13.7: Extreme Values and Saddle Points - Math LibreTexts

In this graph, the origin is a saddle point. This is because the first partial derivatives of f(x,y)=x2−y2 are both equal to zero at this ... Skiptomaincontent CriticalPointsSecondDerivativeTestAbsoluteMaximaandMinimaKeyConceptsKeyEquationsGlossary LearningObjectives Usepartialderivativestolocatecriticalpointsforafunctionoftwovariables. Applyasecondderivativetesttoidentifyacriticalpointasalocalmaximum,localminimum,orsaddlepointforafunctionoftwovariables. Examinecriticalpointsandboundarypointstofindabsolutemaximumandminimumvaluesforafunctionoftwovariables. Oneofthemostusefulapplicationsforderivativesofafunctionofonevariableisthedeterminationofmaximumand/orminimumvalues.Thisapplicationisalsoimportantforfunctionsoftwoormorevariables,butaswehaveseeninearliersectionsofthischapter,theintroductionofmoreindependentvariablesleadstomorepossibleoutcomesforthecalculations.Themainideasoffindingcriticalpointsandusingderivativetestsarestillvalid,butnewwrinklesappearwhenassessingtheresults. CriticalPoints Forfunctionsofasinglevariable,wedefinedcriticalpointsasthevaluesofthevariableatwhichthefunction'sderivativeequalszeroordoesnotexist.Forfunctionsoftwoormorevariables,theconceptisessentiallythesame,exceptforthefactthatwearenowworkingwithpartialderivatives. Definition:CriticalPoints Let\(z=f(x,y)\)beafunctionoftwovariablesthatisdifferentiableonanopensetcontainingthepoint\((x_0,y_0)\).Thepoint\((x_0,y_0)\)iscalledacriticalpointofafunctionoftwovariables\(f\)ifoneofthetwofollowingconditionsholds: \(f_x(x_0,y_0)=f_y(x_0,y_0)=0\) Either\(f_x(x_0,y_0)\;\text{or}\;f_y(x_0,y_0)\)doesnotexist. Example\(\PageIndex{1}\):FindingCriticalPoints Findthecriticalpointsofeachofthefollowingfunctions: \(f(x,y)=\sqrt{4y^2−9x^2+24y+36x+36}\) \(g(x,y)=x^2+2xy−4y^2+4x−6y+4\) Solution a.First,wecalculate\(f_x(x,y)\;\text{and}\;f_y(x,y):\) \[\begin{align*}f_x(x,y)&=\dfrac{1}{2}(−18x+36)(4y^2−9x^2+24y+36x+36)^{−1/2}\\[4pt]&=\dfrac{−9x+18}{\sqrt{4y^2−9x^2+24y+36x+36}}\end{align*}\] \[\begin{align*}f_y(x,y)&=\dfrac{1}{2}(8y+24)(4y^2−9x^2+24y+36x+36)^{−1/2}\\[4pt]&=\dfrac{4y+12}{\sqrt{4y^2−9x^2+24y+36x+36}}\end{align*}.\] Next,weseteachoftheseexpressionsequaltozero: \[\begin{align*}\dfrac{−9x+18}{\sqrt{4y^2−9x^2+24y+36x+36}}&=0\\[4pt]\dfrac{4y+12}{\sqrt{4y^2−9x^2+24y+36x+36}}&=0.\end{align*}\] Then,multiplyeachequationbyitscommondenominator: \[\begin{align*}−9x+18&=0\\[4pt]4y+12&=0.\end{align*}\] Therefore,\(x=2\)and\(y=−3,\)so\((2,−3)\)isacriticalpointof\(f\). Wemustalsocheckforthepossibilitythatthedenominatorofeachpartialderivativecanequalzero,thuscausingthepartialderivativenottoexist.Sincethedenominatoristhesameineachpartialderivative,weneedonlydothisonce: \[4y^2−9x^2+24y+36x+36=0.\label{critical1}\] Equation\ref{critical1}representsahyperbola.Weshouldalsonotethatthedomainof\(f\)consistsofpointssatisfyingtheinequality \[4y^2−9x^2+24y+36x+36≥0.\] Therefore,anypointsonthehyperbolaarenotonlycriticalpoints,theyarealsoontheboundaryofthedomain.Toputthehyperbolainstandardform,weusethemethodofcompletingthesquare: \[\begin{align*}4y^2−9x^2+24y+36x+36&=0\\[4pt]4y^2−9x^2+24y+36x&=−36\\[4pt]4y^2+24y−9x^2+36x&=−36\\[4pt]4(y^2+6y)−9(x^2−4x)&=−36\\[4pt]4(y^2+6y+9)−9(x^2−4x+4)&=−36−36+36\\[4pt]4(y+3)^2−9(x−2)^2&=−36.\end{align*}\] Dividingbothsidesby\(−36\)putstheequationinstandardform: \[\begin{align*}\dfrac{4(y+3)^2}{−36}−\dfrac{9(x−2)^2}{−36}&=1\\[4pt]\dfrac{(x−2)^2}{4}−\dfrac{(y+3)^2}{9}&=1.