Is this result using Lagrange Multipliers a Maximum or is it a Minimum? I have a question about the result I am getting when using the Lagrange multipliers method to find the extrema of $f(x,y)=e^{xy}$ with constraint $g(x,y)=x^3+y^3=16$.
Fist the equation $\nabla f = \lambda \nabla g$ gives me $3$ equations in $3$ unknowns:
$$ \begin{align}
(1)\;\qquad ye^{xy} & = 3\lambda x^2\\
(2)\;\qquad xe^{xy} & = 3\lambda y^2\\
(3)\quad x^3 + y^3 & = 16
\end{align}$$
If I multiply equations $(1)$ by $x$ and equation $(2)$ by $y$ I get 
$$
3\lambda x^3 = 3\lambda y^3 \Rightarrow x=y
$$
If $x=y=0$ it contradicts equation $(3)$ such that the only solution I get is $x=y=2$ which means that the point $(2,2)$ is an extremum.
Hence, I have 
$$
f(2,2)=e^4.
$$
What I can't figure out at this point is whether it is a maximum or a minimum of $f$ under the given constraint. The second derivative test gives me no information since
$$
D=det
    \begin{pmatrix}
    y^2e^{xy} & xye^{xy} \\
    xye^{xy} & x^2e^{xy} \\
    \end{pmatrix}
=0.
$$
Your help is appreciated, thanks!
V.H.
 A: Since the set $A:=\{(x,y) \in \mathbb{R}^2 \mid x^3+y^3=16, x \geq0, y \geq 0\}$ is compact, the restriction of $f$ to $A$ achieves a maximum there. This maximum can't be when $x=0$ or $y=0$, since $f=1$ there, and you know that $f(2,2)=e^4$. Thus, the maximum must be located at $B:=\{(x,y) \in \mathbb{R}^2 \mid x^3+y^3=16, x >0, y > 0\}$. Since Lagrange multipliers gave you only $(2,2)$ and we know that there is a maximum in $B$, it follows that  $(2,2)$ is a point of maximum* (for now, we only know that it is a point of maximum for the restriction to $A$ or $B$, not to the whole restriction $g=16$).
Note now that outside of $A$ (but still under the restriction by $g=16$), we have that $x \leq 0$ or $y \leq 0$ (but not both!), thus $e^{xy} \leq 1$, since $xy \leq 0$. Thus the point $(2,2)$ is indeed a global maximum on the entirety of the original restriction of the problem, not only on $B$.
*This argument is often a source of confusion for students. What Lagrange multipliers essentially tells you is that if you have an open set $U \subset \mathbb{R}^n$ and $g: U \to \mathbb{R}$ such that $\nabla g \neq 0$ in $g^{-1}(c)$, then
$$\mathrm{LocalMax|_{g^{-1}(c)}}(f) \subset \mathrm{Lag}(f),$$
where the left side are the points of local maximum of $f|_{g^{-1}(c)}$ and the right side are the points which satisfy the "Lagrange condition" $\nabla f = \lambda \nabla g$ for some $\lambda$. Note that the points of maximum are contained in the left side, so we obviously have
$$\mathrm{Max}|_{g^{-1}(c)}(f) \subset \mathrm{Lag}(f). $$
This might seem pedantry, but it makes the argument clear now: When you know that there is a maximum, we know that the left side has at least an element. If there is one single element in $\mathrm{Lag}(f)$, it must then follow that both sets are equal, and thus that element is indeed a maximum.
