14.7 Homework-Max. and Min. Values PDF

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10/24/2017

14.7 Homework-Max. and Min. Values

WebAssign 14.7HomeworkMax.andMin.Values(Homework) CurrentScore:–/19

ClaireJenkins APMAE2000,section001,Fall2017 Instructor:DrewYoungren

Due:Thursday,October19201708:40AMEDT

Theduedateforthisassignmentispast.Yourworkcanbeviewedbelow,butnochangescanbemade.  Important!Beforeyouviewtheanswerkey,decidewhetherornotyouplantorequestanextension.YourInstructormaynotgrantyouanextensionifyou haveviewedtheanswerkey.Automaticextensionsarenotgrantedifyouhaveviewedtheanswerkey. RequestExtension

1. –/1pointsSCalcET814.7.VE.002.

Watchthevideobelowthenanswerthequestion.

Thefunction f(x,y) hasarelativemaximumat(x,y)when fx(x,y)=fy(x,y)=0,fxx(x,y)>0, and f

(x,y)fyy(x,y)−[fxy(x,y)]2>0. rue False

SolutionorExplanation False

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2. –/1pointsSCalcET814.7.003.

Usethelevelcurvesinthefiguretopredictthelocationofthecriticalpointsoffandwhetherfhasasaddlepointoralocal maximumorminimumateachcriticalpoint.ThenusetheSecondDerivativesTesttoconfirmyourpredictions.(Orderyour answersbytheirorderedpairs,fromsmallesttolargestx.) f(x,y)=4+x3+y3−3xy (x,y)= (NoResponse) 



 (NoResponse) 



saddlepoint

(x,y)= (NoResponse) 



 (NoResponse) 



localminimum

SolutionorExplanation Inthefigure,apointatapproximately(1,1)isenclosedbylevelcurveswhichareovalinshapeandindicatethataswemove awayfromthepointinanydirectionthevaluesoffareincreasing.Hencewewouldexpectalocalminimumatornear(1,1). Thelevelcurvesnear(0,0)resemblehyperbolas,andaswemoveawayfromtheorigin,thevaluesoffincreaseinsome directionsanddecreaseinothers,sowewouldexpecttofindasaddlepointthere. Toverifyourpredictions,wehave f(x,y)=4+x3+y3−3xy 

 fx(x,y)=3x2−3y,fy(x,y)=3y2−3x.Wehavecritical 2 pointswherethesepartialderivativesareequalto0: 3x −3y=0,3y2−3x=0. Substituting y=x2 fromthefirstequation intothesecondequationgives 3(x2)2−3x=0 

 3x(x3−1)=0 

 x=0orx=1. Thenwehavetwocriticalpoints,

(0,0)and(1,1).Thesecondpartialderivativesare fxx(x,y)=6x,fxy(x,y)=−3,andfyy(x,y)=6y,soD(x,y)=fxx(x,y)fyy(x,y)−[fxy(x,y)]2=(6x)(6y)−(−3)2=36xy−9. Then D(0,0)=36(0)(0)−9=−9,andD(1,1)=36(1)(1)−9=27.SinceD(0,0)0andfxx(1,1)>0, fhasalocalminimumat(1,1).

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3. –/1pointsSCalcET814.7.507.XP.

Findthelocalmaximumandminimumvaluesandsaddlepoint(s)ofthefunction.Ifyouhavethreedimensionalgraphing software,graphthefunctionwithadomainandviewpointthatrevealalltheimportantaspectsofthefunction.(Enteryour answersasacommaseparatedlist.Ifananswerdoesnotexist,enterDNE.) f(x,y)=7−2x+4y−x2−4y2 localmaximumvalue(s)

(NoResponse) 

localminimumvalue(s)

(NoResponse) 

saddlepoint(s)



(x,y,f) = (NoResponse) 

SolutionorExplanation ClicktoViewSolution







4. –/1pointsSCalcET814.7.015.

Findthelocalmaximumandminimumvaluesandsaddlepoint(s)ofthefunction.Ifyouhavethreedimensionalgraphing software,graphthefunctionwithadomainandviewpointthatrevealalltheimportantaspectsofthefunction.(Enteryour answersasacommaseparatedlist.Ifananswerdoesnotexist,enterDNE.) f(x,y)=5excos(y) localmaximumvalue(s)

(NoResponse) 

localminimumvalue(s)

(NoResponse) 

saddlepoint(s)



(x,y,f) = (NoResponse) 

SolutionorExplanation ClicktoViewSolution

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



5. –/1pointsSCalcET814.7.025.

Useagraphorlevelcurvesorbothtofindthelocalmaximumandminimumvaluesandsaddlepointsofthefunction.Thenuse calculustofindthesevaluesprecisely.(Enteryouranswersasacommaseparatedlist.Ifananswerdoesnotexist,enterDNE.) f(x,y)=sin(x)+sin(y)+sin(x+y)+9,0≤x≤2π,0≤y≤2π localmaximumvalue(s) (NoResponse)  

localminimumvalue(s) (NoResponse)  saddlepoint(s)

(x,y,f) =

(NoResponse) 

SolutionorExplanation ClicktoViewSolution

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

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6. –/1pointsSCalcET814.7.505.XP.

