Let #\ \ h(x,y)=x^3-6x^2-3y^2\ \ #. Find all critical points of #\ \ f\ \ #? Then classify each as either a relative maximum, minimum, saddle point, or neither.

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Answer 1 · 2018-06-03T09:25:41

Please see the explanation below

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The function is

#h(x,y)=x^3-6x^2-3y^2#

The first derivatives are

#(delh)/(delx)=3x^2-12x-0#

#(delh)/(delx)=0#, #=>#, #3x^2-12x=3x(x-4)=0#

#<=>#, #{(x=0),(x=4):}#

#(delh)/(dely)=-6y#

#(delh)/(dely)=0#, #=>#, #y=0#

The second derivatives are

#(del^2h)/(delx^2)=6x^2-12#

#(del^2h)/(dely^2)=-6#

#(del^2h)/(delxdely)=0#

#(del^2h)/(delydelx)=0#

The Hessian of #h(x,y)# is

#D^2h(x,y)=|((del^2h)/(delx^2),(del^2h)/(delydelx)),((del^2h)/(delxdely),(del^2h)/(dely^2))|#

#=|(6x^2-12,0),(0,-6)|#

#=-6(6x^2-12)#

Therefore,

#D^2h(0,0)=72# and #(del^2h(0,0))/(delx^2)=-12#

The point #(0,0)# corresponds to a relative maximum.

#D^2h(4,0)=-504# and #(del^2h(4,0))/(delx^2)=84#

The point #(4,0)# corresponds to a saddle pint.