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In Exercises 31–52, find the relative maxima, relative minima, and saddle points for the given functions. Determine whether the function has an absolute maximum or absolute minimum as well.g(x,y)=x3+6x2+6y2−4

Short Answer

Expert verified

Thegivenfunctionhaslocalminimumat0,0withg(0,0)=-4andsaddlepointat(-4,0)withg(-4,0)=28

Step by step solution

01

Step 1. Given information  

A function,g(x,y)=x3+6x2+6y2−4

02

Step 2. Finding the first-order, second-order partial derivatives and determinant of hessian 

Thefirst-orderpartialderivativesofthefunctionare:gx(x,y)=∂g∂x=3x2+12xandgy(x,y)=∂g∂y=12yNow,solvethesystemofequations:3x2+12x=0and12y=0,weget,x(3x+12)=0andy=0⇒x=0,x=-4andy=0Wefindtwostationarypointsofg,namely:(0,0),(-4,0)Thesecond-orderpartialderivativesofthefunctionare:gxx(x,y)=∂2g∂x2=6x+12,gyy(x,y)=∂2g∂y2=12andgxy(x,y)=∂2g∂x∂y=0gxx(0,0)=12,gyy(0,0)=12andgxy(0,0)=0gxx(-4,0)=-24,gyy(-4,0)=12andgxy(-4,0)=0ThedeterminantoftheHessianis:detHgx,y=∂2g∂x2∂2g∂y2-∂2g∂x∂y2detHg0,0=12×12-0=144detHg-4,0=-24×12-0=-288

03

Step 3. Testing and finding relative maximum, relative minimum and saddle points

Ifghasastationarypointat(x0,y0),then(a)ghasarelativemaximumat(x0,y0)ifdet(Hg(x0,y0))>0withgxx(x0,y0)<0orgyy(x0,y0)<0.(b)ghasarelativeminimumat(x0,y0)ifdet(Hg(x0,y0))>0withgxx(x0,y0)>0orgyy(x0,y0)>0.(c)ghasasaddlepointat(x0,y0)ifdet(Hg(x0,y0))<0.(d)Ifdet(Hg(x0,y0))=0,noconclusionmaybedrawnaboutthebehaviorofgat(x0,y0).Inthegivenfunction,detHg0,0=144>0withgxx0,0=12>0Hence,thegivenfunctionhasminimumat0,0withminimumvalue,g0,0=03+602+6(0)2-4=-4Also,detHg-4,0=-288<0Hence,thegivenfunctionhassaddlepointat-4,0withg-4,0=-43+6-42+6(0)2-4=28

04

Step 4. Testing and finding absolute maximum and absolute minimum 

Wheny=0,limx→∞f(x,0)=∞andlimx→-∞f(x,0)=-∞Therefore,thegivenfunctionhaslocalminimumat0,0andsaddlepointat(-4,0)

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