Langrange Multipliers and Minimum Distance

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In summary, Austin was trying to find the points on the ellipse (x^2)+(2y^2)=2 that are closest to and farthest from the line x-y=2. He was using the method of Lagrange multipliers to find the points. He found that the point in the 4th quadrant was closest and the point in the 2nd quadrant was farthest away from the line. Thanks so much for the help!
  • #1
schaefera
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Homework Statement


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.


Homework 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).


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
 
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  • #2
You need a formula for the distance of a point (x,y) to the given line, don't you?
 
  • #3
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?
 
  • #5
But in this case, it still has (xo, yo) which I don't seem to have in the problem statement.
 
  • #6
(x0,x0) in this case is just your point (x,y) on the ellipse.
 
  • #7
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
schaefera said:
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.

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

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
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!
 

1. What are Langrange multipliers?

Langrange multipliers are a method used in mathematical optimization to find the minimum or maximum value of a function subject to constraints. They involve the use of a multiplier parameter to incorporate the constraints into the objective function and solve for the optimal solution.

2. How do Langrange multipliers work?

Langrange multipliers work by finding the critical points of the objective function, which are points where the gradient of the function is equal to the gradient of the constraint function multiplied by the multiplier. These points can then be plugged back into the original objective function to find the optimal solution.

3. What is the importance of Langrange multipliers?

Langrange multipliers are important in optimization problems because they allow for the incorporation of constraints into the objective function, making it possible to find the optimal solution even when there are limitations or restrictions on the variables.

4. How are Langrange multipliers used in real-world applications?

Langrange multipliers have a wide range of applications in fields such as economics, engineering, and physics. They can be used to optimize production processes, minimize costs, and maximize profits. They are also commonly used in physical systems, such as finding the path of least resistance in a circuit.

5. What is the relationship between Langrange multipliers and minimum distance?

Langrange multipliers can be used to find the minimum distance between a point and a curve or surface. This is done by setting up a constraint function that represents the distance between the point and the curve or surface, and then using the method of Langrange multipliers to find the minimum value of this function. This minimum value corresponds to the minimum distance between the point and the curve or surface.

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