Lagrange Multipliers and ConstraintsDate: 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. Sam 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 algebra. 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/