作者JohnMash (Paul)
看板Math
标题Re: [微积] 用Lagrange's multiplier求极值
时间Thu Apr 28 08:05:06 2011
※ 引述《love15 ( )》之铭言:
: 请教以下两题
: 1、已知 x^2-2xy+9y^2=1
: 求x^2+y^2之极值?
Sorry, I don't like Lagrange.
(x-y)^2+8y^2=1
x-y=cosθ,√8 y=sinθ
8x^2+8y^2=(√8 cosθ+sinθ)^2+sin^2θ
=8cos^2θ+2√8 cosθsinθ+2sin^2θ
=5(cos^2θ+sin^2θ)+3(cos^2θ-sin^2θ)+√8 sin2θ
=5+3cos2θ+√8 sin2θ
=5+√17 sin(2θ+α)
max(x^2+y^2)=(5+√17)/8
min(x^2+y^2)=(5-√17)/8
: 2、已知 x+y+z=1
: x^2+y^2+z^2=1
: 求x^3+y^3+z^3之极值?
2(xy+yz+zx)=(x+y+z)^2-(x^2+y+z^2)=0
Hence, x,y,z are the roots of f(u)
where
f(u)=u^3-u^2-xyz
Denote g(u)=u^3-u^2
g'(u)=3u^2-2u
g'(0)=g'(2/3)=0
g(0)=0, g(2/3)=8/27-4/9=-4/27
then
f(u) has three REAL roots if and only if -4/27≦xyz≦0
---------------
x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz
=1+3xyz
max(x^3+y^3+z^3)=1
min(x^3+y^3+z^3)=5/9
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 112.104.96.23
※ 编辑: JohnMash 来自: 112.104.96.23 (04/28 08:05)
1F:推 love15 :感谢 我了解了^^ 04/29 19:32