# 求解二元隐函数的极值

## 二、解析

$$x^{2}+y^{2}+z^{2}-2 x-2 y-4 z-10=0 \Rightarrow$$

$$2 x+2 z \cdot z_{x}^{\prime}-2-4 z_{x}^{\prime}=0 \quad ①$$

$$2 y+2 z \cdot z^{\prime} _{y} -2-4 z^{\prime} _{y} = 0 \quad ②$$

$$\left\{\begin{array}{l}z_{x}^{\prime}=0 \\ z^{\prime} _{y} = 0\end{array} \Rightarrow \left\{\begin{array}{l}2 x-2=0 \\ 2 y-2=0\end{array} \Rightarrow \left\{\begin{array}{l}x=1 \\ y=1\end{array}\right.\right.\right.$$

$$x^{2}+y^{2}+z^{2}-2 x-2 y-4 z-10=0 \Rightarrow$$

$$1+1+z^{2}-2-2-4 z-10=0 \Rightarrow$$

$$z^{2}-4 z-12=0 \Rightarrow$$

$$z=\frac{4 \pm \sqrt{16+48}}{2} \Rightarrow$$

$$z=\frac{4 \pm 8}{2} \Rightarrow$$

$$z>0 \Rightarrow z=\frac{8+4}{2}=6.$$

$$2+2 z_{x}^{\prime} \cdot z_{x}^{\prime}+2 z \cdot z_{x x}^{\prime \prime}-4 z_{x x}^{\prime \prime}=0$$

$$2+(12-4) z_{x x}^{\prime \prime}=0 \Rightarrow$$

$$A=z_{x x}^{\prime \prime}=\frac{-2}{8}=\frac{-1}{4}<0$$

$$z(1, 1) = 6.$$

Next

$$2 z_{y}^{\prime} \cdot z_{x}^{\prime}+2 z \cdot z_{x y}^{\prime \prime}-4 z_{x y}^{\prime \prime}=0 \Rightarrow$$

$$z_{x}^{\prime}=0, z_{y}^{\prime}=0, z=6 \Rightarrow$$

$$B = z_{x y}^{\prime \prime}=0.$$

$$2+2 z^{\prime} _{y} \cdot z_{y}^{\prime}+2 z \cdot z_{y y}^{\prime \prime}-4 z_{y y}^{\prime \prime}=0 \Rightarrow$$

$$z_{y}^{\prime}=0, z=6 \Rightarrow$$

$$2+8 z_{y y}^{\prime \prime}=0 \Rightarrow$$

$$C=z_{y y}^{\prime \prime}=\frac{-1}{4}$$

$$A C-B^{2}=\left(\frac{-1}{4}\right)\left(\frac{-1}{4}\right)-0>0$$