Extrema of $e^{xy}$ under condition $x^2+y^2\leq 2$

Question

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?

Answer 1

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 $x = y, x^2+y^2 = 2 \Rightarrow x = y = \pm 1$

Answer 2

Similarly, $xy\geq -\frac{x^2+y^2}2$, so the lower bound is $e^{-1}$ at $x=-y=\pm1$