let f(x,y,z) = distance^2
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Lagrange multipliers
2010-12-10 4:27 am
回答 (1)
2010-12-10 4:52 am
✔ 最佳答案
2 Let f(x)=x^2+y^2+z^2, g(x)=2x+y-3z-6L(x)=x^2+y^2+z^2+λ(2x+y-3z-6)∂L/∂x= 2x+2λ...(1)
∂L/∂y= 2y+λ...(2)
∂L/∂z= 2z-3λ...(3)
∂L/∂λ= 2x+y-3z-6...(4)
From (1-3) λ=-x=-2y=2z/3. Sub. into (4)
2(-λ)+(-λ/2)-3(3λ/2)-6=0
-7λ=6
λ=-6/7
So x=6/7,y=3/7,z=-9/7 and the point is (6/7,3/7,-9/7)
收錄日期: 2021-04-26 14:01:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101209000051KK01140