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Lagrange Multipliers and Constraints

Date: 11/24/98 at 16:29:56
From: Sam
Subject: LaGrange Multiplier

I have a quick question concerning the setup of problems requiring the
use of the LaGrange Multiplier method. Basically, how do you determine 
which of the two equations given is the "constraint"? For most 
problems, the equation for the function and the constraint is given.  
All you'd have to do is to manipulate them to forms appropriate in 
setting up the Lagrange Multiplier. However, what about those problems 
which aren't so "nice" - where is the constraint explicitly revealed?

For example, here's a question from an assignment I had:

Find the two points, one on each curve, which are as close together as
possible using the Lagrange Multiplier method:

   y = 2x - 3
   y = x^2

Now which one of those is the constraint?

If you can answer that for me I would really appreciate it. Thank you.

Date: 11/24/98 at 17:07:04
From: Doctor Santu
Subject: Re: LaGrange Multiplier

Hello there!

Actually, Samuel, (are you ready for this?) they're BOTH constraints!  
You probably remember that there could be more than one constraint. You 
have to use one multiplier for each constraint.

The word "constraint" means something that prevents you from going 
away. In this case the function that you want to minimize (or optimize) 
is the distance between two points, (x[1],y[1]) and (x[2],y[2]). 
Obviously, we could make them as close as we want, so that the distance 
is zero, except that there are constraints. One point has to lie on one 
curve, and the other has to lie on the other curve.

So you're maximizing sqrt[(x[1]-x[2])^2 + (y[1]-y[2])^2], the distance 
formula, subject to the "constraints" that y[1] = 2x[1] - 3, and 
y[2] = x[2]^2. The restriction is that the two points have to stay on 
their respective curves.

There are other ways of doing it, and some shortcuts. One shortcut is 
that you can forget the "sqrt."  If you find the points so that the 
distance SQUARED is as small as possible, the same two points are the 
ones whose distance is as small as possible. This saves a lot of 

Other things you can do are to solve for the y variables in terms of 
the x variables, and to solve the problem without using Lagrange 
multipliers at all, but that's sort of cheating.

- Doctor Santu, The Math Forum
Associated Topics:
College Calculus
High School Calculus

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