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Lagrange multiplier problem

  1. Apr 23, 2005 #1
    Hey guys, i need some help with this problem. It goes as follow: Find the global max and min valves of the fuction z=x^2+2y^2 on the circle x^2+y^2=1. Ok and here is what i have done. I found the derivites and have done (K is the lagrange constant)
    2x=2xK K=1
    4y=2yK K=2

    Then i set them equal and get 1=2? I know this is not right, but my teacher said do it by lagrange multipliers, so any help is appreciated. Thanks.
     
  2. jcsd
  3. Apr 23, 2005 #2

    arildno

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    Why do you assume that both x and y must be different from zero?
     
  4. Apr 23, 2005 #3

    dextercioby

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    [tex] \{x=0;y=\pm 1\} \mbox{and} \{y=0;x=\pm 1\} [/tex] should be the solution...I think the second delivers the minimum value...

    Daniel.
     
  5. Apr 23, 2005 #4
    From looking at it I can see how you get those values but how would i do this mathamatically? WOuld i just take 1=2=0? Then go back and plug is or what. I am just trying to understand this, thanks for helping me out.
     
  6. Apr 23, 2005 #5
    Just substitute [tex]x^2+y^2 =1[/tex] into [tex]z= x^2+ y^2+y^2= 1+ y^2[/tex]. Then just take the derivative of this in terms of y. Once you know the y value where the maximum happens just solve for x such that [tex]x^2= 1- y^2[/tex]

    Oh and since you are on the unit circle you also need to check the bound [tex]y=\pm 1 [/tex]
     
    Last edited: Apr 23, 2005
  7. Apr 23, 2005 #6
    I got to do this by lagrange multipliers though, that is what is messing me up because I don't see how it can be done by using this method.
     
  8. Apr 23, 2005 #7

    arildno

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    Pepschevy:
    You've got 3 equations in 3 unknowns:

    [tex]2x=2xK, 4y=2yK, x^{2}+y^{2}=1[/tex]
    Agreed?
    Your flaw has been to divide the first equation with 2x, and the second equation with 2y.
    This is only possible to do at the same time if neither x or y is 0!!

    But, furthermore:
    You have proven that if we assume this we get 1=K=2, that is, 1=2 a contradiction.

    Thus, you must conclude:
    Either x or y MUST be zero!
     
    Last edited: Apr 23, 2005
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