FindtheabsolutemaximumandminimumvaluesoffonthesetD. f(x,y)=4x+6y−x2−y2+9,  D={(x,y)|0≤x≤4,0≤y≤5} absolutemaximumvalue (NoResponse)  absoluteminimumvalue (NoResponse) 

 

22  9

SolutionorExplanation ClicktoViewSolution

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

7. –/1pointsSCalcET814.7.037.

FindtheabsolutemaximumandminimumvaluesoffonthesetD. f(x,y)=2x3+y4+3,D={(x,y)|x2+y2≤1}  5 absolutemaximumvalue (NoResponse)    1 absoluteminimumvalue (NoResponse)  SolutionorExplanation ClicktoViewSolution

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8. –/1pointsSCalcET814.7.AE.005.

VideoExample



EXAMPLE5 Findtheshortestdistancefromthepoint (1,0,−3) tothe plane x+2y+z=25.  SOLUTION Thedistancefromanypoint(x,y,z)tothepoint (1,0,−3) is (NoResponse) 

d=

2



+y2+(z+3)2



butif (x,y,z) liesontheplane x+2y+z=25, then andsowehave

z= (NoResponse)  d=

(NoResponse) 



2

+y2+(28−x−2y)2. Wecan



minimizedbyminimizingthesimplerexpression d2=f(x,y)= (NoResponse) 



2

+y2+(28−x−2y)2.



Bysolvingtheequations fx = 2(x−1)−2(28−x−2y)=4x+4y− (NoResponse)  fy = 2y−4(28−x−2y)=4x+10y− (NoResponse) 



wefindthattheonlycriticalpointis (x,y)= (NoResponse) 

 112 

 .

Since fxx=4,  fxy=4, and fyy=10, wehave D(x,y)=fxxfyy−(fxy)2=24>0 and fxx>0, sobytheSecond DerivativesTestfhasalocalminimumat (x,y)= (NoResponse) 

 . Intuitively,wecanseethatthislocal

minimumisactuallyanabsoluteminimumbecausetheremustbeapointon thegivenplanethatisclosestto (1,0,−3). If x= (NoResponse) 

d



11/2 and y= (NoResponse) 

(NoResponse) 

=



2



9  , then

+y2+(28−x−2y)2



=

(NoResponse) 



2 

+ 9

2 

+

9 2 2 

= (NoResponse) 

.

Theshortestdistancefrom (1,0,−3) totheplane x+2y+z=25 is (No

Response) 



.

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9. –/1pointsSCalcET814.7.041.MI.

Findtheshortestdistance,d,fromthepoint (7,0,−6) totheplane x+y+z=6. 

d= (NoResponse) 



SolutionorExplanation ClicktoViewSolution

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





10.–/1pointsSCalcET814.7.043.

Findthepointsonthecone z2=x2+y2 thatareclosesttothepoint (8,2,0).  (x,y,z) =

(NoResponse) 

(x,y,z) =

(NoResponse) 

 (smallerzvalue)  

(largerzvalue)

SolutionorExplanation ClicktoViewSolution

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

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11.–/1pointsSCalcET814.7.045.MI.

Findthreepositivenumberswhosesumis260andwhoseproductisamaximum.(Enteryouranswersasacommaseparated list.)

(NoResponse) 



SolutionorExplanation ClicktoViewSolution

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12.–/1pointsSCalcET814.7.AE.006.

EXAMPLE6 Arectangularboxwithoutalidistobemadefrom12m2of cardboard.Findthemaximumvolumeofsuchabox. SOLUTION Letthelength,width,andheightofthebox(inmeters)bex,y,  VideoExample

andz,asshowninthefigure.Thenthevolumeoftheboxis



V=xyz. WecanexpressVasafunctionofjusttwovariables,xandybyusingthe factthattheareaofthefoursidesandthebottomoftheboxis 2xz+ (NoResponse) 

+xy=12.

Solvingthisequationforz,wegetz= (NoResponse) 

,sothe

expressionforVbecomes

V(x,y)= (NoResponse) 

.

Wecomputethepartialderivatives.

∂V = ∂x 

(NoResponse) 

∂V = ∂y

(NoResponse)  IfVisamaximum,then ∂V/∂x=∂V/∂y=0, but x=0 or y=0 gives V= (NoResponse) 



0  , sowemustsolvetheequations

12−2xy−x2 = 0 12−2xy−y2 = 0. Theseimplythat x2=y2 andso x=y. (Notethatxandymustbothbe positiveinthisproblem.)Ifweput x=y ineitherequationweget 12−3x2=0, whichgives x=2,  y=2, and z=(12−2·2)/[2(2+2)]= (NoResponse) 



1  . Wecouldusethe

SecondDerivativesTesttoshowthatthisgivesalocalmaximumofV,soit mustoccurwhen x=2,  y=2, and z= (NoResponse)  V=2·2· (NoResponse) 

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1  = (NoResponse) 

maximumvolumeoftheboxis (NoResponse) 

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

 

1  . Then 4  , sothe

4  m3.