\end{align*}\] Noticethatpoint\((2,−3)\)isthecenterofthehyperbola. Thus,thecriticalpointsofthefunction\(f\)are\((2,-3)\)andallpointsonthehyperbola,\(\dfrac{(x−2)^2}{4}−\dfrac{(y+3)^2}{9}=1\). b.First,wecalculate\(g_x(x,y)\)and\(g_y(x,y)\): \[\begin{align*}g_x(x,y)&=2x+2y+4\\[4pt]g_y(x,y)&=2x−8y−6.\end{align*}\] Next,weseteachoftheseexpressionsequaltozero,whichgivesasystemofequationsin\(x\)and\(y\): \[\begin{align*}2x+2y+4&=0\\[4pt]2x−8y−6&=0.\end{align*}\] Subtractingthesecondequationfromthefirstgives\(10y+10=0\),so\(y=−1\).Substitutingthisintothefirstequationgives\(2x+2(−1)+4=0\),so\(x=−1\). Therefore\((−1,−1)\)isacriticalpointof\(g\).Therearenopointsin\(\mathbb{R}^2\)thatmakeeitherpartialderivativenotexist. Figure\(\PageIndex{1}\)showsthebehaviorofthesurfaceatthecriticalpoint. Figure\(\PageIndex{1}\):Thefunction\(g(x,y)\)hasacriticalpointat\((−1,−1,5)\). Exercise\(\PageIndex{1}\) Findthecriticalpointofthefunction\(f(x,y)=x^3+2xy−2x−4y.\) Hint Calculate\(f_x(x,y)\)and\(f_y(x,y)\),thensetthemequaltozero. Answer Theonlycriticalpointof\(f\)is\((2,−5)\). Themainpurposefordeterminingcriticalpointsistolocaterelativemaximaandminima,asinsingle-variablecalculus.Whenworkingwithafunctionofonevariable,thedefinitionofalocalextremuminvolvesfindinganintervalaroundthecriticalpointsuchthatthefunctionvalueiseithergreaterthanorlessthanalltheotherfunctionvaluesinthatinterval.Whenworkingwithafunctionoftwoormorevariables,weworkwithanopendiskaroundthepoint. Definition:GlobalandLocalExtrema Let\(z=f(x,y)\)beafunctionoftwovariablesthatisdefinedandcontinuousonanopensetcontainingthepoint\((x_0,y_0).\)Then\(f\)hasalocalmaximumat\((x_0,y_0)\)if \[f(x_0,y_0)≥f(x,y)\] forallpoints\((x,y)\)withinsomediskcenteredat\((x_0,y_0)\).Thenumber\(f(x_0,y_0)\)iscalledalocalmaximumvalue.Iftheprecedinginequalityholdsforeverypoint\((x,y)\)inthedomainof\(f\),then\(f\)hasaglobalmaximum(alsocalledanabsolutemaximum)at\((x_0,y_0).\) Thefunction\(f\)hasalocalminimumat\((x_0,y_0)\)if \[f(x_0,y_0)≤f(x,y)\] forallpoints\((x,y)\)withinsomediskcenteredat\((x_0,y_0)\).Thenumber\(f(x_0,y_0)\)iscalledalocalminimumvalue.Iftheprecedinginequalityholdsforeverypoint\((x,y)\)inthedomainof\(f\),then\(f\)hasaglobalminimum(alsocalledanabsoluteminimum)at\((x_0,y_0)\). If\(f(x_0,y_0)\)iseitheralocalmaximumorlocalminimumvalue,thenitiscalledalocalextremum(seethefollowingfigure). Figure\(\PageIndex{2}\):Thegraphof\(z=\sqrt{16−x^2−y^2}\)hasamaximumvaluewhen\((x,y)=(0,0)\).Itattainsitsminimumvalueattheboundaryofitsdomain,whichisthecircle\(x^2+y^2=16.\) InCalculus1,weshowedthatextremaoffunctionsofonevariableoccuratcriticalpoints.Thesameistrueforfunctionsofmorethanonevariable,asstatedinthefollowingtheorem. Fermat’sTheoremforFunctionsofTwoVariables Let\(z=f(x,y)\)beafunctionoftwovariablesthatisdefinedandcontinuousonanopensetcontainingthepoint\((x_0,y_0)\).Suppose\(f_x\)and\(f_y\)eachexistat\((x_0,y_0)\).Iffhasalocalextremumat\((x_0,y_0)\),then\((x_0,y_0)\)isacriticalpointof\(f\). SecondDerivativeTest Considerthefunction\(f(x)=x^3.