多元极值问题

倪举鹏

实数满足a+b+c+d+e=8,a^2+b^2+c^2+d^2+e^2=16,求e的最大值。这里五个未知数换成其他多个未知数呢？

mathe

根据平均不等式有
\(\frac{16-e^2}4=\frac{a^2+b^2+c^2+d^2}4\ge(\frac{a+b+c+c}4)^2=(\frac{8-e}4)^2\)
于是得出\(5e^2-16e\le 0\),所以\(0\le e\le \frac{16}5\)

倪举鹏

W = e + \[Lambda]*(8 - a - b - c - d - e) + \[Mu]*(16 - a^2 - b^2 -
     c^2 - d^2 - e^2)
Solve[{\!\ (
\*SubscriptBox[\ (\ [PartialD]\), \ (a\)]W\) == 0, \!\ (
\*SubscriptBox[\ (\ [PartialD]\), \ (b\)]W\) == 0, \!\ (
\*SubscriptBox[\ (\ [PartialD]\), \ (c\)]W\) == 0, \!\ (
\*SubscriptBox[\ (\ [PartialD]\), \ (d\)]W\) == 0, \!\ (
\*SubscriptBox[\ ( \ [PartialD]\), \ (e\)]W\) == 0,
  8 - a - b - c - d - e == 0,
  16 - a^2 - b^2 - c^2 - d^2 - e^2 == 0}, {a, b, c, d,
  e, \ [Lambda], \ [Mu]}]
{{a -> 2, b -> 2, c -> 2, d -> 2,
  e -> 0, \ [Lambda] -> 1, \ [Mu] -> -(1/4)}, {a -> 6/5, b -> 6/5,
  c -> 6/5, d -> 6/5, e -> 16/5, \ [Lambda] -> -(3/5), \ [Mu] -> 1/4}}