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Lagrangian multipliers

  1. Mar 6, 2009 #1
    How does lagrange multipliers work?
    i was able to work out this proof of the idea, but its only true for a function with two independent variables and one dependent variable.

    Rn=the space that is the independent variables.

    x[Rn]=x
    C[Rn]=C=constant.


    dx/d[Rn]=grad(x)*v; v is a unit vector
    dC/d[Rn]=grad(c)*v

    because C is held constant, dC/d[Rn]=0 everywhere.
    because cos(pi/2)=0, **grad(C) is perpendicular to v.**

    In order for extrema to exist, dx/d[Rn]=0. grad(x)*v is zero meaning **grad(x) is perpendicular to v.**

    in the case Rn=R2:
    both grad(x) and grad(c) are perpendiular to v, it means grad(x) must be parallel to grad(c).

    That is the requirement given by the system
    grad(x)=L*grad(c)
    C[Rn]=C
    where L is the scalar multiplier (upside down y).

    but it seems as though this is only true for the R2 case. in 3 dimnensions, if both grad(x) and grad(c) are perpendicular to v, it doesnt necessarily mean grad(x) is parallel to grad(c). It seems like im missing something.


    How do i extend this to more than two independent variables?
     
  2. jcsd
  3. Mar 6, 2009 #2

    HallsofIvy

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    What do you mean by "dx/d[Rn]" where Rn is n dimensional Euclidean space? I don't believe that is standard notation.
     
  4. Mar 6, 2009 #3
    i know it isnt
    [Rn]=x,y,z, etc
    basically i meant the directional derivative by dx/d[Rn]"
     
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