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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!