a question about lagrange multipliers?

Question

can the langrange multiplier itself (i.e. Lambda) assume the value of 0? or do we assume the lamda is always a none zero value.

but in many cases u have to divide by lambda to obtain some relation between the variables. in that case u have to always make a case of lambda not equal to 0 and equal to 0 so my point is should i always be bother to check the case of lambda=0

Answer 1

If you put lambda =0 you would find the unrestricted critical points of the function which normally don´t satisfy the restrictions

Answer 2

I know this is 7 years old, but I ended up here and others might too. Yes, lambda can be zero. As stated in wikipedia while trying to give a visualization of the Lagrange problem setup: "Suppose we walk along the contour line with g = c. We are interested in finding points where f does not change as we walk, since these points might be maxima. There are two ways this could happen: First, we could be following a contour line of f, since by definition f does not change as we walk along its contour lines. This would mean that the contour lines of f and g are parallel here. The second possibility is that we have reached a "level" part of f, meaning that f does not change in any direction."

The first possibility has lambda nonzero and the gradients of f and g are multiples of each other. In the second possibility, the constraint simply passes over a flat spot of f, in which case we need lambda =0.

A simple example of this is finding the minimum of f(x,y)=xy constrained to y-x=0. This results in a triple zero solution (x,y,lambda) = (0,0,0) which may lead to more confusion. So look at a slightly more complicated problem:

Minimize f(x,y)=(x-1)(y-1) = xy - x - y +1 on g(x, y) = y-x = 0. (Using k=lambda)

F(x, y, k) = xy - x - y + 1 - k(y - x)

F_x = 0 : y - 1 - k = 0 (A)
F_y = 0 : x - 1 + k = 0 (B)
g = 0 : y - x = 0 (C)

Adding (A) and (B) gives: x + y - 2 = 0 or y = 2 - x
Subbing that into (C) gives: (2 - x) - x = 0 or x = 1
Subbing x = 1 into y = 2 - x: y = 2 - 1 = 1
And we have the critical point (1,1), we use geometric arguments or some other logic to say this is a absolute minimum and move on to the next problem.

However, if you go back to (A) and (B) and solve to lambda if x = y = 1, you see that lambda = 0.

The moral of this story is that we can not assume that lambda is non-zero.

I ran across this in a HW problem from J Stewart's book. Max/Min of f(x,y,z) = xyz over x^2 + 2y^2 + 3z^2 = 6. The solutions manual incorrectly states that if any of x, y, or z = 0 then they all are zero, which then violates the constraint. However, the solution (indirectly) discounts the fact that lambda CAN be zero, so their logic is flawed there. As a result, 6 additional critical points surface (in addition to the 8 already found in the solution, yes...that is 14 points of interest...). None of the additional critical points end up being max's or min's however in that particular problem (unlike the above example).

Note, the bounded-ness or closed-ness of the constraint is not at issue here either. A zero lambda min can occur in a problem like the following:
max/min of f(x,y) = x^2 + y^2 over (x-1)^2+y^2 = 1
The minimum occurs at (0,0) despite needing lambda = 0 there. The max will have a non-zero lambda.

Answer 3

In most instances of the use of Lagrange multipliers, you are not concerned with the value of lambda, you only use lambda to find relations among the original independent variables. However, taking lambda equal to zero would usually not be helpful.