/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 38 Extrema: Find the local maxi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Extrema:Findthelocalmaxima,localminima,andsaddlepointsofthegivenfunctionsf(x,y)=x3−12xy−y3.

Short Answer

Expert verified

Criticalpointas(0,0),(-4,-4)andthelocalminimapointforthefunctionis-4,-4

Step by step solution

01

Step 1. Given 

f(x,y)=x3−12xy−y3.

02

Step 2.  Calculating saddle point .

Givenfunctionisf(x,y)=x3−12xy−y3.,whichisapolynomialandhencedifferentiable.Thegradientofthisfunctionis∇f(x,y)=∂f∂xi+∂f∂yj=(3x2-12y)i+(-12x-3y2)jAtthecriticalpointsthegradientofafunctionvanishes,thatis∇f(x,y)=0.fromtheabovethecriticalpointsaregivenby3x2-12y=0...(1)and-12x-3y2=0.....(2)from(1)y=x24,usein(2)gives4x+(x24)2=0⇒64x+x4=0simplifying⇒x(x+4)(x2-4x+16)=0factorising⇒x=0,-4.theonlyrealroot.Whenx=0,y=0andwhenx=-4,y=-4.Solvingthesetwogivescriticalpointas(0,0),(-4,-4).

03

Step 3. Finding maxima and minima .

Nowthesecondorderderivativesofthegivenfunctionare∂2f∂x2=6x,∂2f∂y2=-6y,∂2f∂y∂x=-12.HencethediscriminateofthegivenfunctionisHf(x,y)=∂2f∂x2∂2f∂y2-(∂2f∂y∂x)2=6x(-6y)-(-12)2=-36xy-144.At(0,0),Hf(0,0)=-144<0Sothecriticalpoint(0,0)isasaddlepoint.At(-4,-4),Hf(-4,-4)=432>0alsoat(-4,-4)∂2f∂x2=24>0,∂2f∂y2=24>0Thereforethelocalminimapointforthefunctionis-4,-4

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