Find the extreme values of $f(x,y)=e^{-xy}$ on the region described by $x^2+25y^2\leq 4$

Question

I have been stuck on this question for a very long time. I have tried to use lagrange multipliers but the equation seems nearly impossible to solve as the derivative of $f$ with respect to $x$ and the derivative of $f$ with respect to $y$ has both algebraic and exponential components, making it hard to express lamba in terms of a single variable $x$ or $y$.

I am supposed to find an absolute min and an absolute max, but the only critical point I can seem to find is $(0,0)$ which gives me a value of $1$ which is apparently wrong. Likewise, I have tried to evaluate the edges and the boundary by plugging in the domain $-2<x<2$ and $-\frac{2}{5}<y<\frac{2}{5}$ but cannot find a critical point on the boundary. Furthermore, the edges all return a value of 1.

The coordinates of my edges are $(2,0)$, $(-2,0)$, $(0,-\frac{2}{5})$ and $(0,\frac{2}{5})$.

Please help me.

Answer 1

Since $f(x,y)=e^{-xy}$ is a continuous function and $A:=\{(x,y) \in \mathbb{R}^2 \ \text{s.t.} \ x^2+25y^2\le 4\}$ is a compact set, for the extreme value theorem there exist the maximum and the minimum of $f$ on $A$.

The interior points of extrema satisfy $$\nabla f(x,y)=(0,0) \iff(-ye^{-xy},-xe^{-xy})=(0,0) \iff (x,y)=(0,0)$$ hence $(0,0)$ is the only possible interior point that could maximum or minimum, because the gradient always exists.

Any other possible point of extrema must lie on the boundary of $A$: notice that $e^{-t}$ is a decreasing function for any $t\in\mathbb{R}$, so you can evaluate the extrema of $g(x,y)=xy$ on $\{(x,y) \in \mathbb{R}^2 \ \text{s.t.} \ x^2+25y^2\le 4\}$ and then the minimum of $g$ will be the maximum of $f$ and vice-versa because of the monotonicity; again, the maximum and the minimum of $g$ exist on $A$ because $g$ is continuous and $A$ is compact.

The product $xy$ is minimum on $A$ when $(x,y)=(\sqrt{2},-\sqrt{2}/5)$ and $(x,y)=(-\sqrt{2},\sqrt{2}/5)$ and it is maximum when $(x,y)=(\sqrt{2},\sqrt{2}/5)$ and $(x,y)=(-\sqrt{2},-\sqrt{2}/5)$. All of these points lie on the boundary of $A$.

Hence, after doing the calculations for all these five points, the maximum of $f$ is at $(x,y)=(\sqrt{2},-\sqrt{2}/5)$ and $(x,y)=(-\sqrt{2},\sqrt{2}/5)$ and the minimum of $f$ is at $(x,y)=(\sqrt{2},\sqrt{2}/5)$ and $(x,y)=(-\sqrt{2},-\sqrt{2}/5)$. Explicitly, the maximum is $e^{2/5}$ and the minimum is $e^{-2/5}$.

Answer 2

For convenience, we substitute $x$ and $y$ into the polar coordinate system by such a system of equations: $$x = 2r\cos\theta \\ y=\frac25r\sin\theta \\ 0 \le r \le 1, 0\le \theta \lt 2\pi$$

Thus, $f(x, y) = e^{-xy} = e^{-\frac45r^2\sin\theta \cos\theta} =e^{-\frac{2}{5}r^2\sin2\theta}$, thus the question can be reduced to finding the extreme values of $g(r, \theta)={r^2\sin2\theta}(0 \le r \le 1, 0\le \theta \lt 2\pi)$.

It is not difficult to calculate that the maximum of $g(r, \theta)$ is $1$ when $(r, \theta) = (1, \frac{\pi}4)$ or $(1, \frac{5\pi}{4})$, and the minimum is $-1$ when $(r, \theta) = (1, \frac{3\pi}4)$ or $(1, \frac{7\pi}{4})$.

Since $f(x, y) =\left(e^{-\frac{2}{5}}\right)^{r^2\sin2\theta}=\left(e^{-\frac{2}{5}}\right)^{g(r, \theta)}$, the maximum of $f(x)$ is $e^{\frac25}$ and the minimum is $e^{-\frac25}$. Furthermore, by putting the values of $(r, \theta)$ into the equations above, we can get the coordinates we want. $f(x, y) = e^{\frac25} $ when $(x, y) = (\sqrt2, -{\frac{\sqrt2}5})$ or $(-\sqrt2, {\frac{\sqrt2}5})$, $f(x, y) = e^{-\frac{2}5} $ when $(x, y) = (\sqrt2, {\frac{\sqrt2}5})$ or $(-\sqrt2, -{\frac{\sqrt2}5})$.

Hope this answer is helpful to you.