What is the Maximum Inscribed Sphere for Ellipsoid?

[Haftred]

I'm trying to find the largest sphere that be inscribed inside the ellipsoid with equation 3x^2 + 2y^2 + z^2 = 6.

Homework Equations

I know I will need at least 2 equations. One of them is the constraining equation (f(x) = a, where 'a' is a constant) and the other is the equation you want to maximize. I need to use Lagrange multipliers to find a lambda such that (del) F = lambda (del G).

The Attempt at a Solution

My first attempt was to maximize the function x^2 + y^2 + z^2 = f(x,y,z) and use the constraining equation for the ellipsoid (3x^2 + 2y^2 + z^2 = 6). However, there is no lambda that will made del (F) = L (del [G]). I think I'm using the wrong equations. Another idea is to find the point(s) on the ellipsoid that lies closest to the origin and use that as the radius for my sphere. However, that boils down to the same equations I already unsuccesfully used.

Any help appreciated.

 

[AlephZero]

You can do it by Lagrange multipliers if you really want to, but the easy way is to see what the equation of the ellipsoid means geometrically. What are its principal axes? What is the length of the shortest principal axis? Then you know enough to just write down the answer.

 

[Haftred]

yeah, but the problem makes you use lagrange multipliers =(

 

[AlephZero]

Find the extrema of f(x,y,z) = x^2 + y^2 + z^2
subject to g(x,y,z) = 3x^2 + 2y^2 + z^2 - 6 = 0

The Lagrange equations are the components of the gradient of f(x) + L.g(x) = 0
2x + L.6x = 0 (1)
2y + L.4y = 0 (2)
2z + L.2z = 0 (3)
and 3x^2 + 2y^2 + z^2 = 6 (4)

(3) gives L = -1 or z = 0
if L = -1:
(1) and (2) give x = 0 and y = 0
(4) gives z^2 = 6 so the extreme radius = sqrt(6). But (by inspection) that's a maximum not a minumum.

If z = 0:
(1) and (2) both give L = 0
(4) gives 3x^2 + 2y^2 = 6

So I suppose you now have to find the extrema of x^2 + y^2 subject to 3x^2 + 2y^2 = 6. Obviously that will lead to the answer but I don't recall having this situation happen with Lagrange multipliers before.

 

[AlephZero]

Sorry, I couldn't see what was in front of my nose in the previous post till after I logged off the forum!

2x + L.6x = 0 (1)
2y + L.4y = 0 (2)
2z + L.2z = 0 (3)

There are THREE sets of solutions of those equations with L non equal to zero:

L = -1/3 y = 0 z = 0
L = -1/2 x = 0 z = 0
L = -1 x = 0 y = 0

And using (4) you get the three extreme values of the function, corresponding to the three principal axes of the ellipse.