Maximum and minimum distance (lagrange multipliers)

[Gregg]

Homework Statement

A point lies on the plane

x-y+z=0

and on the ellipsoid

$$x^2 +\frac{y^2}{4} + \frac{z^2}{4} = 1$$

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

The Attempt at a Solution

The two contraints

$$g = x-y+z =h= x^2 +\frac{y^2}{4} + \frac{z^2}{4}-1=0$$

$$f = x^2+y^2+z^2$$



$$2x = \lambda + 2\mu x$$
$$2y = -\lambda + \frac{1}{2}\mu y$$
$$2z = \lambda + \frac{1}{2}\mu z$$

So y=-z?


From g:

$$x-y+z=0 \therefore x=2y=-2z$$

from h:

$$4y^2+\frac{1}{2}y^2=1\Rightarrow y = \pm\frac{\sqrt{2}}{3}$$

$$z=\mp\frac{\sqrt{2}}{3}$$

$$x=\pm\frac{2\sqrt{2}}{3}$$

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!

 

[vela]

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

$$y = -\frac{\lambda}{2(1-\mu/4)}$$

$$z = \frac{\lambda}{2(1-\mu/4)}$$

and came to your conclusion that y=-z. This logic works as long as $\mu \ne 4$, so you still have to consider what happens when $\mu=4$. If you do that, you'll find the other solutions you were looking for.

 

[Gregg]

That is what I did, yes. If mu is 4 then from the 3 equations

$$2x = \lambda + 8x$$

$$2y=-\lambda + 2y$$

$$2z=\lambda + 2z$$


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

from g: z=y

from h: $$z=y=\sqrt{2}$$

A distance of 2 as opposed to the other extreme which is 14/9

 

[vela]

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).

 

[Gregg]

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?

 

[vela]

Just in calculating the distance.