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Can Lagrange's Undetermined Multiplier Have Imaginary Points?

  1. Aug 23, 2012 #1
    Hey guys, new here and this is my first post. Wondering if anyone could help me.
    So I've encountered a problem on Lagrange's undetermined multiplier. Usually i have no problem with these, but this one caught me off a little.

    g(x,y) = x^2 + y^2 - 4xy - 6 = 0
    Find the points closest to the origin.
    With this in mind:
    f(x,y) = x^2 + y^2
    Using the formula:
    d(f + λg) = 0
    Let (f + g) = F
    F_x = 2x + 2λx -λ4y
    F_y = 2y + 2λy -λ4x
    By inspection you can see x = y, so into g(x,y) and...
    -2x^2-6=0
    x^2 = -3

    ∴ x = √-3

    I haven't encountered an imaginary point yet in this type of question, and since i can't quite make sense of it in my mind, i was wondering if anyone could help me? Have I made an error somewhere, or can you have imaginary points closest to the origin?

    Cheers.
     
  2. jcsd
  3. Aug 23, 2012 #2

    AlephZero

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    That's your error. Check the signs!
     
  4. Aug 23, 2012 #3
    I only say by inspection to save writing out the rest of the calculation but I did try it.

    2x + λ(2x - 4y) = 0
    2y + λ(2y - 4x) = 0

    λ= -2x/(2x - 4y) = -2y/(2y - 4x)
    [-2x/(2x - 4y) = -2y/(2y - 4x)] *-1
    2x/(2x - 4y) = 2y/(2y - 4x)
    [2x/(2x - 4y) = 2y/(2y - 4x)]/2
    x/(2x - 4y) = y/(2y - 4x)
    rearrange:
    x(2y - 4x) = y(2x - 4y)
    2xy -4x^2 = 2xy -4y^2
    -4x^2=-4y^2
    y^2 = x^2
    y = x.

    Unless I'm making a basic floor I'm not seeing.
     
  5. Aug 23, 2012 #4

    AlephZero

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    ##y = x## is not the only solution of ##y^2 = x^2##.
     
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