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13.–/1pointsSCalcET814.7.047.

Findthemaximumvolumeofarectangularboxthatisinscribedinasphereofradiusr.

(NoResponse) 



SolutionorExplanation ClicktoViewSolution

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14.7 Homework-Max. and Min. Values

14.–/1pointsSCalcET814.7.053.

Acardboardboxwithoutalidistohaveavolumeof10,976cm3.Findthedimensionsthatminimizetheamountofcardboard used.(Letx,y,andzbethedimensionsofthecardboardbox.) (x,y,z)= (NoResponse) 

 

SolutionorExplanation ClicktoViewSolution

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

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14.7 Homework-Max. and Min. Values

15.–/1pointsSCalcET814.7.TEC.001.

Concept

Threefunctionsaregraphedalongwiththeircontourmaps.Foreachsurface,usethecontourmaptoidentifythe approximatelocationsofcriticalpoints.Thenlookatthegraphtoclassifyeachcriticalpointasalocalmaximumor minimumorsaddlepoint.Afterward,youcanverifyyourresultsanalytically. Instructions

Selectoneofthethreesurfacesfromthepulldownmenu.Theequationofthefunctionisshownattheright.Inthe exercises,youwillusethecontourmaponthelefttoidentifythelocationsofanycriticalpoints.Thendeterminewhether eachcriticalpointisalocalmaximum,localminimum,orsaddlepointbyexaminingthegraphofthefunction.Youmay wishtorotatethe3Daxeswiththemousefordifferentviewsofthesurface. Simulation

ClickheretoaccesstheTECsimulation. Exercise

CompletethefollowingforSurfaceA: z=x4+2y2−4xy. (a)Usethecontourmaptoestimatethelocationofeachcriticalpointvisually.Byexaminingthegraphofthefunction, determinewhethereachcriticalpointisalocalmaximum,localminimum,orsaddlepoint.(Orderyouranswersbytheir orderedpairs,fromsmallesttolargestx.) (x,y)= (NoResponse) 



 (NoResponse) 



localminimum

(x,y)= (NoResponse) 



 (NoResponse) 



saddlepoint

(x,y)= (NoResponse) 



 (NoResponse) 



localminimum

(b)UsetheSecondDerivativesTesttodeterminethelocalmaximum,localminimum,andsaddlepoint(s)ofthefunction. Howaccuratewereyourestimatesfrompart(a)? (NoResponse) Key:Answersmayvary. Thisanswerhasnotbeengradedyet.

SolutionorExplanation (a)Thecriticalpointsare(1,1),(−1,−1)and(0,0),where(1,1)and(−1,−1)arelocalminimumsand(0,0)isasaddle point. Thelevelcurvesnear(1,1)and(−1,−1)areovalinshapewhichindicatesthatthesepointsareeitheralocalmaximumor minimum.Ifthevalueofzincreasesaswemoveawayfromthesepoints,thentheyarelocalminimumsandviceversa.The valuesofzarenotdepictedonthecontourmap.Therefore,wecannotdeterminewhetherthesepointsarealocalmaximumor minimumbyjustusingthecontourmap.However,bylookingatthegraphofthefunctionweseethattheyarelocalminimums. Ontheotherhand,thelevelcurvesnear(0,0)resemblehyperbolas.Movingawayfromthispoint,weseethatthevaluesofz decreaseinonedirection,butincreaseinotherdirections.Therefore,(0,0)isasaddlepoint. (b)Theestimatesinpart(a)werefoundaccurately.Inordertofindthecriticalpointsofthefunction z=x4+2y2−4xy, we havetofindthesolutionstothefollowingequations: d d z=4x3−4y=0, z=4y−4x=0.  dx dy Thesolutionsarecalledcriticalpoints,whichare(1,1),(−1,−1)and(0,0).Thenextstepistoevaluatethefunction D=zxxzyy−(zxy)2=(12x2)(4)−(−4)2=48x2−16  forallthecriticalpoints. Notethat D(1,1)=32>0 and zxx(1,1)=12>0. Therefore,bythesecondderivativetest,(1,1)isalocalminimum.At (−1,−1),wehave D(−1,−1)=32>0 and zxx(−1,−1)=12>0. Therefore,(−1,−1)isalocalminimum.Finally, D(0,0)=−160 and zxx(0,0)=4>0. Therefore,bythesecondderivativetest,(0,0)isalocalminimum.Atthe 1 1 point 1, wehave D 1, 1 =−12...