题号:1419    题型:解答题    来源:2004年全国硕士研究生招生考试试题
设 $z=z(x, y)$ 是由 $x^{2}-6 x y+10 y^{2}-2 y z-z^{2}+18=0$ 确定的函数, 求 $z=z(x, y)$ 的极值点 和极限
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答案:
因为 $x^{2}-6 x y+10 y^{2}-2 y z-z^{2}+18=0$, 所以
两边对 $x$ 求导: $2 x-6 y-2 y \frac{\partial z}{\partial x}-2 z \frac{\partial z}{\partial x}=0$,
两边对 $y$ 求导: $-6 x+20 y-2 z-2 y \frac{\partial z}{\partial y}-2 z \frac{\partial z}{\partial y}=0$.

根据极值点存在的充分条件, 令 $\left\{\begin{array}{l}\frac{\partial z}{\partial x}=0 \\ \frac{\partial z}{\partial y}=0\end{array}\right.$, 得 $\left\{\begin{array}{c}x-3 y=0 \\ -3 x+10 y-z=0\end{array}\right.$, 故 $\left\{\begin{array}{c}x=3 y, \\ z=y .\end{array}\right.$ 对照极值点存在的充分条件, 为判别两点是否为极值点, 再(1)分别对 $x, y$ 求偏导数, (2) 分别对 $x, y$ 求偏导数
(1)式对 $x$ 求导: $\quad 2-2 y \frac{\partial^{2} z}{\partial x^{2}}-2\left(\frac{\partial z}{\partial x}\right)^{2}-2 z \frac{\partial^{2} z}{\partial x^{2}}=0$,
(2)式对 $x$ 求导: $-6-2 \frac{\partial z}{\partial x}-2 y \frac{\partial^{2} z}{\partial x \partial y}-2 \frac{\partial z}{\partial y} \cdot \frac{\partial z}{\partial x}-2 z \frac{\partial^{2} z}{\partial x \partial y}=0$,
(1)式对 $y$ 求导: $\quad-6-2 \frac{\partial z}{\partial x}-2 y \frac{\partial^{2} z}{\partial x \partial y}-2 \frac{\partial z}{\partial y} \cdot \frac{\partial z}{\partial x}-2 z \frac{\partial^{2} z}{\partial x \partial y}=0$,
(2)式对 $y$ 求导: $\quad 20-2 \frac{\partial z}{\partial y}-2 \frac{\partial z}{\partial y}-2 y \frac{\partial^{2} z}{\partial y^{2}}-2\left(\frac{\partial z}{\partial y}\right)^{2}-2 z \frac{\partial^{2} z}{\partial y^{2}}=0$,

将 $\left\{\begin{array}{l}x=9, \\ y=3, \\ z=3\end{array}\left\{\begin{array}{l}\frac{\partial z}{\partial x}=0, \\ \frac{\partial z}{\partial y}=0\end{array}\right.\right.$ 代入, 于是 $A=\left.\frac{\partial^{2} z}{\partial x^{2}}\right|_{(9,3,3)}=\frac{1}{6}, B=\left.\frac{\partial^{2} z}{\partial x \partial y}\right|_{(9,3,3)}=-\frac{1}{2},$, $C=\left.\frac{\partial^{2} z}{\partial y^{2}}\right|_{(9,3,3)}=\frac{5}{3}$, 故 $A C-B^{2}=\frac{1}{36} > 0$, 又 $A=\frac{1}{6} > 0$, 从而点 $(9,3)$ 是 $z(x, y)$ 的极小 值点, 极小值为 $z(9,3)=3$. $B=\left.\frac{\partial^{2} z}{\partial x \partial y}\right|_{(-9,-3,-3)}=\frac{1}{2}, C=\left.\frac{\partial^{2} z}{\partial y^{2}}\right|_{(-9,-3,-3)}=-\frac{5}{3}$, 可知 $A C-B^{2}=\frac{1}{36} > 0$, 又 $A=-\frac{1}{6} < 0$, 从而点 $(-9,-3)$ 是 $z(x, y)$ 的极大值点, 极大值为 $z(-9,-3)=-3$.
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