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：

$$\{\begin{array}{c}\frac{\partial z}{\partial x}=0;\\ \frac{\partial z}{\partial y}=0.\end{array}$$

则：

$$\{\begin{array}{c}2xz+2=0;\\ 2yz+2=0.\end{array}\Rightarrow$$

$$\{\begin{array}{c}x=\frac{-1}{z};\\ y=\frac{-1}{z}.\end{array}\mathrm{①}$$

将 $\mathrm{①}$ 式代入到 $(x^2+$ $y^2)z+$ $\ln z+$ $2(x+y+1)=0$, 可得：

$$\frac{2z}{z^2}+\ln z+\frac{-4}{z}+2=0\Rightarrow$$

$$\frac{2}{z}+\ln z+\frac{-4}{z}+2=0\Rightarrow$$

$$\ln z–\frac{2}{z}+2=0\Rightarrow$$

$$z=1.\mathrm{②}$$

结合 $\mathrm{①}$ 式和 $\mathrm{②}$ 式，可得：

$$\{\begin{array}{c}x=-1;\\ y=-1.\end{array}$$

接着：

$$2xz+(x^2+y^2)\frac{\partial z}{\partial x}+\frac{1}{z}\frac{\partial z}{\partial x}+2=0\Rightarrow$$

$$\mathrm{继}\mathrm{续}\mathrm{对}x\mathrm{求}\mathrm{偏}\mathrm{导}\Rightarrow$$

$$2(z+x\frac{\partial z}{\partial x})+2x\frac{\partial z}{\partial x}+(x^2+y^2)\frac{{\partial }^{2}z}{\partial x^2}–\frac{1}{z^2}(\frac{\partial z}{\partial x}{)}^2+\frac{1}{z}\frac{{\partial }^{2}z}{\partial x^2}=0\Rightarrow$$

$$2z+4x\frac{\partial z}{\partial x}+(x^2+y^2)\frac{{\partial }^{2}z}{\partial x^2}–\frac{1}{z^2}(\frac{\partial z}{\partial x}{)}^2+\frac{1}{z}\frac{{\partial }^{2}z}{\partial x^2}=0.\mathrm{③}$$

$$2xz+(x^2+y^2)\frac{\partial z}{\partial x}+\frac{1}{z}\frac{\partial z}{\partial x}+2=0\Rightarrow$$

$$\mathrm{继}\mathrm{续}\mathrm{对}y\mathrm{求}\mathrm{偏}\mathrm{导}\Rightarrow$$

$$2x\frac{\partial z}{\partial y}+2y\frac{\partial z}{\partial x}+(x^2+y^2)\frac{{\partial }^{2}z}{\partial x\partial y}–\frac{1}{z^2}\frac{\partial z}{\partial y}\frac{\partial z}{\partial x}+\frac{1}{z}\frac{{\partial }^{2}z}{\partial x\partial y}=0.\mathrm{④}$$

$$2yz+(x^2+y^2)\frac{\partial z}{\partial y}+\frac{1}{z}\frac{\partial z}{\partial y}+2=0\Rightarrow$$

$$\mathrm{继}\mathrm{续}\mathrm{对}y\mathrm{求}\mathrm{偏}\mathrm{导}\Rightarrow$$

$$2(z+y\frac{\partial z}{\partial y})+2y\frac{\partial z}{\partial y}+(x^2+y^2)\frac{{\partial }^{2}z}{\partial y\partial y}–\frac{1}{z^2}(\frac{\partial z}{\partial y}{)}^2+\frac{1}{z}\frac{{\partial }^{2}z}{\partial y\partial y}=0\Rightarrow$$

$$2z+4y\frac{\partial z}{\partial y}+(x^2+y^2)\frac{{\partial }^{2}z}{\partial y\partial y}–\frac{1}{z^2}(\frac{\partial z}{\partial y}{)}^2+\frac{1}{z}\frac{{\partial }^{2}z}{\partial y\partial y}=0.\mathrm{⑤}$$

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 $\mathrm{③}$, $\mathrm{④}$, $\mathrm{⑤}$ 式，可得：

$$\{\begin{array}{c}2z+4x\frac{\partial z}{\partial x}+(x^2+y^2)\frac{{\partial }^{2}z}{\partial x^2}–\frac{1}{z^2}(\frac{\partial z}{\partial x}{)}^2+\frac{1}{z}\frac{{\partial }^{2}z}{\partial x^2}=0;\\ 2x\frac{\partial z}{\partial y}+2y\frac{\partial z}{\partial x}+(x^2+y^2)\frac{{\partial }^{2}z}{\partial x\partial y}–\frac{1}{z^2}\frac{\partial z}{\partial y}\frac{\partial z}{\partial x}+\frac{1}{z}\frac{{\partial }^{2}z}{\partial x\partial y}=0;\\ 2z+4y\frac{\partial z}{\partial y}+(x^2+y^2)\frac{{\partial }^{2}z}{\partial y\partial y}–\frac{1}{z^2}(\frac{\partial z}{\partial y}{)}^2+\frac{1}{z}\frac{{\partial }^{2}z}{\partial y\partial y}=0.\end{array}\Rightarrow$$

$$\{\begin{array}{c}2+3\cdot \frac{{\partial }^{2}z}{\partial x^2}=0;\\ 3\cdot \frac{{\partial }^{2}z}{\partial x\partial y}=0;\\ 2+3\cdot \frac{{\partial }^{2}z}{\partial y\partial y}=0.\end{array}\Rightarrow$$

$$\{\begin{array}{c}A=\frac{{\partial }^{2}z}{\partial x^2}=\frac{-2}{3};\\ B=\frac{{\partial }^{2}z}{\partial x\partial y}=0;\\ C=\frac{{\partial }^{2}z}{\partial y\partial y}=\frac{-2}{3}.\end{array}$$

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：

$$z=1.$$