F(x,y) = x^2+y^2+x^2y+6 find the absolute max and min

[tex]f(x,y)=x^2+y^2+x^2y+6[/tex] has critical points where the partial derivatives simultaneously vanish:

[tex]f_x=2x+2xy=2x(1+y)=0\implies x=0\text{ or }y=-1[/tex]
[tex]f_y=2y+x^2=0[/tex]
[tex]x=0\implies 2y=0\implies y=0[/tex]
[tex]y=-1\implies-2+x^2=0\implies x=\pm\sqrt2[/tex]

So we have three critical points to consider, [tex](0,0)[/tex], [tex](-\sqrt2,-1)[/tex], and [tex](\sqrt2,-1)[/tex].

The function has Hessian

[tex]\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2+2y&2x\\2x&2\end{bmatrix}[/tex]

At the critical points, we have

[tex]|\mathbf H(0,0)|=\begin{vmatrix}2&0\\0&2\end{vmatrix}=4>0[/tex]
[tex]f_{xx}(0,0)=2>0[/tex]

which means there is a minimum at (0, 0) of [tex]f(0,0)=6[/tex];

[tex]|\mathbf H(-\sqrt2,-1)|=\begin{vmatrix}0&-2\sqrt2\\-2\sqrt2&2\end{vmatrix}=-8<0[/tex]

which means [tex](-\sqrt2,-1)[/tex] is a saddle point; and

[tex]|\mathbf H(\sqrt2,-1)|=\begin{vmatrix}0&2\sqrt2\\2\sqrt2&2\end{vmatrix}=-8<0[/tex]

which means [tex](\sqrt2,-1)[/tex] is also a saddle point.