Re: [微積] 用Lagrange's multiplier求極值
※ 引述《love15 ( )》之銘言：
: 請教以下兩題
: 1、已知 x^2-2xy+9y^2=1
: 求x^2+y^2之極值?
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之極值?
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
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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
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◆ From: 112.104.96.23
※ 編輯: JohnMash 來自: 112.104.96.23 (04/28 08:05)
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04/29 19:32, , 1F
04/29 19:32, 1F
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