| Author |
Message
|
| bit4bit |
 Posted: Tue May 13, 2008 10:45 am Post subject: LaGrange multipliers |
 |
|
Forum Ph.D.
Joined: 14 Jul 2007
Posts: 625
|
When using LaGrange multipliers, why do the maximums on the bounday, ∂D, occur only at places where the boundary is tangential to a level curve (perpendicular to the gradient) of the function f(x,y)?
_________________
Chance favours the prepared mind. |
|
| Back to top |
|
 |
| river_rat |
| Posted: Tue May 13, 2008 10:52 pm Post subject: |
|
|
Forum Ph.D.
Joined: 01 Jun 2006
Posts: 1055
Location: South Africa
|
I'll give you a hint - you try it 
Consider a curve g : R -> R^2 and look what happens when you follow f along that curve.
PS. I think i have a geometry post somewhere here where i went through this...
_________________
As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong. |
|
| Back to top |
|
|
| bit4bit |
| Posted: Wed May 14, 2008 6:07 am Post subject: |
|
|
Forum Ph.D.
Joined: 14 Jul 2007
Posts: 625
|
When you say g:R -> R2 are you talking about a parameterization?
_________________
Chance favours the prepared mind. |
|
| Back to top |
|
|
| river_rat |
| Posted: Wed May 14, 2008 10:24 pm Post subject: |
|
|
Forum Ph.D.
Joined: 01 Jun 2006
Posts: 1055
Location: South Africa
|
of the curve, yes. I gave the explanation in my geometry thread (page 2) - it should be relatively easy to find with search.
_________________
As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong. |
|
| Back to top |
|
|
| bit4bit |
| Posted: Sun May 18, 2008 10:47 am Post subject: |
|
|
Forum Ph.D.
Joined: 14 Jul 2007
Posts: 625
|
Hi, river_rat, I've been busy with exams and work, so I haven't looked at this for a while. I found your geometry thread, but I couldn't really follow the proof...I'm not really that comfortable talking about things in terms of sets. I've been doing some reading about it today, and from what you said, I've got a few ideas, but can't seem to put them together.
So g:R -> R2 is a parameterized curve in R2, so you have g(t)=(x(t),y(t)). Then to follow the curve over f, you have the composition:
f(g(t))=f(x(t),y(t))
which is really a function f:R -> R
I thought about taking an interval of this function, [ta, tb], and comparing the value of the function, to the value of the derivatives at the endpoints....i.e. comparing:
f(g(ta)) and (d/dt)f(g(ta))
with
f(g(tb)) and (d/dt)f(g(tb))
I know that (d/dt)f(g(t)) = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt) , but without knowing exactly what the functions f(g), and g(t) are, I can't see how to go about analysing the enpoints as above.
I also know that ∇f=<∂f/∂x,∂f/∂y>, and points in the maximum rate of change of f, at any point, and that for the directional derivative V.∇f=0, then V points along the direction of the level curve.
I can't seem to tie it all together though. Am I at least getting warm? Thanks
_________________
Chance favours the prepared mind. |
|
| Back to top |
|
|
| river_rat |
| Posted: Mon May 19, 2008 11:59 am Post subject: |
|
|
Forum Ph.D.
Joined: 01 Jun 2006
Posts: 1055
Location: South Africa
|
Hi bit4bit
Please explain to me why you think you need an interval. Remember we are assuming that we already have the maximal point.
_________________
As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong. |
|
| Back to top |
|
|
| bit4bit |
| Posted: Mon May 19, 2008 1:07 pm Post subject: |
|
|
Forum Ph.D.
Joined: 14 Jul 2007
Posts: 625
|
Well cause I thought to find the absolute maximum of the curve we are parameterizing (∂D), we would have to check its endpoints. We can look for a local maximum with the second derivative test, having:
(d/dt)f(g(t)) = (∂f/∂x)(dx/dt)+(∂f/∂y)(dy/dt) = 0
(d2/dt2)f(g(t)) = (∂2f/∂x2)(d2x/dt2)+(∂2f/∂y2)(d2y/dt2) < 0
but we might need to look at the endpoints for the absolute maximum of f on ∂D. But now I've thought this might not be the case anyway (?), since ∂D is a closed curve (it closes the domain).
Assuming we have the maximum on ∂D already, at a point (x0,y0), then the above conditions hold for this point, but I'm not sure what else to really say about it. we can't say ∇f(x0,y0)=0, or use an equation for a second partials test, because it might not be a local maximum (with 0 gradient).
_________________
Chance favours the prepared mind. |
|
| Back to top |
|
|
| bit4bit |
| Posted: Wed May 21, 2008 4:34 am Post subject: |
|
|
Forum Ph.D.
Joined: 14 Jul 2007
Posts: 625
|
I figured this out now. If g:R -> R2 parameterizes the boundary of D, ∂D, then g'(t) = <x'(t), y'(t)> , which is a vector in R2 tangent to ∂D. When ∂D is tangent to a level curve, it is perpendicular to the gradient, and so:
g'(t)⋅∇f = 0
<x'(t),y'(t)>⋅<∂f/∂x, ∂f/∂y> = 0
(∂f/∂x)(dx/dt)+(∂f/∂y)(dy/dt) = 0
This is the derivative of the composition f(g(t)), and since it equals 0, then we have a critical point (MAX/MIN) at this point.
_________________
Chance favours the prepared mind. |
|
| Back to top |
|
|
|
|