\)Thisfunctionhasacriticalpointat\(x=0\),since\(f'(0)=3(0)^2=0\).However,\(f\)doesnothaveanextremevalueat\(x=0\).Therefore,theexistenceofacriticalvalueat\(x=x_0\)doesnotguaranteealocalextremumat\(x=x_0\).Thesameistrueforafunctionoftwoormorevariables.Onewaythiscanhappenisatasaddlepoint.Anexampleofasaddlepointappearsinthefollowingfigure. Figure\(\PageIndex{3}\):Graphofthefunction\(z=x^2−y^2\).Thisgraphhasasaddlepointattheorigin. Inthisgraph,theoriginisasaddlepoint.Thisisbecausethefirstpartialderivativesoff\((x,y)=x^2−y^2\)arebothequaltozeroatthispoint,butitisneitheramaximumnoraminimumforthefunction.Furthermoretheverticaltracecorrespondingto\(y=0\)is\(z=x^2\)(aparabolaopeningupward),buttheverticaltracecorrespondingto\(x=0\)is\(z=−y^2\)(aparabolaopeningdownward).Therefore,itisbothaglobalmaximumforonetraceandaglobalminimumforanother. Definition:SaddlePoint Giventhefunction\(z=f(x,y),\)thepoint\(\big(x_0,y_0,f(x_0,y_0)\big)\)isasaddlepointifboth\(f_x(x_0,y_0)=0\)and\(f_y(x_0,y_0)=0\),but\(f\)doesnothavealocalextremumat\((x_0,y_0).\) Thesecondderivativetestforafunctionofonevariableprovidesamethodfordeterminingwhetheranextremumoccursatacriticalpointofafunction.Whenextendingthisresulttoafunctionoftwovariables,anissuearisesrelatedtothefactthatthereare,infact,fourdifferentsecond-orderpartialderivatives,althoughequalityofmixedpartialsreducesthistothree.Thesecondderivativetestforafunctionoftwovariables,statedinthefollowingtheorem,usesadiscriminant\(D\)thatreplaces\(f''(x_0)\)inthesecondderivativetestforafunctionofonevariable. SecondDerivativeTest Let\(z=f(x,y)\)beafunctionoftwovariablesforwhichthefirst-andsecond-orderpartialderivativesarecontinuousonsomediskcontainingthepoint\((x_0,y_0)\).Suppose\(f_x(x_0,y_0)=0\)and\(f_y(x_0,y_0)=0.\)Definethequantity \[D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−\big(f_{xy}(x_0,y_0)\big)^2.\] Then: If\(D>0\)and\(f_{xx}(x_0,y_0)>0\),thenfhasalocalminimumat\((x_0,y_0)\). If\(D>0\)and\(f_{xx}(x_0,y_0)<0\),thenfhasalocalmaximumat\((x_0,y_0)\). If\(D<0\),then\(f\)hasasaddlepointat\((x_0,y_0)\). If\(D=0\),thenthetestisinconclusive. SeeFigure\(\PageIndex{4}\). Figure\(\PageIndex{4}\):Thesecondderivativetestcanoftendeterminewhetherafunctionoftwovariableshaslocalminima(a),localmaxima(b),orasaddlepoint(c). Toapplythesecondderivativetest,itisnecessarythatwefirstfindthecriticalpointsofthefunction.Thereareseveralstepsinvolvedintheentireprocedure,whichareoutlinedinaproblem-solvingstrategy. Problem-SolvingStrategy:UsingtheSecondDerivativeTestforFunctionsofTwoVariables Let\(z=f(x,y)\)beafunctionoftwovariablesforwhichthefirst-andsecond-orderpartialderivativesarecontinuousonsomediskcontainingthepoint\((x_0,y_0).\)Toapplythesecondderivativetesttofindlocalextrema,usethefollowingsteps: Determinethecriticalpoints\((x_0,y_0)\)ofthefunction\(f\)where\(f_x(x_0,y_0)=f_y(x_0,y_0)=0.\)Discardanypointswhereatleastoneofthepartialderivativesdoesnotexist. Calculatethediscriminant\(D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−\big(f_{xy}(x_0,y_0)\big)^2\)foreachcriticalpointof\(f\). Applythefourcasesofthetesttodeterminewhethereachcriticalpointisalocalmaximum,localminimum,orsaddlepoint,orwhetherthetheoremisinconclusive. Example\(\PageIndex{2}\):UsingtheSecondDerivativeTest Findthecriticalpointsforeachofthefollowingfunctions,andusethesecondderivativetesttofindthelocalextrema: \(f(x,y)=4x^2+9y^2+8x−36y+24\) \(g(x,y)=\dfrac{1}{3}x^3+y^2+2xy−6x−3y+4\) Solution a.Step1oftheproblem-solvingstrategyinvolvesfindingthecriticalpointsof\(f\).Todothis,wefirstcalculate\(f_x(x,y)\)and\(f_y(x,y)\),thenseteachofthemequaltozero: \[\begin{align*}f_x(x,y)&=8x+8\\[4pt]f_y(x,y)&=18y−36.\end{align*}\] Settingthemequaltozeroyieldsthesystemofequations \[\begin{align*}8x+8&=0\\[4pt]18y−36&=0.