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Lagrange Multiplier Method

  1. Apr 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the stationary value of

    $$ f(u,v,w) = \left( \frac{c}{u} \right)^m + \left( \frac{d}{v} \right)^m + \left( \frac{e}{w} \right)^m $$

    Constraint: $$ u^2 + v^2 + w^2 = t^2 $$

    Note: $$ u, v, w > 0 $$. $$ c,d, e, t > 0 $$. $$ m > 0 $$ and is a positive integer.


    2. Relevant equations

    I have found the auxiliary function:

    $$ F = \left( \frac{c}{u} \right)^m + \left( \frac{d}{v} \right)^m + \left( \frac{e}{w} \right)^m - \lambda ( u^2 + v^2 + w^2 - t^2) $$

    and

    $$ F_u = \frac{- m c^m}{ u^{m+1}} - 2\lambda u = 0 $$

    $$ F_v = \frac{- m d^m}{ v^{m+1}} - 2\lambda v = 0 $$

    $$ F_w = \frac{- m e^m}{ w^{m+1}} - 2\lambda w = 0 $$



    3. The attempt at a solution

    Its after this I am having difficult. I don't understand what to do. I assume I have to rearrange the above equations to get $$ u^2, v^2, w^2 $$ and substitute into the constraint, but I can't see how to do that because I get expressions for $$ u^{m + 2} , v^{m+2} , w^{m+2} $$ . Help please.







    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 6, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Okay, that's good. What I would do is switch all of those "[itex]\lambda[/itex]" terms to the right:
    [itex]\frac{-m c^m}{u^{m+1}}= 2\lambda u[/itex]
    [itex]\frac{-m d^m}{v^{m+1}}= 2\lambda v[/itex]
    [itex]\frac{-m d^m}{w^{m+1}}= 2\lambda w[/itex]

    (In fact, to find extrema of f with constraint g= constant, I tend to think of the Lagrange multiplier condition as [itex]\nabla f= \lambda \nabla g[/itex] rather than using the "auxiliary function" [itex]f+ \lambda g[/itex]. It gives the same result, of course.)

    Now, since a particular value of [itex]\lambda[/itex] is not part of the solution start by eliminating [itex]\lambda[/itex] by dividing one equation by another. Dividing the first equation by the second, for example, gives
    [tex]\left(\frac{c}{d}\right)^m\left(\frac{v}{u}\right)^{m+1}= \frac{u}{v}[/tex]
    which gives
    [tex]\left(\frac{v}{u}\right)^{m+2}= \left(\frac{d}{c}\right)^m[/tex]
    and similarly for the other equations.
     
  4. Apr 6, 2012 #3

    Ray Vickson

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    Homework Helper

    You have expressions for [itex] u^{m + 2} , v^{m+2} , w^{m+2}[/itex] in terms of λ. You can take (m+2) roots, to get [itex] u, v, w[/itex]. You can then substitute these expressions into the constraint, to get a single, simple equation for λ. Actually, it is easier to re-write everything in terms of μ = -λ; that will eliminate a lot of minus signs in the equations and solutions.

    RGV
     
  5. Apr 6, 2012 #4
    Thank you both. I'm understanding this problem now.
     
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