접평면과 선형근사

Let $(x_0,y_0,z_0)$ be any point on the surface. A vector normal to the surface at $(x_0,y_0,z_0)$ is

$$\begin{aligned} &{ \boldsymbol{\nabla} }\Big(xy e^{-(x^2+y^2)/2}-z\Big)\bigg|_{(x_0,y_0,z_0)} \\& =\left< y_0 e^{-(x_0^2+y_0^2)/2}-x_0^2y_0 e^{-(x_0^2+y_0^2)/2}, x_0 e^{-(x_0^2+y_0^2)/2}-x_0y_0^2 e^{-(x_0^2+y_0^2)/2},-1\right> \end{aligned}$$

The tangent plane to the surface at $(x_0,y_0,z_0)$ is horizontal if and only if this vector is vertical, which is the case if and only if its $x$- and $y$-components are zero, which in turn is the case if and only if

$$\begin{aligned} &y_0(1-x_0^2)=0\text{ and }x_0(1-y_0^2)=0\\ &\iff\big\{y_0=0\text{ or }x_0=1\text{ or }x_0=-1\big\} \text{ and }\big\{x_0=0\text{ or }y_0=1\text{ or }y_0=-1\big\}\\ &\iff (x_0,y_0)=(0,0)\text{ or }(1,1)\text{ or }(1,-1) \text{ or }(-1,1)\text{ or }(-1,-1) \end{aligned}$$

The values of $z_0$ at these points are $0$, $e^{-1}$, $-e^{-1}$, $-e^{-1}$ and $e^{-1}$, respectively. So the horizontal tangent planes are $z=0$, $z=e^{-1}$ and $z=-e^{-1}$. At the highest and lowest points of the surface, the tangent plane is horizontal. So the largest and smallest values of $z$ are $e^{-1}$ and $-e^{-1}$, respectively.