# It appears that the Lagrange Multiplier is failing due to different values of Lamda.

1. Oct 22, 2011

### xdrgnh

1. The problem statement, all variables and given/known data
Use Lagrange multipliers to ¯nd the maximum and mini-
mum value(s), if they exist, of
f(x; y; z) = x^2 -2y + 2z^2
subject to the constraint x^2+y^2+z^2

2. Relevant equations

3. The attempt at a solution

Basically after I find the gradient of the functions I get this.

2x=2x lamda
-2=2y lamda
4z=2z lamda.

One case lamda equals 1 while the other it equals 2. Does this mean that the Lagrange Mult can't be used?

2. Oct 22, 2011

### Dick

Re: It appears that the Lagrange Multiplier is failing due to different values of Lam

2x*lambda=2x doesn't necessarily mean lambda=1. It might mean x=0. You have to start splitting it into cases.

3. Oct 22, 2011

### xdrgnh

Re: It appears that the Lagrange Multiplier is failing due to different values of Lam

I got it thanks a lot Dick.

4. Oct 22, 2011

### dynamicsolo

Re: It appears that the Lagrange Multiplier is failing due to different values of Lam

When you have equations of the form $2x = \lambda \cdot 2x$ , it is dangerous to simply "cancel out" the "x" common to both sides of the equation. By removing the factor, you may be removing a solution to the equation along with it. Generally speaking, one should not divide out a factor unless one is sure it cannot be zero.

It is safer to re-write these equations in the form $2x - \lambda \cdot 2x = 0 \Rightarrow 2x ( 1 - \lambda ) = 0$ and solve the two implied equations from there.