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Bonus (Unexpected) solution to lagrange equation?

  1. Jul 25, 2012 #1
    1. The problem statement, all variables and given/known data

    The lagrange equations are obtained as in the picture. I am only showing the final part of the solution, where they consider the final case of x≠y≠z.

    2. Relevant equations

    The equation at the second paragraph is obtained by subtracting: (5.34 - 5.35).

    The final equations are obtained by dividing 5.37 by (x-y) throughout, same for the other 2. (Which is ok, since x - y ≠ 0)

    3. The attempt at a solution

    I understand their method, but why can't I just do this:

    (5.34) + (5.35) + (5.36)

    3(x2 + y2 + z2) + 2λ(x + y + z) + 3μ = 0

    Using the constraints,

    3(1) + 0 + 3λ = 0

    λ = -1

    Not sure if this is a appropriate solution..
     

    Attached Files:

  2. jcsd
  3. Jul 25, 2012 #2

    ehild

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    You mixed lambda with mu. μ=-1.

    ehild
     
  4. Jul 26, 2012 #3
    I see..is it possible to prove that μ=-1 leads to the equations being inconsistent?
     
  5. Jul 26, 2012 #4

    ehild

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    No, μ=-1 does not matter in the argument which proves that x,y,z can not be all different.

    Two of them can be equal and in this case, you would use μ=-1 to get λ and the possible values of x,y,z.

    ehild
     
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