题目
编号：A2016217
已知函数 $z =$ $z(x,y)$ 由方程 $(x^{2} +$ $y^{2}) z +$ $\ln z +$ $2(x + y + 1) = 0$ 确定，求 $z = z(x,y)$ 的极值.
解析
对式子 $(x^{2} +$ $y^{2}) z +$ $\ln z +$ $2(x + y + 1) = 0$ 中的变量 $x$ 求偏导，可得：
$$
2xz + (x^{2} + y^{2}) \frac{\partial z}{\partial x} + \frac{1}{z} \frac{\partial z}{\partial x} + 2 = 0.
$$
对式子 $(x^{2} +$ $y^{2}) z +$ $\ln z +$ $2(x + y + 1) = 0$ 中的变量 $y$ 求偏导，可得：
$$
2yz + (x^{2} + y^{2}) \frac{\partial z}{\partial y} + \frac{1}{z} \frac{\partial z}{\partial y} + 2 = 0.
$$
令：
$$
\left\{\begin{matrix}
\frac{\partial z}{\partial x} = 0;\\
\frac{\partial z}{\partial y} = 0.
\end{matrix}\right.
$$
则：
$$
\left\{\begin{matrix}
2xz + 2 = 0;\\
2yz + 2 = 0.
\end{matrix}\right.
\Rightarrow
$$
$$
\left\{\begin{matrix}
x = \frac{-1}{z};\\
y = \frac{-1}{z}.
\end{matrix}\right. ①
$$
将 $①$ 式代入到 $(x^{2} +$ $y^{2}) z +$ $\ln z +$ $2(x + y + 1) = 0$, 可得：
$$
\frac{2z}{z^{2}} + \ln z + \frac{-4}{z} + 2 = 0 \Rightarrow
$$
$$
\frac{2}{z} + \ln z + \frac{-4}{z} + 2 = 0 \Rightarrow
$$
$$
\ln z – \frac{2}{z} + 2 = 0 \Rightarrow
$$
$$
z = 1. ②
$$
结合 $①$ 式和 $②$ 式，可得：
$$
\left\{\begin{matrix}
x = -1;\\
y = -1.
\end{matrix}\right.
$$
接着：
$$
2xz + (x^{2} + y^{2}) \frac{\partial z}{\partial x} + \frac{1}{z} \frac{\partial z}{\partial x} + 2 = 0 \Rightarrow
$$
$$
继续对 x 求偏导 \Rightarrow
$$
$$
2(z + x \frac{\partial z}{\partial x}) + 2x \frac{\partial z}{\partial x} + (x^{2} + y^{2}) \frac{\partial ^{2} z}{\partial x ^{2}} – \frac{1}{z^{2}} (\frac{\partial z}{\partial x})^{2} + \frac{1}{z} \frac{\partial ^{2} z}{\partial x ^{2}} = 0 \Rightarrow
$$
$$
2z + 4x \frac{\partial z}{\partial x} + (x^{2} + y^{2}) \frac{\partial ^{2} z}{\partial x ^{2}} – \frac{1}{z^{2}} (\frac{\partial z}{\partial x})^{2} + \frac{1}{z} \frac{\partial ^{2} z}{\partial x ^{2}} = 0. ③
$$
$$
2xz + (x^{2} + y^{2}) \frac{\partial z}{\partial x} + \frac{1}{z} \frac{\partial z}{\partial x} + 2 = 0 \Rightarrow
$$
$$
继续对 y 求偏导 \Rightarrow
$$
$$
2x \frac{\partial z}{\partial y} + 2y \frac{\partial z}{\partial x} + (x^{2} + y^{2}) \frac{\partial ^{2} z}{\partial x \partial y} – \frac{1}{z^{2}} \frac{\partial z}{\partial y} \frac{\partial z}{\partial x} + \frac{1}{z} \frac{\partial ^{2} z}{\partial x \partial y} = 0. ④
$$
$$
2yz + (x^{2} + y^{2}) \frac{\partial z}{\partial y} + \frac{1}{z} \frac{\partial z}{\partial y} + 2 = 0 \Rightarrow
$$
$$
继续对 y 求偏导 \Rightarrow
$$
$$
2(z + y \frac{\partial z}{\partial y}) + 2y \frac{\partial z}{\partial y} + (x^{2} + y^{2}) \frac{\partial ^{2} z}{\partial y \partial y} – \frac{1}{z^{2}} (\frac{\partial z}{\partial y})^{2} + \frac{1}{z} \frac{\partial ^{2} z}{\partial y \partial y} = 0 \Rightarrow
$$
$$
2z + 4y \frac{\partial z}{\partial y} + (x^{2} + y^{2}) \frac{\partial ^{2} z}{\partial y \partial y} – \frac{1}{z^{2}} (\frac{\partial z}{\partial y})^{2} + \frac{1}{z} \frac{\partial ^{2} z}{\partial y \partial y} = 0. ⑤
$$
将 $x = -1$, $y = -1$, $\frac{\partial z}{\partial x} = 0$, $\frac{\partial z}{\partial y} = 0$, $z = 1$ 分别代入上面得到的 $③$, $④$, $⑤$ 式，可得：
$$
\left\{\begin{matrix}
2z + 4x \frac{\partial z}{\partial x} + (x^{2} + y^{2}) \frac{\partial ^{2} z}{\partial x ^{2}} – \frac{1}{z^{2}} (\frac{\partial z}{\partial x})^{2} + \frac{1}{z} \frac{\partial ^{2} z}{\partial x ^{2}} = 0;\\
2x \frac{\partial z}{\partial y} + 2y \frac{\partial z}{\partial x} + (x^{2} + y^{2}) \frac{\partial ^{2} z}{\partial x \partial y} – \frac{1}{z^{2}} \frac{\partial z}{\partial y} \frac{\partial z}{\partial x} + \frac{1}{z} \frac{\partial ^{2} z}{\partial x \partial y} = 0;\\
2z + 4y \frac{\partial z}{\partial y} + (x^{2} + y^{2}) \frac{\partial ^{2} z}{\partial y \partial y} – \frac{1}{z^{2}} (\frac{\partial z}{\partial y})^{2} + \frac{1}{z} \frac{\partial ^{2} z}{\partial y \partial y} = 0.
\end{matrix}\right.
\Rightarrow
$$
$$
\left\{\begin{matrix}
2 + 3 \cdot \frac{\partial ^{2} z}{\partial x^{2}} = 0;\\
3 \cdot \frac{\partial ^{2} z}{\partial x \partial y} = 0;\\
2 + 3 \cdot \frac{\partial ^{2} z}{\partial y \partial y} = 0.
\end{matrix}\right.
\Rightarrow
$$
$$
\left\{\begin{matrix}
A = \frac{\partial ^{2} z}{\partial x^{2}} = \frac{-2}{3};\\
B = \frac{\partial ^{2} z}{\partial x \partial y} = 0;\\
C = \frac{\partial ^{2} z}{\partial y \partial y} = \frac{-2}{3}.
\end{matrix}\right.
$$
由于 $AC- B^{2} = \frac{4}{9} > 0$, 于是，点 $(-1, -1)$ 是函数 $z(x, y)$ 的一个极值点。
又由于 $A = \frac{-2}{3} < 0$, 因此，点 $(-1, -1)$ 是函数 $z(x, y)$ 的一个极大值点，且极大值为：
$$
z = 1.
$$