\end{align*}\] Thesolutiontothissystemis\(x=−1\)and\(y=2\).Therefore\((−1,2)\)isacriticalpointof\(f\). Step2oftheproblem-solvingstrategyinvolvescalculating\(D.\)Todothis,wefirstcalculatethesecondpartialderivativesof\(f:\) \[\begin{align*}f_{xx}(x,y)&=8\\[4pt]f_{xy}(x,y)&=0\\[4pt]f_{yy}(x,y)&=18.\end{align*}\] Therefore,\(D=f_{xx}(−1,2)f_{yy}(−1,2)−\big(f_{xy}(−1,2)\big)^2=(8)(18)−(0)^2=144.\) Step3statestoapplythefourcasesofthetesttoclassifythefunction'sbehavioratthiscriticalpoint. Since\(D>0\)and\(f_{xx}(−1,2)>0,\)thiscorrespondstocase1.Therefore,\(f\)hasalocalminimumat\((−1,2)\)asshowninthefollowingfigure. Figure\(\PageIndex{5}\):Thefunction\(f(x,y)\)hasalocalminimumat\((−1,2,−16).\)Notethescaleonthe\(y\)-axisinthisplotisinthousands. b.Forstep1,wefirstcalculate\(g_x(x,y)\)and\(g_y(x,y)\),thenseteachofthemequaltozero: \[\begin{align*}g_x(x,y)&=x^2+2y−6\\[4pt]g_y(x,y)&=2y+2x−3.\end{align*}\] Settingthemequaltozeroyieldsthesystemofequations \[\begin{align*}x^2+2y−6&=0\\[4pt]2y+2x−3&=0.\end{align*}\] Tosolvethissystem,firstsolvethesecondequationfor\(y\).Thisgives\(y=\dfrac{3−2x}{2}\).Substitutingthisintothefirstequationgives \[\begin{align*}x^2+3−2x−6&=0\\[4pt]x^2−2x−3&=0\\[4pt](x−3)(x+1)&=0.\end{align*}\] Therefore,\(x=−1\)or\(x=3\).Substitutingthesevaluesintotheequation\(y=\dfrac{3−2x}{2}\)yieldsthecriticalpoints\(\left(−1,\frac{5}{2}\right)\)and\(\left(3,−\frac{3}{2}\right)\). Step2involvescalculatingthesecondpartialderivativesof\(g\): \[\begin{align*}g_{xx}(x,y)&=2x\\[4pt]g_{xy}(x,y)&=2\\[4pt]g_{yy}(x,y)&=2.\end{align*}\] Then,wefindageneralformulafor\(D\): \[\begin{align*}D(x_0,y_0)&=g_{xx}(x_0,y_0)g_{yy}(x_0,y_0)−\big(g_{xy}(x_0,y_0)\big)^2\\[4pt]&=(2x_0)(2)−2^2\\[4pt]&=4x_0−4.\end{align*}\] Next,wesubstituteeachcriticalpointintothisformula: \[\begin{align*}D\left(−1,\tfrac{5}{2}\right)&=(2(−1))(2)−(2)^2=−4−4=−8\\[4pt]D\left(3,−\tfrac{3}{2}\right)&=(2(3))(2)−(2)^2=12−4=8.\end{align*}\] Instep3,wenotethat,applyingNotetopoint\(\left(−1,\frac{5}{2}\right)\)leadstocase\(3\),whichmeansthat\(\left(−1,\frac{5}{2}\right)\)isasaddlepoint.Applyingthetheoremtopoint\(\left(3,−\frac{3}{2}\right)\)leadstocase\(1\),whichmeansthat\(\left(3,−\frac{3}{2}\right)\)correspondstoalocalminimumasshowninthefollowingfigure. Figure\(\PageIndex{6}\):Thefunction\(g(x,y)\)hasalocalminimumandasaddlepoint. Exercise\(\PageIndex{2}\) Usethesecondderivativetesttofindthelocalextremaofthefunction \[f(x,y)=x^3+2xy−6x−4y^2.\nonumber\] Hint Followtheproblem-solvingstrategyforapplyingthesecondderivativetest. Answer \(\left(\frac{4}{3},\frac{1}{3}\right)\)isasaddlepoint,\(\left(−\frac{3}{2},−\frac{3}{8}\right)\)isalocalmaximum. AbsoluteMaximaandMinima Whenfindingglobalextremaoffunctionsofonevariableonaclosedinterval,westartbycheckingthecriticalvaluesoverthatintervalandthenevaluatethefunctionattheendpointsoftheinterval.Whenworkingwithafunctionoftwovariables,theclosedintervalisreplacedbyaclosed,boundedset.Asetisboundedifallthepointsinthatsetcanbecontainedwithinaball(ordisk)offiniteradius.First,weneedtofindthecriticalpointsinsidethesetandcalculatethecorrespondingcriticalvalues.Then,itisnecessarytofindthemaximumandminimumvalueofthefunctionontheboundaryoftheset.Whenwehaveallthesevalues,thelargestfunctionvaluecorrespondstotheglobalmaximumandthesmallestfunctionvaluecorrespondstotheabsoluteminimum.First,however,weneedtobeassuredthatsuchvaluesexist.Thefollowingtheoremdoesthis. ExtremeValueTheorem