Q:

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

Accepted Solution

A:
$$f(x,y)=x^2+y^2+x^2y+6$$ 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 $$f(0,0)=6$$;

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

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

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

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