## Thomas' Calculus 13th Edition

Saddle Point of $f(0,0)=15$ and Local Maximum $f(1,1 )=14$
$$f_x(x,y)=3x^2-3y=0$$ and $$f_y(x,y)=3y^2-3x=0$$ Simplify the above two equations to get the critical points. Thus, the critical point are: $(1,1 )$ and $(0,0)$ Apply the second derivative test for $(0,0)$. $D=f_{xx}f_{yy}-f^2_{xy}=(0)(0)-(-3)^2=0-9=-9 \lt 0$ Saddle Point of $f(0,0)=15$ Apply the second derivative test for $(1,1)$. $D=f_{xx}f_{yy}-f^2_{xy}=(6)(6)-(-3)^2=36-9=27 \gt 0$ Local Maximum of $f(1,1 )=14$