Acontinuousfunction\(f(x,y)\)onaclosedandboundedset\(D\)intheplaneattainsanabsolutemaximumvalueatsomepointof\(D\)andanabsoluteminimumvalueatsomepointof\(D\). Nowthatweknowanycontinuousfunction\(f\)definedonaclosed,boundedsetattainsitsextremevalues,weneedtoknowhowtofindthem. FindingExtremeValuesofaFunctionofTwoVariables Assume\(z=f(x,y)\)isadifferentiablefunctionoftwovariablesdefinedonaclosed,boundedset\(D\).Then\(f\)willattaintheabsolutemaximumvalueandtheabsoluteminimumvalue,whichare,respectively,thelargestandsmallestvaluesfoundamongthefollowing: Thevaluesof\(f\)atthecriticalpointsof\(f\)in\(D\). Thevaluesof\(f\)ontheboundaryof\(D\). TheproofofthistheoremisadirectconsequenceoftheextremevaluetheoremandFermat’stheorem.Inparticular,ifeitherextremumisnotlocatedontheboundaryof\(D\),thenitislocatedataninteriorpointof\(D\).Butaninteriorpoint\((x_0,y_0)\)of\(D\)that’sanabsoluteextremumisalsoalocalextremum;hence,\((x_0,y_0)\)isacriticalpointof\(f\)byFermat’stheorem.Thereforetheonlypossiblevaluesfortheglobalextremaof\(f\)on\(D\)aretheextremevaluesof\(f\)ontheinteriororboundaryof\(D\). Problem-SolvingStrategy:FindingAbsoluteMaximumandMinimumValues Let\(z=f(x,y)\)beacontinuousfunctionoftwovariablesdefinedonaclosed,boundedset\(D\),andassume\(f\)isdifferentiableon\(D\).Tofindtheabsolutemaximumandminimumvaluesof\(f\)on\(D\),dothefollowing: Determinethecriticalpointsof\(f\)in\(D\). Calculate\(f\)ateachofthesecriticalpoints. Determinethemaximumandminimumvaluesof\(f\)ontheboundaryofitsdomain. Themaximumandminimumvaluesof\(f\)willoccuratoneofthevaluesobtainedinsteps\(2\)and\(3\). Findingthemaximumandminimumvaluesof\(f\)ontheboundaryof\(D\)canbechallenging.Iftheboundaryisarectangleorsetofstraightlines,thenitispossibletoparameterizethelinesegmentsanddeterminethemaximaoneachofthesesegments,asseeninExample\(\PageIndex{3}\).Thesameapproachcanbeusedforothershapessuchascirclesandellipses. Iftheboundaryoftheset\(D\)isamorecomplicatedcurvedefinedbyafunction\(g(x,y)=c\)forsomeconstant\(c\),andthefirst-orderpartialderivativesof\(g\)exist,thenthemethodofLagrangemultiplierscanproveusefulfordeterminingtheextremaof\(f\)ontheboundarywhichisintroducedinLagrangeMultipliers. Example\(\PageIndex{3}\):FindingAbsoluteExtrema Usetheproblem-solvingstrategyforfindingabsoluteextremaofafunctiontodeterminetheabsoluteextremaofeachofthefollowingfunctions: \(f(x,y)=x^2−2xy+4y^2−4x−2y+24\)onthedomaindefinedby\(0≤x≤4\)and\(0≤y≤2\) \(g(x,y)=x^2+y^2+4x−6y\)onthedomaindefinedby\(x^2+y^2≤16\) Solution a.Usingtheproblem-solvingstrategy,step\(1\)involvesfindingthecriticalpointsof\(f\)onitsdomain.Therefore,wefirstcalculate\(f_x(x,y)\)and\(f_y(x,y)\),thensetthemeachequaltozero: \[\begin{align*}f_x(x,y)&=2x−2y−4\\[4pt]f_y(x,y)&=−2x+8y−2.\end{align*}\] Settingthemequaltozeroyieldsthesystemofequations \[\begin{align*}2x−2y−4&=0\\[4pt]−2x+8y−2&=0.\end{align*}\] Thesolutiontothissystemis\(x=3\)and\(y=1\).Therefore\((3,1)\)isacriticalpointof\(f\).Calculating\(f(3,1)\)gives\(f(3,1)=17.\) Thenextstepinvolvesfindingtheextremaof\(f\)ontheboundaryofitsdomain.Theboundaryofitsdomainconsistsoffourlinesegmentsasshowninthefollowinggraph: Figure\(\PageIndex{7}\):Graphofthedomainofthefunction\(f(x,y)=x^2−2xy+4y^2−4x−2y+24.\) \(L_1\)isthelinesegmentconnecting\((0,0)\)and\((4,0)\),anditcanbeparameterizedbytheequations\(x(t)=t,y(t)=0\)for\(0≤t≤4\).Define\(g(t)=f\big(x(t),y(t)\big)\).Thisgives\(g(t)=t^2−4t+24\).Differentiating\(g\)leadsto\(g′(t)=2t−4.