可以，但没必要，甚至更耗时间。
第一题\begin{aligned}
\text{let } f(x,y) &= \frac 4 {x+y} +\lambda (x+4y-xy-0) \\
\therefore \frac{ \partial f(x,y)} {\partial x} &= \frac{-4}{(x+y)^2} + \lambda(1-y) \\
\frac{ \partial f(x,y)} {\partial y} &= \frac{-4}{(x+y)^2} + \lambda(4-x) \\
\therefore &
\left\{
\begin{aligned}
\frac{-4}{(x+y)^2} + \lambda(1-y) &= 0 \\
\frac{-4}{(x+y)^2} + \lambda(4-x) &= 0 \\
x+4y-xy &= 0
\end{aligned}
\right. \\
\Rightarrow & \left\{
\begin{aligned}
1-y=4-x \\
x+4y-xy &= 0 \\
\end{aligned}
\right. \\
\therefore &
\left\{
\begin{aligned}
x & = 6 \\
y & = 3
\end{aligned}
\right. \qquad
\left\{
\begin{aligned}
x & = 2 \\
y & = -1
\end{aligned}
\right. \\
\text{ps}&\text{，应该求出}\lambda\text{但是我懒}\\
\because &x,y > 0 \\
\therefore &x = 6,\quad y =3 \\
\text{the } &\text{max is } \frac 4 9
\end{aligned}第二题第二题就当成练习，格式已经写好，填空就能得到正确答案。这样有助于对于拉格朗日乘子法的计算量有个理解。\begin{aligned}
\text{let } f(x,y) &= A +\lambda (B-C) \\
\therefore \frac{ \partial f(x,y)} {\partial x} &= \\
\frac{ \partial f(x,y)} {\partial y} &= \\
\therefore &
\left\{
\begin{aligned}
F_x:\quad &\frac{ \partial f(x,y)} {\partial x} &=&\ 0 \\
F_y:\quad &\frac{ \partial f(x,y)} {\partial y} &=&\ 0 \\
F_{\lambda}: \quad & B-C &=&\ 0
\end{aligned}
\right. \\
\therefore &
\left\{
\begin{aligned}
x & =
\\
y & =
\\
\lambda &=
\end{aligned}
\right.
\\
\text{the } &\text{max/min is }
\end{aligned}然后对于第二题来说，正常方法，即题目所说的常数替换，会简单到只需要一行：\displaystyle \frac 1 m + \frac 1 n =(2m+n)(\frac {1}{m} + \frac{1}{n}) = 3 + \frac n m + \frac m n \ge 3+2\sqrt2<--
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