（一）拉格朗日乘子法——分析推导

如果z=f(x,y)$z=f(x,y)$在条件g(x,y)=0$g(x,y)=0$的条件下，在点(x0,y0)$(x_0,y_0)$取得极值，如下图所示。

那么，f(x,y)$f(x,y)$的梯度与g(x,y)$g(x,y)$的梯度平行，即向量(fx′(x0,y0),fy′(x0,y0))$({f_x}^{\prime }(x_0,y_0),{f_y}^{\prime }(x_0,y_0))$与向量(gx′(x0,y0),gy′(x0,y0))$({g_x}^{\prime }(x_0,y_0),{g_y}^{\prime }(x_0,y_0))$平行。所以有

fx′(x0,y0)gx′(x0,y0)=fy′(x0,y0)gy′(x0,y0)=−λ0(1)$$\frac{{f_x}^{\prime }(x_0,y_0)}{{g_x}^{\prime }(x_0,y_0)}=\frac{{f_y}^{\prime }(x_0,y_0)}{{g_y}^{\prime }(x_0,y_0)}=-{\lambda }_0(1)$$

即

⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪fx′(x0,y0)+λ0gx′(x0,y0)=0fy′(x0,y0)+λ0gy′(x0,y0)=0g(x0,y0)=0(2)$$\{\begin{array}{c}{f_x}^{\prime }(x_0,y_0)+{\lambda }_0{g_x}^{\prime }(x_0,y_0)=0\\ \\ {f_y}^{\prime }(x_0,y_0)+{\lambda }_0{g_y}^{\prime }(x_0,y_0)=0\\ \\ g(x_0,y_0)=0\end{array}(2)$$

引入辅助函数L(x,y,λ)=f(x,y)+λg(x,y)$L(x,y,\lambda )=f(x,y)+\lambda g(x,y)$，对该函数求偏导得

⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪∂L(x,y,λ)∂x=fx′(x,y)+λgx′(x,y)=0∂L(x,y,λ)∂y=fy′(x,y)+λgy′(x,y)=0∂L(x,y,λ)∂λ=g(x,y)=0(3)$$\{\begin{array}{c}\frac{\mathrm{\partial }L(x,y,\lambda )}{\mathrm{\partial }x}={f_x}^{\prime }(x,y)+\lambda {g_x}^{\prime }(x,y)=0\\ \\ \frac{\mathrm{\partial }L(x,y,\lambda )}{\mathrm{\partial }y}={f_y}^{\prime }(x,y)+\lambda {g_y}^{\prime }(x,y)=0\\ \\ \frac{\mathrm{\partial }L(x,y,\lambda )}{\mathrm{\partial }\lambda }=g(x,y)=0\end{array}(3)$$

求偏导的结果显示，公式(3)实际上就是公式(2)，这意味着求条件极值问题可以转化为无条件极值问题求解。其中，辅助函数L(x,y,λ)=f(x,y)+λg(x,y)$L(x,y,\lambda )=f(x,y)+\lambda g(x,y)$是拉格朗日函数。即求z=f(x,y)$z=f(x,y)$在条件g(x,y)=0$g(x,y)=0$的条件下，在点(x0,y0)$(x_0,y_0)$取得极值的问题可以转化为求解拉格朗日函数L(x,y,λ)=f(x,y)+λg(x,y)$L(x,y,\lambda )=f(x,y)+\lambda g(x,y)$偏导数为0的解.