# Homework Help: Maximum distance from point on ellipsoid

1. May 22, 2012

### stefaneli

1. The problem statement, all variables and given/known data

Find the point on an given ellipsoid that is the farthest to a given surface.(Distance between point on ellipsoid and surface should be max).

2. Relevant equations

ellipsoid: $\left(x-3\right)^{2}\over{3}$+$y^{2}\over{4}$+$z^{2}\over{5}$ $= 1$
surface: $3x$+$4y^{2}$+$6z$ + $6=0$

3. The attempt at a solution

Last edited: May 22, 2012
2. May 22, 2012

### vela

Staff Emeritus
Using Lagrange multipliers sounds good. What's the function you're trying to maximize?

3. May 22, 2012

### stefaneli

I don't know. Finding the constraint maximum of ellipsoid function won't do the job, won't find the point(where constraint is function of surface). That's why I need help.

4. May 22, 2012

### vela

Staff Emeritus
Try rereading what you wrote in parentheses in your original post and then try to answer my question again.

5. May 22, 2012

### stefaneli

I thought about maximizing distance from one point to another :
$\sqrt{(x-x_{0})^2+(y-y_0)^2)+(z-z_0)^2}$ where $(x_0,y_0,z_0)$ is a point on surface. But I don't have a specific point.
Constraint would be ellipsoid function.

6. May 22, 2012

### D H

Staff Emeritus
You don't want to do that. This is essentially a maximin problem. The distance from some given point (x,y,z) on the ellipsoid to the surface is the minimum value over all points on the surface of the distance between the given point on the ellipsoid and that other point on the surface. You want to find the point on the ellipsoid that maximizes this minimum distance.

7. May 22, 2012

### stefaneli

That's a new way of looking at the problem, but I don't see how can I solve this. How to find that minimum distance that I'm gonna maximize?

8. May 22, 2012

### vela

Staff Emeritus
Is your equation for the surface correct? You mentioned a plane in the original post, but the equation for the surface isn't that of a plane.

9. May 22, 2012

### stefaneli

Oh, sorry. Distance between point on ellipsoid and surface should be max. The equation is correct.

10. May 23, 2012

### D H

Staff Emeritus
Rather than solving this with Lagrange multipliers, it might be easier to use geometric reasoning.

Imagine some point on the ellipsoid and the point on the parabolic cylinder (that's the name for that other surface) that is closest to this point on the ellipsoid. The line connecting the two points will be normal to the parabolic cylinder at the point on the parabolic cylinder. What if that line is also normal to the ellipsoid? That's a stationary point of some kind: A minimum, maximum, or inflection point. So solve for this condition. You will have to worry about local versus global mimima. Then again, that problem will pop up with a Lagrange multiplier approach too.

11. May 25, 2012

### stefaneli

ellipsoid: $\left(x-3\right)^{2}\over{3}$+$y^{2}\over{4}$+$z^{2}\over{5}$ $= 1$
surface: $3x$+$4y^{2}$+$6z$ + $6=0$

Can I use substitution $u = y^{2}$? Ellipsoid will than transform to paraboloid and parabolic cylinder to plane. And after that I would maximize (Lagrange multipliers):

$|Ax+Bu+Cz+D|\over{\sqrt{A^{2}+B^{2}+C^{2}}}$

where $(x,u,z)$ is a point on paraboloid (ellipsoid before transf.) and $(A,B,C,D)$ are parameters of the plane(cylinder before transf.). And constraint will be:
$\left(x-3\right)^{2}\over{3}$+$u\over{4}$+$z^{2}\over{5}$ - $1$.

Last edited: May 25, 2012
12. May 25, 2012

### D H

Staff Emeritus
If you insist on using Lagrange multipliers you are going to have to use two of them because this is a maximin problem. And it is messy. That's why I suggested using geometric reasoning. Formulate this right and it's a one-liner to Wolfram alpha. From that you just have to pick out which of the multiple solutions offered is the correct one.

13. May 25, 2012

### stefaneli

I wouldn't insist on using L.m., if I don't have to. I agree that your approach is better, but I must use L.m. So the substitution isn't the right way? It's not correct? Thanks.

14. May 25, 2012

### D H

Staff Emeritus
I don't know if the substitution would help. I haven't tried it. This is your nasty problem, not mine.

I solved your problem geometrically (with the aid of Wolfram alpha) and did a quick sanity check on the obvious solution. (It offered several families of real solutions which obviously represented the intersection of the ellipsoid and the surface, a plethora of imaginary solutions which obviously weren't "real", plus two distinct real solutions. One of those two was obviously the correct one.)

If you insist on using (or if you are told to use) Lagrange multipliers, I don't see how you can escape using two of them.

15. May 25, 2012

### stefaneli

I didn't want you to try, but to give the opinion if the substitution could work or not. Can you post the solution?

16. May 25, 2012

### D H

Staff Emeritus
That substitution might work, but only because (big hint) the y value of the point on the ellipse happens to be the same as the y value of the closest point on the surface. In fact (Mt. Kilimanjaro-sized hint), the y values are rather trivial. I'll leave it up to you to show this is the case (which can be done geometrically).

This reduces the problem to a 2 dimensional problem, the point on an ellipse that is furthest from a line. Now you don't need a pair of Lagrange multipliers because the distance from a point on a plane to a line on that plane is easily calculated.

17. May 25, 2012

### stefaneli

Thanks. (Mt. Kilimanjaro-size thanks)