Extrema of $e^{xy}$ under condition $x^2+y^2\leq 2$ Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ such that $f(x, y)=e^{xy}$.
Find the extrema of the function under the restriction $(x, y) \in \mathcal{D}$ where $\mathcal{D}=\{(x,y)\mid x^2+y^2\leq 2\}$.

Ok.
I proved using the gradient that in the internal  there are no extrema values since the gradient is not equal zero for any $x, y$. Now what about the frontier?
I cannot have a visualization (graph) of the function to use it in order to evaluate /see the extrema on the frontier. And I cannot use the Lagrange theorem. 
Any suggestions?
 A: Since $xy \leq \dfrac{x^2+y^2}{2} \Rightarrow e^{xy} \leq e^{\frac{x^2+y^2}{2}}\leq e\Rightarrow f_{\text{max}} = e$. The equality holds when $x = y, x^2+y^2 = 2 \Rightarrow x = y = \pm 1$
A: Similarly, $xy\geq -\frac{x^2+y^2}2$, so the lower bound is $e^{-1}$ at $x=-y=\pm1$
A: Unconditional extrema of $e^{xy}$ doesn't exist in the area, so it requires looking for conditional extrema on the area bound. Using method of Lagrange multiplyers, one can get:
$$e^{xy}+\lambda(x^2+y^2-2)\to \mathrm{extr},$$
$$
\begin{cases}
ye^{xy}+2x\lambda=0\\
xe^{xy}+2y\lambda=0\\
x^2+y^2-2=0
\end{cases}\rightarrow
\begin{cases}
(x^2+y^2)e^{xy}+4xy\lambda=0\\
2xye^{xy}+2(x^2+y^2)\lambda=0\\
x^2+y^2=2
\end{cases}\rightarrow
\begin{cases}
e^{xy}+2xy\lambda=0\\
xye^{xy}+2\lambda=0\\
x^2+y^2=2
\end{cases}$$
$$\begin{cases}
(x^2y^2-1)e^{xy}=0\\
x^2+y^2=2
\end{cases}\rightarrow
\begin{cases}
x^2y^2=1\\
x^2+y^2=2
\end{cases}$$
$$x^2=y^2=1.$$
There are conditional maxima $e$ when $$x=\pm1,\quad y=x$$ and conditional minima $e^{-1}$ when $$x=\pm1,\quad y=-x.$$
A: The exponential function is a strictly increasing function of its argument, so finding the extrema the product $xy$ is sufficient. Since you have established that the extrema must be on the boundary of $\mathcal{D}$, we have $y = \pm \sqrt{2-x^2}$, and the problem reduces to finding the extrema of the function
$$
g(x) = \pm x\sqrt{2-x^2}.
$$
Differentiating gives
$$
g'(x) = \pm \sqrt{2-x^2} \mp \frac{x^2}{\sqrt{2-x^2}} = \pm 2\frac{1-x^2}{\sqrt{2-x^2}},
$$
which is clearly zero for $x = \pm 1$. Back substitution then gives $y = \pm 1$, and so the extrema of $f$ in $\mathcal{D}$ occur at $x = \pm 1, y = \pm 1$. You can probably work out which are maxima and which are minima.
A: The symmetry of the variables in the expression and the inequality, together with the nature of the exponential, suggest that maxima occur when $x = y$ at $(1, 1)$ and $(-1, -1)$ and minima occur when $x = -y$ at $(1, -1)$ and $(-1, 1)$. There is also a saddle point at the origin.
