Question

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

Answer 1

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

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

So we have three critical points to consider, $(0,0)$, $(-\sqrt2,-1)$, and $(\sqrt2,-1)$.

The function has Hessian

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

At the critical points, we have

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

which means there is a minimum at (0, 0) of $f(0,0)=6$;

$|\mathbf H(-\sqrt2,-1)|=\begin{vmatrix}0&-2\sqrt2\\-2\sqrt2&2\end{vmatrix}=-8$

which means $(-\sqrt2,-1)$ is a saddle point; and

$|\mathbf H(\sqrt2,-1)|=\begin{vmatrix}0&2\sqrt2\\2\sqrt2&2\end{vmatrix}=-8$

which means $(\sqrt2,-1)$ is also a saddle point.