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Maximum and minimum distance (lagrange multipliers)

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data

    A point lies on the plane

    x-y+z=0

    and on the ellipsoid

    [tex] x^2 +\frac{y^2}{4} + \frac{z^2}{4} = 1 [/tex]

    Find the minimum and maximum distances from the origin of this point.


    3. The attempt at a solution

    The two contraints

    [tex] g = x-y+z =h= x^2 +\frac{y^2}{4} + \frac{z^2}{4}-1=0 [/tex]

    [tex] f = x^2+y^2+z^2 [/tex]



    [tex] 2x = \lambda + 2\mu x [/tex]
    [tex] 2y = -\lambda + \frac{1}{2}\mu y [/tex]
    [tex] 2z = \lambda + \frac{1}{2}\mu z [/tex]

    So y=-z?


    From g:

    [tex]x-y+z=0 \therefore x=2y=-2z [/tex]

    from h:

    [tex] 4y^2+\frac{1}{2}y^2=1\Rightarrow y = \pm\frac{\sqrt{2}}{3}[/tex]

    [tex] z=\mp\frac{\sqrt{2}}{3}[/tex]

    [tex] x=\pm\frac{2\sqrt{2}}{3}[/tex]

    I'm not sure if this is correct.

    The question asks for minimum and maximum but the distance function will give identical answers regardless of whether they are positive or negative. So I think I have done it wrong!
     
  2. jcsd
  3. Jun 12, 2010 #2

    vela

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    You found the minimum. I take it you did something like

    [tex]y = -\frac{\lambda}{2(1-\mu/4)}[/tex]

    [tex]z = \frac{\lambda}{2(1-\mu/4)}[/tex]

    and came to your conclusion that y=-z. This logic works as long as [itex]\mu \ne 4[/itex], so you still have to consider what happens when [itex]\mu=4[/itex]. If you do that, you'll find the other solutions you were looking for.
     
  4. Jun 13, 2010 #3
    That is what I did, yes. If mu is 4 then from the 3 equations

    [tex]2x = \lambda + 8x [/tex]

    [tex]2y=-\lambda + 2y [/tex]

    [tex]2z=\lambda + 2z [/tex]


    I think that this implies that lambda is 0, and that x = 0.

    from g: z=y

    from h: [tex] z=y=\sqrt{2}[/tex]

    A distance of 2 as opposed to the other extreme which is 14/9
     
  5. Jun 13, 2010 #4

    vela

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    You also have the negative solutions, which results in the same distance of 2.

    You calculated the minimum distance incorrectly. It should be sqrt(4/3).
     
  6. Jun 13, 2010 #5
    are the x,y and z values wrong for the minimum distance or have I made an error in the distance from correct x,y,z?
     
  7. Jun 13, 2010 #6

    vela

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    Just in calculating the distance.
     
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