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Lagrange Multipliers to Find Extreme Values of a Multi-Variable Function

  1. Jul 14, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to find the extrema of [itex]f(x,y) = 3x^{2} + y^{2}[/itex] given the constraint [itex]x^{2} + y^{2} = 1[/itex]

    2. Relevant equations

    I'm not sure what goes here. I've been trying to solve it with this:

    ∇f(x,y) = λ∇g(x,y)

    3. The attempt at a solution

    [itex]f(x,y) = 3x^{2} + y^{2}[/itex]
    [itex]g(x,y) = x^{2} + y^{2} = 1[/itex]

    [itex]∇f(x,y) = <6x, 2y>[/itex]
    [itex]∇g(x,y) = <2x, 2y>[/itex]

    With the previous equation, we get a system of equations:

    [itex]6x = 2xλ[/itex]

    [itex]2y = 2yλ[/itex]

    and the constraint

    [itex]x^{2} + y^{2} = 1[/itex]

    Solving for x and y, I get x=y=0. But that fails the final constraint. I've used the above procedure for other Lagrange problems without much issue, but this one's stumping me.

    So I tried finding a solution on Wolfram: Maxes at (-1, 0) and (1, 0), and Mins at (0, -1) and (0, 1). Okay, nice and simple looking solutions. Right?

    But I just can't figure this out. What am I missing?

    Thanks
     
  2. jcsd
  3. Jul 14, 2012 #2

    Ray Vickson

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    Science Advisor
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    [tex] 6x = 2x \lambda \Longrightarrow x = 0 \text{ or } \lambda = 3\\
    2y = 2y \lambda \Longrightarrow y = 0 \text{ or } \lambda = 1.[/tex]
    Either x or y must = 0, and then the other one = ± 1. One value of λ corresponds to the constrained min and the other to the constrained max.

    RGV
     
  4. Jul 14, 2012 #3
    Ooh, I see how that works now. Thank you very much!
     
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