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Lagrange multipliers with two constraints

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data
    By using the Lagrange multipliers find the extrema of the following function:
    subject to the constraints:


    2. The attempt at a solution
    Using lambda = 1/(2x) I got x=y-z and y=1-z
    plugging that into the first constraint, I got:
    6y^2-6y+1=0 which makes y=0.5+-(31/2/6)

    I got the same thing when solving for z, which means x=0 and lambda = infinity, which doesn't make sense.
  2. jcsd
  3. Dec 10, 2009 #2


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    You've also got z=1-y. So if you choose the root y=(3+sqrt(3))/2 you have to choose z=(3-sqrt(3))/2 not the other root for z. You can't mix and match any two roots with each other.
  4. Dec 10, 2009 #3
    Ah, I forgot to distribute the negative! I hate when that happens...it's all worked out now, thanks a lot!
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