【例8】设f(x,y)的全微分df(x,y)=-(1+ae^y)sinxdx+e^ay(acosx-1-ay)dy,且f(0,0)=2,(a≠0).(I)求a的值及f(x,y);(II)求f(x,y)的极值.

**解：** **(I) 求 $ a $ 的值及 $ f(x, y) $** 由全微分条件： $$ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \Rightarrow -ae^y \sin x = -ae^{ay} \sin x \Rightarrow a = 1 $$ 代入得： $$ df = -(1 + e^y) \sin x \, dx + e^y (\cos x - 1 - y) \, dy $$ 积分求 $ f(x, y) $： $$ f(x, y) = (1 + e^y) \cos x - y e^y + C $$ 由 $ f(0, 0) = 2 $，得 $ C = 0 $，故： $$ f(x, y) = (1 + e^y) \cos x - y e^y $$ **(II) 求极值** 求偏导并令为零： $$ f_x = -(1 + e^y) \sin x = 0 \Rightarrow x = k\pi $$ $$ f_y = e^y (\cos x - 1 - y) = 0 \Rightarrow y = \cos x - 1 $$ 解得临界点：$ (2n\pi, 0) $ 和 $ ((2n+1)\pi, -2) $ 二阶偏导： $$ f_{xx} = -(1 + e^y) \cos x, \quad f_{xy} = -e^y \sin x, \quad f_{yy} = e^y (\cos x - 2 - y) $$ 对于 $ (2n\pi, 0) $： $$ AC - B^2 = 2 > 0, \quad A < 0 \Rightarrow \text{极大值 } 2 $$ 对于 $ ((2n+1)\pi, -2) $： $$ AC - B^2 < 0 \Rightarrow \text{非极值点} $$ **答案：** \[ \boxed{ \begin{array}{ll} \text{(I)} & a = 1, \quad f(x, y) = (1 + e^y) \cos x - y e^y \\ \text{(II)} & \text{极大值 } 2 \text{，在点 } (2k\pi, 0) \text{ 处取得，} k \in \mathbb{Z} \end{array} } \]