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Interpreting Lagrange Multipliers

Date: 04/13/2001 at 02:28:15
From: Derek
Subject: Lagrange Multiplier

I have a question on an assignment and the answer is something I've 
never come across before, and I don't know its significance:

Use Lagrange multipliers to find the maximum and minimum values of the 
function f(x,y) = 6x^2 + 9y^2 subject to constraint x^2 + y^2 = 1. 
After calculating the partial derivatives I get lambda = -6 for 
partial derivative with respect to x and lambda = -9 for partial 
derivative with respect to y. What does that signify? Does it mean 
there is no minimum or maximum?

Date: 04/13/2001 at 04:51:06
From: Doctor Mitteldorf
Subject: Re: Lagrange Multiplier

Dear Derek,

First, let's think very concretely about this question. The constraint 
x^2 + y^2 = 1 says "look only at points on the unit circle." The 
function 6x^2 + 9y^2 gets larger as you go away from the origin, but 
it gets larger fastest in the +y or -y directions, and slowest in the 
x directions. So without any fancy Lagrange methods, we know right 
away what the answer is: maxima at (0,1) and (0,-1); minima at (1,0) 
and (-1,0).

Now let's go back and see how to get this answer using the Lagrange 
method: The differential of the function f(x,y) = 6x^2 + 9y^2 can be 
written df = (12x)dx + (18y)dy. We must compare this to the 
differential of the constraint g(x) = x^2 + y^2; dg = (2x)dx + (2y)dy.

Think of dx as the unit vector in the x direction and dy as the unit 
vector in the y direction. The essence of the Lagrange method is to 
say that the two vectors df and dg must point in the same direction, 
i.e., that df + lambda * dg = 0. The equation reduces to:

     (12x + lambda*2x) dx + (18y + lambda*2y) dy = 0 

This equation needs to be identically true, that is, the dx part and 
the dy part must separately be equal to zero. The only way to make 
this happen is for lambda to be -6 while y = 0, or lambda to be -9 
while x = 0. These conditions lead us to the four answers that we know 
are correct.

I think you went astray when you divided through by x, implicitly 
assuming that x was not 0 (and the same for y).

- Doctor Mitteldorf, The Math Forum
Associated Topics:
College Calculus

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