2026年考研数学一第17题

答案: 见解析

解析:

【解】由题可知，$f_{x}^{\prime}=4 x \mathrm{e}^{x}+\left(2 x^{2}-y^{2}\right) \mathrm{e}^{x}, f_{y}^{\prime}=-2 y \mathrm{e}^{x}$ ．

令 $\left{\begin{array}{l}f_{x}^{\prime}=0 \ f_{y}^{\prime}=0\end{array} \Rightarrow\left{\begin{array}{l}x=0, \ y=0,\end{array}\right.\right.$ 或 $\left{\begin{array}{l}x=-2, \ y=0 .\end{array}\right.$

又 $A=f_{x x}^{\prime \prime}=\mathrm{e}^{x}\left(2 x^{2}+8 x+4-y^{2}\right), B=f_{x y}^{\prime \prime}=-2 y \mathrm{e}^{y}, C=f_{y y}^{\prime \prime}=-2 \mathrm{e}^{x}$ ，故 $A C-\left.B^{2}\right|{(0,0)}=-8<0, A C-\left.B^{2}\right|{(-2,0)}=8 \mathrm{e}^{-4}>0$ 且 $A_{(-2,0)}=-4 \mathrm{e}^{-2}<0, \mid$则 $(0,0)$ 为非极值点，$(-2,0)$ 为极大值点且极大值 $f(-2,0)=\displaystyle\frac{8}{\mathrm{e}^{2}}$ ．