andsolve.Checkallcriticalvaluesandendpointstodeterminethe
extremevalue.
Exercises
1. Letf(x)=
1+4x−x
2
forx≤3
(x+5)/2
forx>3
Findthemaximumvalueandminimumvalues off(x)for xin[0,4]. . Graphf(x)tocheck
2. Findthedimensionsoftherectangleoflargestareahavingﬁxedperimeter100.
3. FindthedimensionsoftherectangleoflargestareahavingﬁxedperimeterP.
4. Aboxwithsquarebaseandnotopistoholdavolume100.Findthedimensionsofthebox
thatrequirestheleastmaterialfortheﬁvesides. Alsoﬁndtheratioofheighttosideofthe
base.
5. Aboxwithsquarebaseistoholdavolume200. . Thebottomandtopareformedbyfolding
inﬂapsfromallfoursides,sothat thebottomandtopconsist oftwolayers ofcardboard.
Findthedimensionsoftheboxthatrequirestheleastmaterial. Alsoﬁndtheratioofheight
tosideofthebase.
6. AboxwithsquarebaseandnotopistoholdavolumeV.Find(intermsofV)thedimensions
oftheboxthatrequirestheleastmaterialfortheﬁvesides. Alsoﬁndtheratioofheightto
sideofthebase.(ThisratiowillnotinvolveV.)
7. Youhave100feetoffencetomakearectangularplayareaalongsidethewallofyourhouse.
Thewallofthehouseboundsoneside. Whatisthelargestsizepossible(insquarefeet)for
theplayarea?
8. Youhavel l feet offencetomakea rectangularplay areaalongside the wallofyourhouse.
Thewallofthehouseboundsoneside. Whatisthelargestsizepossible(insquarefeet)for
theplayarea?
9. Marketingtellsyouthatifyousetthepriceofanitemat\$10thenyouwillbeunabletosell
it,butthatyoucansell500itemsforeachdollarbelow\$10thatyousettheprice. Suppose
yourﬁxedcoststotal\$3000,andyourmarginalcostis\$2peritem. Whatisthemostproﬁt
youcanmake?
10. Findtheareaofthe e largest rectangle thatﬁts inside asemicircle ofradius10(oneside of
therectangleisalongthediameterofthesemicircle).
rectangleisalongthediameterofthesemicircle).
116
Chapter6 Applications s oftheDerivative
12. Foracylinderwithsurfacearea50,includingthetopandthebottom,ﬁndtheratioofheight
13. ForacylinderwithgivensurfaceareaS,includingthetopandthebottom,ﬁndtheratioof
14. Youwanttomakecylindricalcontainerstohold1literusingtheleastamountofconstruction
nomaterialwasted. However,thetopandbottomarecut t fromsquares ofside2r,sothat
2(2r)
2
=8r
2
ofmaterialisneeded(ratherthan2πr
2
,whichisthetotalareaofthetopand
bottom). Findthedimensionsofthecontainerusingtheleastamountofmaterial,andalso
15. You u want to o make e cylindricalcontainers of a givenvolume e V V using g the least amount t of
bedonewithnomaterialwasted. However,thetopandbottomarecutfromsquaresofside
2r,sothat 2(2r)
2
=8r
2
ofmaterialisneeded(ratherthan2πr
2
,whichisthetotalareaof
16. Givenarightcircularcone,youputanupside-downconeinsideitsothatitsvertexisatthe
centerofthebaseofthelargerconeanditsbaseisparalleltothebaseofthelargercone.If
youchoose the upside-downconetohave the largestpossiblevolume,whatfractionofthe
volumeof thelargerconedoesitoccupy? ? (LetH H andRbetheheight t andbaseradius of
similartrianglestogetanequationrelatinghandr.)
17. Inexample6.12,whathappensifw≥v(i.e.,yourspeedonsandisatleastyourspeedon
18. Acontainer r holdingaﬁxedvolumeis beingmadeintheshape ofacylinder withahemi-
unitareaofthetopistwiceasgreatasthecostperunitareaoftheside,andthecontainer
circular bottom,for whichthecostperunitareais1.5times thecostper unitareaof the
side.
19. Apieceofcarboardis1 1 meterby 1/2meter. . Asquareis s tobe cut fromeachcorner and
the sides folded up to make an open-topbox. . What t are e the dimensions s of f the box x with
maximumpossiblevolume?
20. (a)Asquarepieceofcardboardofside e ais usedtomakeanopen-topboxby cuttingout
asmallsquare fromeachcorner andbendingupthe sides. . Howlarge e asquareshouldbe
cutfromeachcornerinorderthattheboxhavemaximumvolume? (b)Whatifthepieceof
cardboardusedtomaketheboxisarectangleofsidesaandb?
21. Awindowconsistsofarectangularpiece e ofclear glass withasemicircularpieceofcolored
glassontop;thecoloredglasstransmitsonly1/2asmuchlightperunitareaasthetheclear
glass. Ifthe e distance fromtoptobottom(across boththerectangleandthesemicircle)is
2metersandthewindowmaybenomorethan1.5meterswide,ﬁndthedimensionsofthe
rectangularportionofthewindowthatletsthroughthemostlight.
22. Awindowconsistsofarectangularpiece e ofclear glass withasemicircularpieceofcolored
glassontop. Supposethat t thecoloredglasstransmits only ktimesasmuchlightperunit