\)Therefore,\(g\)hasacriticalvalueat\(t=2\),whichcorrespondstothepoint\((2,0)\).Calculating\(f(2,0)\)givesthe\(z\)-value\(20\). \(L_2\)isthelinesegmentconnecting\((4,0)\)and\((4,2)\),anditcanbeparameterizedbytheequations\(x(t)=4,y(t)=t\)for\(0≤t≤2.\)Again,define\(g(t)=f\big(x(t),y(t)\big).\)Thisgives\(g(t)=4t^2−10t+24.\)Then,\(g′(t)=8t−10\).ghasacriticalvalueat\(t=\frac{5}{4}\),whichcorrespondstothepoint\(\left(0,\frac{5}{4}\right).\)Calculating\(f\left(0,\frac{5}{4}\right)\)givesthe\(z\)-value\(27.75\). \(L_3\)isthelinesegmentconnecting\((0,2)\)and\((4,2)\),anditcanbeparameterizedbytheequations\(x(t)=t,y(t)=2\)for\(0≤t≤4.\)Again,define\(g(t)=f\big(x(t),y(t)\big).\)Thisgives\(g(t)=t^2−8t+36.\)Thecriticalvaluecorrespondstothepoint\((4,2).\)So,calculating\(f(4,2)\)givesthe\(z\)-value\(20\). \(L_4\)isthelinesegmentconnecting\((0,0)\)and\((0,2)\),anditcanbeparameterizedbytheequations\(x(t)=0,y(t)=t\)for\(0≤t≤2.\)Thistime,\(g(t)=4t^2−2t+24\)andthecriticalvalue\(t=\frac{1}{4}\)correspondtothepoint\(\left(0,\frac{1}{4}\right)\).Calculating\(f\left(0,\frac{1}{4}\right)\)givesthe\(z\)-value\(23.75.\) Wealsoneedtofindthevaluesof\(f(x,y)\)atthecornersofitsdomain.Thesecornersarelocatedat\((0,0),(4,0),(4,2)\)and\((0,2)\): \[\begin{align*}f(0,0)&=(0)^2−2(0)(0)+4(0)^2−4(0)−2(0)+24=24\\[4pt]f(4,0)&=(4)^2−2(4)(0)+4(0)^2−4(4)−2(0)+24=24\\[4pt]f(4,2)&=(4)^2−2(4)(2)+4(2)^2−4(4)−2(2)+24=20\\[4pt]f(0,2)&=(0)^2−2(0)(2)+4(2)^2−4(0)−2(2)+24=36.\end{align*}\] Theabsolutemaximumvalueis\(36\),whichoccursat\((0,2)\),andtheglobalminimumvalueis\(20\),whichoccursatboth\((4,2)\)and\((2,0)\)asshowninthefollowingfigure. Figure\(\PageIndex{8}\):Thefunction\(f(x,y)\)hastwoglobalminimaandoneglobalmaximumoveritsdomain. b.Usingtheproblem-solvingstrategy,step\(1\)involvesfindingthecriticalpointsof\(g\)onitsdomain.Therefore,wefirstcalculate\(g_x(x,y)\)and\(g_y(x,y)\),thensetthemeachequaltozero: \[\begin{align*}g_x(x,y)&=2x+4\\[4pt]g_y(x,y)&=2y−6.\end{align*}\] Settingthemequaltozeroyieldsthesystemofequations \[\begin{align*}2x+4&=0\\[4pt]2y−6&=0.\end{align*}\] Thesolutiontothissystemis\(x=−2\)and\(y=3\).Therefore,\((−2,3)\)isacriticalpointof\(g\).Calculating\(g(−2,3),\)weget \[g(−2,3)=(−2)^2+3^2+4(−2)−6(3)=4+9−8−18=−13.\] Thenextstepinvolvesfindingtheextremaofgontheboundaryofitsdomain.Theboundaryofitsdomainconsistsofacircleofradius\(4\)centeredattheoriginasshowninthefollowinggraph. Figure\(\PageIndex{9}\):Graphoftherestricteddomainofthefunction\(g(x,y)=x^2+y^2+4x−6y\). Theboundaryofthedomainof\(g\)canbeparameterizedusingthefunctions\(x(t)=4\cost,\,y(t)=4\sint\)for\(0≤t≤2π\).Define\(h(t)=g\big(x(t),y(t)\big):\) \[\begin{align*}h(t)&=g\big(x(t),y(t)\big)\\[4pt]&=(4\cost)^2+(4\sint)^2+4(4\cost)−6(4\sint)\\[4pt]&=16\cos^2t+16\sin^2t+16\cost−24\sint\\[4pt]&=16+16\cost−24\sint.\end{align*}\] Setting\(h′(t)=0\)leadsto \[\begin{align*}−16\sint−24\cost&=0\\[4pt]−16\sint&=24\cost\\[4pt]\dfrac{−16\sint}{−16\cost}&=\dfrac{24\cost}{−16\cost}\\[4pt]\tant&=−\dfrac{3}{2}.\end{align*}\] Thisequationhastwosolutionsovertheinterval\(0≤t≤2π\).Oneis\(t=π−\arctan(\frac{3}{2})\)andtheotheris\(t=2π−\arctan(\frac{3}{2})\).Forthefirstangle, \[\begin{align*}\sint&=\sin(π−\arctan(\tfrac{3}{2}))=\sin(\arctan(\tfrac{3}{2}))=\dfrac{3\sqrt{13}}{13}\\[4pt]\cost&=\cos(π−\arctan(\tfrac{3}{2}))=−\cos(\arctan(\tfrac{3}{2}))=−\dfrac{2\sqrt{13}}{13}.