What are the values and types of the critical points, if any, of #f(x, y) = x^3+y^3-3*x*y-7#?

1 Answer
Nov 12, 2017

There is a saddle point at #(0,0)# and a relative minimum at #(1,1)#

Explanation:

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Here, we have

#f(x,y)=x^3+y^3-3xy-7#

Compute the partial derivatives

#f_x=3x^2-3y#

#f_y=3y^2-3x#

#f_x=0#, #=>#, #x^2=y#

#f_y=0#, #=>#, #y^2=x#

#x^4=y^2=x#

#x^4-x=0#

#x(x^3-1))=0#

#x(x-1)(x^2+x+1)=0#

#x=0# and #x=1#

#y=0# and #y=1#

The critical point are #(0,0)# and #(1,1)#

Now, compute the second partial derivatives

#f_(x x)=6x#

#f_(yy)=6y#

#f_(xy)=-3#

#f_(yx)=-3#

Therefore,

#D(x,y)=f_(x x) f_(yy)-f_(xy)^2=(6x)(6y)-(-3)^2=36xy-9#

#D(0,0)=-9#

This point #(0,0)# corresponds to a saddle point

#D(1,1)=36-9)=27# and #f_(x x)(1,1)=6#

There is a relative min at #(1,1)#