극값과 안장점

The maximum must be either on the boundary of $D$ or in the interior of $D$.

• On the circular part of the boundary $r=2$, $0\le\theta\le\frac{\pi}{2}$ (in polar coordinates) so that $f=r^2\cos\theta\sin\theta e^{-r^2/2}=2\sin(2\theta)e^{-2}$. This has a maximum value of $2e^{-2}$ at $\theta=\frac{\pi}{4}$ or $x=y=\sqrt{2}$.
• On the two flat parts of the boundary $x=0$ or $y=0$ so that $f=0$.
• If the maximum is in the interior of $D$, it must be at a critical point of $f(x,y)$. Since

$$\begin{aligned} f_x(x,y)=e^{-(x^2 + y^2) / 2}\big[y-x^2y\big]\qquad f_y(x,y)=e^{-(x^2 + y^2) / 2}\big[x-xy^2\big] \end{aligned}$$

$(x,y)$ is a critical point if and only if

$$\begin{aligned} &y(1-x^2)=0\text{ and }x(1-y^2)=0 \\ &\iff \{y=0\text{ or }x=1\text{ or }x=-1\}\text{ and } \{x=0\text{ or }y=1\text{ or }y=-1\} \end{aligned}$$

There are two critical points with $x,y\ge 0$, namely $(0,0)$ and $(1,1)$. The first of these is on the boundary of $D$ and the second is in the interior of $D$.

Here is a table giving all candidates for the maximum:

$$\begin{array}{|c|c|}\hline (x,y) & g(x,y) \\ \hline (\sqrt{2},\sqrt{2}) & 2e^{-2}\approx0.271 \\ \hline (x,0) & 0 \\ \hline (0,y) & 0 \\ \hline (1,1) & e^{-1}\approx0.368 \\ \hline \end{array}$$

Since $e>2$, we have that $2e^{-2}=e^{-1}\frac{2}{e}<e^{-1}$ and the largest value is $e^{-1}$.