\end{align*}\] Therefore,\(x(t)=4\cost=−\frac{8\sqrt{13}}{13}\)and\(y(t)=4\sint=\frac{12\sqrt{13}}{13}\),so\(\left(−\frac{8\sqrt{13}}{13},\frac{12\sqrt{13}}{13}\right)\)isacriticalpointontheboundaryand \[\begin{align*}g\left(−\tfrac{8\sqrt{13}}{13},\tfrac{12\sqrt{13}}{13}\right)&=\left(−\tfrac{8\sqrt{13}}{13}\right)^2+\left(\tfrac{12\sqrt{13}}{13}\right)^2+4\left(−\tfrac{8\sqrt{13}}{13}\right)−6\left(\tfrac{12\sqrt{13}}{13}\right)\\[4pt]&=\frac{144}{13}+\frac{64}{13}−\frac{32\sqrt{13}}{13}−\frac{72\sqrt{13}}{13}\\[4pt]&=\frac{208−104\sqrt{13}}{13}≈−12.844.\end{align*}\] Forthesecondangle, \[\begin{align*}\sint&=\sin(2π−\arctan(\tfrac{3}{2}))=−\sin(\arctan(\tfrac{3}{2}))=−\dfrac{3\sqrt{13}}{13}\\[4pt]\cost&=\cos(2π−\arctan(\tfrac{3}{2}))=\cos(\arctan(\tfrac{3}{2}))=\dfrac{2\sqrt{13}}{13}.\end{align*}\] Therefore,\(x(t)=4\cost=\frac{8\sqrt{13}}{13}\)and\(y(t)=4\sint=−\frac{12\sqrt{13}}{13}\),so\(\left(\frac{8\sqrt{13}}{13},−\frac{12\sqrt{13}}{13}\right)\)isacriticalpointontheboundaryand \[\begin{align*}g\left(\tfrac{8\sqrt{13}}{13},−\tfrac{12\sqrt{13}}{13}\right)&=\left(\tfrac{8\sqrt{13}}{13}\right)^2+\left(−\tfrac{12\sqrt{13}}{13}\right)^2+4\left(\tfrac{8\sqrt{13}}{13}\right)−6\left(−\tfrac{12\sqrt{13}}{13}\right)\\[4pt]&=\dfrac{144}{13}+\dfrac{64}{13}+\dfrac{32\sqrt{13}}{13}+\dfrac{72\sqrt{13}}{13}\\[4pt]&=\dfrac{208+104\sqrt{13}}{13}≈44.844.\end{align*}\] Theabsoluteminimumof\(g\)is\(−13,\)whichisattainedatthepoint\((−2,3)\),whichisaninteriorpointof\(D\).Theabsolutemaximumof\(g\)isapproximatelyequalto44.844,whichisattainedattheboundarypoint\(\left(\frac{8\sqrt{13}}{13},−\frac{12\sqrt{13}}{13}\right)\).Thesearetheabsoluteextremaof\(g\)on\(D\)asshowninthefollowingfigure. Figure\(\PageIndex{10}\):Thefunction\(f(x,y)\)hasalocalminimumandalocalmaximum. Exercise\(\PageIndex{3}\): Usetheproblem-solvingstrategyforfindingabsoluteextremaofafunctiontofindtheabsoluteextremaofthefunction \[f(x,y)=4x^2−2xy+6y^2−8x+2y+3\nonumber\] onthedomaindefinedby\(0≤x≤2\)and\(−1≤y≤3.\) Hint Calculate\(f_x(x,y)\)and\(f_y(x,y)\),andsetthemequaltozero.Then,calculate\(f\)foreachcriticalpointandfindtheextremaof\(f\)ontheboundaryof\(D\). Answer Theabsoluteminimumoccursat\((1,0):f(1,0)=−1.\) Theabsolutemaximumoccursat\((0,3):f(0,3)=63.\) Example\(\PageIndex{4}\):ProfitableGolfBalls Pro-\(T\)companyhasdevelopedaprofitmodelthatdependsonthenumber\(x\)ofgolfballssoldpermonth(measuredinthousands),andthenumberofhourspermonthofadvertising\(y\),accordingtothefunction \[z=f(x,y)=48x+96y−x^2−2xy−9y^2,\] where\(z\)ismeasuredinthousandsofdollars.Themaximumnumberofgolfballsthatcanbeproducedandsoldis\(50,000\),andthemaximumnumberofhoursofadvertisingthatcanbepurchasedis\(25\).Findthevaluesof\(x\)and\(y\)thatmaximizeprofit,andfindthemaximumprofit. Figure\(\PageIndex{11}\):(credit:modificationofworkbyoatsy40,Flickr) Solution Usingtheproblem-solvingstrategy,step\(1\)involvesfindingthecriticalpointsof\(f\)onitsdomain.Therefore,wefirstcalculate\(f_x(x,y)\)and\(f_y(x,y),\)thensetthemeachequaltozero: \[\begin{align*}f_x(x,y)&=48−2x−2y\\[4pt]f_y(x,y)&=96−2x−18y.\end{align*}\] Settingthemequaltozeroyieldsthesystemofequations \[\begin{align*}48−2x−2y&=0\\[4pt]96−2x−18y&=0.\end{align*}\] Thesolutiontothissystemis\(x=21\)and\(y=3\).Therefore\((21,3)\)isacriticalpointof\(f\).Calculating\(f(21,3)\)gives\(f(21,3)=48(21)+96(3)−21^2−2(21)(3)−9(3)^2=648.\) Thedomainofthisfunctionis\(0≤x≤50\)and\(0≤y≤25\)asshowninthefollowinggraph. Figure\(\PageIndex{12}\):Graphofthedomainofthefunction\(f(x,y)=48x+96y−x^2−2xy−9y^2.\) \(L_1\)isthelinesegmentconnecting\((0,0)\)and\((50,0),\)anditcanbeparameterizedbytheequations\(x(t)=t,y(t)=0\)for\(0≤t≤50.