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It appears that the Lagrange Multiplier is failing due to different values of Lamda.

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Use Lagrange multipliers to ¯nd the maximum and mini-
    mum value(s), if they exist, of
    f(x; y; z) = x^2 -2y + 2z^2
    subject to the constraint x^2+y^2+z^2



    2. Relevant equations



    3. The attempt at a solution

    Basically after I find the gradient of the functions I get this.

    2x=2x lamda
    -2=2y lamda
    4z=2z lamda.

    One case lamda equals 1 while the other it equals 2. Does this mean that the Lagrange Mult can't be used?
     
  2. jcsd
  3. Oct 22, 2011 #2

    Dick

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    Re: It appears that the Lagrange Multiplier is failing due to different values of Lam

    2x*lambda=2x doesn't necessarily mean lambda=1. It might mean x=0. You have to start splitting it into cases.
     
  4. Oct 22, 2011 #3
    Re: It appears that the Lagrange Multiplier is failing due to different values of Lam

    I got it thanks a lot Dick.
     
  5. Oct 22, 2011 #4

    dynamicsolo

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    Re: It appears that the Lagrange Multiplier is failing due to different values of Lam

    When you have equations of the form [itex]2x = \lambda \cdot 2x[/itex] , it is dangerous to simply "cancel out" the "x" common to both sides of the equation. By removing the factor, you may be removing a solution to the equation along with it. Generally speaking, one should not divide out a factor unless one is sure it cannot be zero.

    It is safer to re-write these equations in the form [itex]2x - \lambda \cdot 2x = 0 \Rightarrow 2x ( 1 - \lambda ) = 0 [/itex] and solve the two implied equations from there.
     
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