题目

14.求函数 (x,y)=(y+dfrac ({x)^3}(3))(e)^x+y 的极值.

题目解答

答案

解析

本题考查多元函数的偏导数以及多元函数的极值求解。解题思路是先求出函数的一阶偏导数，令一阶偏导数为零，解出驻点，再求出函数的二阶偏导数，根据二阶偏导数在驻点处的值判断判断驻点是否为极值点，并求出极值。

  1. 求一阶偏导数：
    • 对$f(x,y)=(y+\dfrac {{x}^{3}}{3}){e}^{xx+y}$关于$x$求偏导数，根据乘积法则$(uv)^\prime = u^\prime v + uv^\prime$，可得：
      $\dfrac {\partial f}{\partial x} = \dfrac{\partial}{\partial x}[(y+\dfrac {{x}^{3}}{3}){e}^{x+y}] = (y+\dfrac {3x^{2}}{3}){e}^{x+y}+(y+\dfrac {{x}^{3}}{3}){e}^{x+y} = ({x}^{2}+y+\dfrac {{x}^{3}}{3}){e}^{x+y}$
    • 对$f(x,y)$关于$y$求偏导数，同理可得：
      $\dfrac {\partial f}{\partial y} = \dfrac{\partial}{\partial y}[(y+\dfrac {{x}^{3}}{3}){e}^{x+y}] = {e}^{x+y}+(y+\dfrac {{x}^{3}}{3}){e}^{x+y} = (1+y+\dfrac {{x}^{3}}{3}){e}^{x+y}$
  2. 求驻点：
    令$\left \{ \begin{matrix} \dfrac {\partial f}{\partial x}=0\\ \rac {\partial f}{\partial y}=0\end{matrix} \right.$，即$\left \{ \begin{matrix} {x}^{2}+y+\dfrac {{x}^{3}}{3}=0\\ 1+y+\dfrac {{x^{2}}{3}=0.\end{matrix} \right.$
    • 由$1+y+\dfrac{x^{2}}{3}}{3}=0$可得$y = -1 - \dfrac{x^{2}}{3}$，将其代入${x}^{2}+y+\dfrac {{x}^{3}}{3}=0$中，得到：
      ${x}^{2}-1 - \dfrac{x^{2}}{3}+\dfrac {{x}^{3}}{3}=0$，即$\dfrac{2x^{2}}{3}-1 + \dfrac {{x}^{3}}{3}=0$，进一步化简为$2x^{2}-3 + x^{3}=0$。
    • 因式分解可得$(x + 1)(2x - 3 + x^{2}) = 0$，解得$x = -1$或$x = 1$。
    • 当$x = -1$时，$y = -1 - \dfrac{1}{3}=-\dfrac{4}{3}$；当$1) \(x = -1$时，$y = -1 - \dfrac{(-1)^{2}}{3}=-\dfrac{2}{3}$。
  3. 求二阶偏导数：
    • $\dfrac {{\partial }^{2}f}{\partial {x}^{2}} = \dfrac{\partial}{\partial x}[({x}^{2}+y+\dfrac {{x}^{3}}{3}){e}^{x+y}] = {e}^{x+y}(y+\dfrac {{x}^{3}}{3}+2{x}^{2}+2x)$
    • $\dfrac {{\partial }^{2}f}{\partial x\partial y} = \dfrac{\partial}{\partial y}[({x}^{2}+y+\dfrac {{x}^{3}}{}){e}^{x+y}] = {e}^{x+y}(y+\dfrac {{x}^{3}}{3}+{x}^{2}+1)$
    • $\dfrac {{\partial }^{2}f}{\partial {y}^{2}} = \dfrac{\partial}{\partial y}[(1+y+\dfrac {{x}^{3}}{}){e}^{x+y}] = {e}^{x+y}(y+\dfrac {{x}^{3}}{3}+2)$
  4. 判断驻点是否为极值点：
    • 当$x = -1$，$y = -\dfrac{2}{3}$时：
      $A = \dfrac {{\partial }^{2}f}{\partial {x}^{2}}\big|_{x=-1,y=-\frac{2\}} = -e^{-\frac{5}{3}}$，$B = \dfrac {{\partial }^{2}f}{\partial x\partial y}\big|_{x=-1,y=-\frac{2}{3}} = e^{-\frac{5}{3}}$，$C = \dfrac {{\partial }^{2}f}{\partial {y}^{2}}\big|_{x=-1,y=-\frac{2\}} = e^{-\frac{5}{3}}$
      $AC - B^{2} = (-e^{-\frac{5}{3}}) \cdot e^{-\frac{5}{3}} - (e^{-\frac{5}{3}})^2 = -e^{-\frac{10}{3} - e^{-\frac{10}{3}} = -2e^{-\frac{10}{3}} < 0}$，所以点$(-1,-\dfrac{2}{3})$不是$f(x,y)$的极值点。
    • 当$x = 1$，$y = -\dfrac{4}{3}$时
      $2) \(A = \dfrac {{\partial }^{2}f}{\partial {x}^{2}}\big|_{x=1,y=-\frac{4}{3}} = 3e^{-\frac{1}{3}}$，$B = \dfrac {{\partial }^{2}f}{\partial x\partial y}\big|_{x=1,y=-\frac{4}{3}} = e^{-\frac{1}{3}{}}$，$C = \dfrac {{\partial }^{2}f}{\partial {y}^{2}}\big|_{x=1,y=-\frac{4}{3}} = e^{-\frac{1}{3}}$
      $AC - B^{2} = 3e^{-\frac{2}{3}} \cdot 3e^{-\frac{1}{3}} - (e^{-\frac{1}{3}})^2 = 3e^{-\frac{2}{3}} - e^{-\frac{2}{3}} = 2e^{-\frac{2\}} > 0$，且$A = 3e^{-\frac{1}{3}} > 0$，所以点$(1,-\dfrac{4}{3})$是$f(x,y)$的极小值点。
  5. **求极小值：
    将$x = 1$，$y = -\dfrac{4}{3}$代入$f(x,y)$中，可得：
    $f(1,-\dfrac{4}{3}) = (-\dfrac{44}{}{3}+\dfrac{1}{3})e^{1 - \frac{4}{3}} = -e^{-\frac{1}{3}}$}})