求函数 $f(x, y)=\left(y-x^2\right)\left(y-x^3\right)$ 的极值.
【答案】 $\left\{\begin{array}{l}f_x^{\prime}=-x\left(2 y+3 x y-5 x^3\right)=0 \\ f_y^{\prime}=2 y-x^2-x^3=0\end{array}\right.$, 得驻点为 $(0,0),(1,1),\left(\frac{2}{3}, \frac{10}{27}\right)$.
$$
f_{x x}^{\prime \prime}=-\left(2 y+3 x y-5 x^3\right)-x\left(3 y-15 x^2\right), f_{x y}^{\prime \prime}=-x(2+3 x), f_{y y}^{\prime \prime}=2 \text {. }
$$
代入 $(0,0),\left\{\begin{array}{l}A=f_{x x}^{\prime \prime}=0 \\ B=f_{x y}^{\prime \prime}=0, \\ C=f_{y y}^{\prime \prime}=2\end{array}\right.$,则 $A C-B^2=0$, 故充分条件失效, 当 $x \rightarrow 0$ 时, 取
$$
y=x^2+k x^3(k > 0), \quad f(x, y)=\left(y-x^2\right)\left(y-x^3\right)=k x^3\left[x^2+(k-1) x^3\right]=k x^5+o\left(x^5\right),
$$

则 $\lim _{x \rightarrow 0} \frac{f(x, y)}{x^5}=\lim _{x \rightarrow 0} \frac{k x^5+o\left(x^5\right)}{x^5}=k > 0$, 由极限的局部保号性: 存在 $\delta > 0$, 当 $x \in(-\delta, 0)$ 时, $\frac{f(x, y)}{x^5} > 0, f(x, y) < 0=f(0,0)$, 当 $x \in(0, \delta)$ 时, $\frac{f(x, y)}{x^5} > 0$, $f(x, y) > 0=f(0,0)$, 故 $(0,0)$ 不是极值点;
代入 $(1,1),\left\{\begin{array}{l}A=f_{x x}^{\prime \prime}=12 \\ B=f_{x y}^{\prime \prime}=-5, \\ C=f_{y y}^{\prime \prime}=2\end{array}\right.$ 则 $A C-B^2 < 0$, 故 $(1,1)$ 不是极值点;
代入 $\left(\frac{2}{3}, \frac{10}{27}\right)$ 的 $\left\{\begin{array}{l}A=f_{x=}^{\prime \prime}=\frac{100}{27} \\ B=f_{w y}^{\prime \prime}=-\frac{8}{3} \\ C=f_{y y}^{\prime \prime}=2\end{array}\right.$, 则 $A C-B^2 > 0$ 且 $A > 0$, 故 $\left(\frac{2}{3}, \frac{10}{27}\right)$ 是极小值点;
故 $f\left(\frac{2}{3}, \frac{10}{27}\right)=-\frac{4}{729}$ 为极小值.


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