# Langrange Multipliers and Minimum Distance

1. ### schaefera

208
1. The problem statement, all variables and given/known data
Use the method of Lagrange multipliers to find the points on the ellipse (x^2)+(2y^2)=2 that are closest to and farthest from the line x-y=2.

2. Relevant equations
I know that to do the lagrange I have to take the gradient of a function f and set it equal to some unknown constant times the gradient of my constraint (function g).

3. The attempt at a solution
I believe my constraint, g(x,y) should be the ellipse itself, so g(x,y)=(x^2)+(2y^2)-2 and [gradient of g]= {2x, 4y}.

What I don't know, however, is what is the function f I use? I think it has something to do with the distance formula, but I can't quite get my finger on what. Any help is greatly appreciated-- I feel like I'm very close to finding the answer, but am just missing one important step.

Thanks!
Austin

2. ### Dick

25,853
You need a formula for the distance of a point (x,y) to the given line, don't you?

3. ### schaefera

208
Yes exactly! And I think that THAT is the equation I would want to then make f and use it as the primary equation whose gradient I compare to grad(g).

Distance= sqrt((x-xo)^2 + (y-yo)^2))... but this isn't from a line to a point, it's from a point (x1, y1) on the ellipse to a point (x2, y2) on the line given. But how do I use this to get a function in the form sqrt ((x2-x1)^2+(y2-y1)^2)-- or something of this sort?

25,853
5. ### schaefera

208
But in this case, it still has (xo, yo) which I don't seem to have in the problem statement.

6. ### Dick

25,853
(x0,x0) in this case is just your point (x,y) on the ellipse.

7. ### schaefera

208
So for me, distance= (x-y-2)/(sqrt(2))?

I guess I'm still a little confused by how I can let their (x0, y0) be my (x, y). Wouldn't that change the vector I'm dealing with?

Can you maybe explain the though process of that link as related to the problem-- I can't quite picture it in my case.

8. ### Dick

25,853
(x0,y0) is just some point. (x,y) is just some point. All that's different is the labelling. And yes, your answer is basically right. Except it's negative for all the points on the ellipse, since you rightly got rid of the absolute value. Just change the sign.

9. ### schaefera

208
So what would the vectors that I am dotting be? And how can distance be negative, I would think it needs to be positive?

10. ### Dick

25,853
The geometry is described in the link. And according to the formula in the link, it should be |x-y-2|/sqrt(2), right? That's always nonnegative. But you don't want absolute values around when you are differentiating. -(x-y-2)/sqrt(2) is always positive for points on the ellipse. I wouldn't obsess about this, the problem is about lagrange multipliers, not distance from a line to point formulas.

11. ### schaefera

208
Ok thanks, I get it now! So I end up finding that there are 4 critical points: (+/- 2/sqrt(6), +/- 1/(sqrt6)). Clearly, the point in the 4th quadrant is closest and the point in the 2nd quadrant is farthest away from the line. Thanks so much for the help!