Answer
Saddle Point of $f(0,0)=15$
Local Maximum $f(1,1 )=14$
Work Step by Step
We have $f_x(x,y)=3x^2-3y=0; f_y(x,y)=3y^2-3x=0$
Now, we simplify the above two equations to get the critical points: $(1,1 )$ and $(0,0)$
We will take the help of the second derivative test, for $(0,0)$ as follows:
$D=f_{xx}f_{yy}-f^2_{xy}=(0)(0)-(-3)^2=-9 \lt 0$
So, Saddle Point of $f(0,0)=15$
We will take the help of the second derivative test, for $(1,1)$ as follows:
$D=f_{xx}f_{yy}-f^2_{xy}=(6)(6)-(-3)^2=27 \gt 0$
So, Local Maximum $f(1,1 )=14$
