Maximum and minimum distance (lagrange multipliers)

  • Thread starter Gregg
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Homework Statement



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.


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!
 

Answers and Replies

  • #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.
 
  • #3
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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
 
  • #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).
 
  • #5
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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?
 
  • #6
vela
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Just in calculating the distance.
 

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