Lagrange Multipliers

By Light Of Orion on Saturday, May 04, 2002 - 10:20 am:

Hello,

I was working on this problem using Lagrange multipliers.

Find the points in the hyperbola x² -y² =1 which is closest to the point (4,0).

I worked through to lambda =(x-4)/2xf=y/(-2yf), where f is the function of distance between a point in the hyperbola and (4,0).

Then, 2fy=fxy
fxy-2fy=0
fy(x-2)=0
f=0, y=0, x=2

Now, i understand that x=2 is the solution, but what does that y=0 and f=0 mean, maybe geometrically?

Thank you.

Regards.

By Ian Short on Saturday, May 04, 2002 - 11:37 am:

Hi, your working is slightly wrong here. If

f = 0

really was the distance between the point and hyperbola then the point would have to lie on the hyperbola. Maybe this is why you asked about the geometry?

The Lagrange multiplier method would consist of minimising $(x - 4)^{2} + y^{2}$ subject to $x^{2} - y^{2} = 1$ . That is, set,

$F = (x - 4)^{2} + y^{2} + λ (x^{2} - y^{2} - 1)$

and differentiate with respect to $x$ , $y$ and $λ$ .

$F_{x} = 2 (x - 4) + 2 λ x = 0$

$F_{y} = 2 y - 2 λ y = 0$

$F_{λ} = x^{2} - y^{2} - 1 = 0$

has solutions with $x = 1$ and $y = 0$ BUT it also, like you say has solution with $λ = 1$ ( $x = 2$ and $y = \sqrt{3}$ ). This is the minimal solution, as you observed.

For an alternate geometric interpretation, rotate your axis so that they coincide with the original asymptotes, explicitly put $X = (x - y) / \sqrt{2}$ and $Y = (x + y) / \sqrt{2}$ so that the equation is $X Y = 1 / 2$ . Draw a picture- the simpler equation means finding the minimum distance will be easier by differentiation.

Ian

By Light Of Orion on Saturday, May 04, 2002 - 07:24 pm:

HI!

Then, what does x=1 and y=0 mean? They are not extreme points i guess?

By Ian Short on Sunday, May 05, 2002 - 08:42 am:

Hi- I think (1,0) is a local maximum. Obviously there is no overall maximum value for the distance. (2 , +/- 3^1/2 ) are unique minima as any minimum point would have to solve the 3 partial differential equations I listed.

Try drawing a 3D graph with the hyperbola on the floor in the xy-plane and the distance from the point to the hyperbola plotted with the z-axis.

Ian

By Light Of Orion on Sunday, May 05, 2002 - 10:59 am:

Yeah, got it. Thanks a lot Ian.