\)Wethendefine\(g(t)=f(x(t),y(t)):\) \[\begin{align*}g(t)&=f(x(t),y(t))\\[4pt]&=f(t,0)\\[4pt]&=48t+96(0)−y^2−2(t)(0)−9(0)^2\\[4pt]&=48t−t^2.\end{align*}\] Setting\(g′(t)=0\)yieldsthecriticalpoint\(t=24,\)whichcorrespondstothepoint\((24,0)\)inthedomainof\(f\).Calculating\(f(24,0)\)gives\(576.\) \(L_2\)isthelinesegmentconnecting\((50,0)\)and\((50,25)\),anditcanbeparameterizedbytheequations\(x(t)=50,y(t)=t\)for\(0≤t≤25\).Onceagain,wedefine\(g(t)=f\big(x(t),y(t)\big):\) \[\begin{align*}g(t)&=f\big(x(t),y(t)\big)\\[4pt]&=f(50,t)\\[4pt]&=48(50)+96t−50^2−2(50)t−9t^2\\[4pt]&=−9t^2−4t−100.\end{align*}\] Thisfunctionhasacriticalpointat\(t=−\frac{2}{9}\),whichcorrespondstothepoint\((50,−29)\).Thispointisnotinthedomainof\(f\). \(L_3\)isthelinesegmentconnecting\((0,25)\)and\((50,25)\),anditcanbeparameterizedbytheequations\(x(t)=t,y(t)=25\)for\(0≤t≤50\).Wedefine\(g(t)=f\big(x(t),y(t)\big)\): \[\begin{align*}g(t)&=f\big(x(t),y(t)\big)\\[4pt]&=f(t,25)\\[4pt]&=48t+96(25)−t^2−2t(25)−9(25^2)\\[4pt]&=−t^2−2t−3225.\end{align*}\] Thisfunctionhasacriticalpointat\(t=−1\),whichcorrespondstothepoint\((−1,25),\)whichisnotinthedomain. \(L_4\)isthelinesegmentconnecting\((0,0)\)to\((0,25)\),anditcanbeparameterizedbytheequations\(x(t)=0,y(t)=t\)for\(0≤t≤25\).Wedefine\(g(t)=f\big(x(t),y(t)\big)\): \[\begin{align*}g(t)&=f\big(x(t),y(t)\big)\\[4pt]&=f(0,t)\\[4pt]&=48(0)+96t−(0)^2−2(0)t−9t^2\\[4pt]&=96t−9t^2.\end{align*}\] Thisfunctionhasacriticalpointat\(t=\frac{16}{3}\),whichcorrespondstothepoint\(\left(0,\frac{16}{3}\right)\),whichisontheboundaryofthedomain.Calculating\(f\left(0,\frac{16}{3}\right)\)gives\(256\). Wealsoneedtofindthevaluesof\(f(x,y)\)atthecornersofitsdomain.Thesecornersarelocatedat\((0,0),(50,0),(50,25)\)and\((0,25)\): \[\begin{align*}f(0,0)&=48(0)+96(0)−(0)^2−2(0)(0)−9(0)^2=0\\[4pt]f(50,0)&=48(50)+96(0)−(50)^2−2(50)(0)−9(0)^2=−100\\[4pt]f(50,25)&=48(50)+96(25)−(50)^2−2(50)(25)−9(25)^2=−5825\\[4pt]f(0,25)&=48(0)+96(25)−(0)^2−2(0)(25)−9(25)^2=−3225.\end{align*}\] Themaximumvalueis\(648\),whichoccursat\((21,3)\).Therefore,amaximumprofitof\($648,000\)isrealizedwhen\(21,000\)golfballsaresoldand\(3\)hoursofadvertisingarepurchasedpermonthasshowninthefollowingfigure. Figure\(\PageIndex{13}\):Theprofitfunction\(f(x,y)\)hasamaximumat\((21,3,648)\). KeyConcepts Acriticalpointofthefunction\(f(x,y)\)isanypoint\((x_0,y_0)\)whereeither\(f_x(x_0,y_0)=f_y(x_0,y_0)=0\),oratleastoneof\(f_x(x_0,y_0)\)and\(f_y(x_0,y_0)\)donotexist. Asaddlepointisapoint\((x_0,y_0)\)where\(f_x(x_0,y_0)=f_y(x_0,y_0)=0\),but\(f(x_0,y_0)\)isneitheramaximumnoraminimumatthatpoint. Tofindextremaoffunctionsoftwovariables,firstfindthecriticalpoints,thencalculatethediscriminantandapplythesecondderivativetest. KeyEquations Discriminant \(D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−(f_{xy}(x_0,y_0))^2\) Glossary criticalpointofafunctionoftwovariables thepoint\((x_0,y_0)\)iscalledacriticalpointof\(f(x,y)\)ifoneofthetwofollowingconditionsholds: 1.\(f_x(x_0,y_0)=f_y(x_0,y_0)=0\) 2.Atleastoneof\(f_x(x_0,y_0)\)and\(f_y(x_0,y_0)\)donotexist discriminant thediscriminantofthefunction\(f(x,y)\)isgivenbytheformula\(D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−(f_{xy}(x_0,y_0))^2\) saddlepoint giventhefunction\(z=f(x,y),\)thepoint\((x_0,y_0,f(x_0,y_0))\)isasaddlepointifboth\(f_x(x_0,y_0)=0\)and\(f_y(x_0,y_0)=0\),but\(f\)doesnothavealocalextremumat\((x_0,y_0)\)