2016年考研数二第17题解析:利用偏导数求函数极值

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:

$$
z = 1.
$$


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