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Homework Help: Help Please! Studying for test : Lagrange Multipliers

  1. Dec 6, 2005 #1
    Help Please! Studying for test : Lagrange Multipliers!!

    Good morning all. I am having trouble with the next step to the following problem:

    Q.Find all realtive extrema of x^2y^2 subject to the constraint 4x^2 + y^2 = 8.

    g(x)= x^2y^2 f(x) = 4x^2 + y^2 = 8.

    the gradiant of f = <8x,2y>
    the gradiant of g = <2xy^2, 2x^2y>

    therefore..

    2xy^2 + 2x^2y = lamda (8x+2y)
    lamda(8x) = 2xy^2 & lamda(2y)= 2x^2y
    lamda = 2xy^2/8x & lamda = 2x^2y/2y
    2x^2y/8x = 2xy^2/2y
    4x^2 = y^2


    I'm confused about how to find the extremas.
    Can anyone help me?
    Thanks all :smile:
     
  2. jcsd
  3. Dec 6, 2005 #2

    benorin

    User Avatar
    Homework Helper

    From

    [tex]\vec{\nabla g}(x,y) =\lambda \vec{\nabla f}(x,y)[/tex]

    we know that [itex]\left< 2xy^2, 2x^{2}y\right> = \lambda \left< 8x, 2y\right> [/itex]

    cancel a 2, and get this system

    [tex]xy^2 = \lambda 4x, x^{2}y = \lambda y[/tex]

    multiply the 1st by x, and the 2nd by y to get

    [tex]x^{2}y^2 = \lambda 4x^2, x^{2}y^2 = \lambda y^2 [/tex]

    so [itex] \lambda 4x^2 = \lambda y^2 [/itex] so that if [itex] \lambda \neq 0[/itex], we have [itex] 4x^2 =y^2 [/itex].

    then use the constraint [itex] 4x^2+ y^2 = 8 \Rightarrow y^2+ y^2 = 8[/itex]. Hence [itex]y =\pm \sqrt{4} = \pm 2[/itex].

    From [itex] 4x^2 =y^2 [/itex], we have [itex] x =\pm \sqrt{\frac {y^2}{4}} =\pm \frac {y}{2} = \pm 1[/itex]
     
    Last edited: Dec 6, 2005
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