Bonus (Unexpected) solution to lagrange equation?

[unscientific]

Homework Statement

The lagrange equations are obtained as in the picture. I am only showing the final part of the solution, where they consider the final case of x≠y≠z.

Homework Equations

The equation at the second paragraph is obtained by subtracting: (5.34 - 5.35).

The final equations are obtained by dividing 5.37 by (x-y) throughout, same for the other 2. (Which is ok, since x - y ≠ 0)

The Attempt at a Solution

I understand their method, but why can't I just do this:

(5.34) + (5.35) + (5.36)

3(x² + y² + z²) + 2λ(x + y + z) + 3μ = 0

Using the constraints,

3(1) + 0 + 3λ = 0

λ = -1

Not sure if this is a appropriate solution..

 

[ehild]

Homework Statement

The lagrange equations are obtained as in the picture. I am only showing the final part of the solution, where they consider the final case of x≠y≠z.

Homework Equations

The equation at the second paragraph is obtained by subtracting: (5.34 - 5.35).

The final equations are obtained by dividing 5.37 by (x-y) throughout, same for the other 2. (Which is ok, since x - y ≠ 0)

The Attempt at a Solution

I understand their method, but why can't I just do this:

(5.34) + (5.35) + (5.36)

3(x² + y² + z²) + 2λ(x + y + z) + 3μ = 0

Using the constraints,

3(1) + 0 + 3λ = 0

[STRIKE]λ = -1[/STRIKE]

Not sure if this is a appropriate solution..

You mixed lambda with mu. μ=-1.

ehild

 

[unscientific]

You mixed lambda with mu. μ=-1.

ehild

I see..is it possible to prove that μ=-1 leads to the equations being inconsistent?

 

[ehild]

I see..is it possible to prove that μ=-1 leads to the equations being inconsistent?

No, μ=-1 does not matter in the argument which proves that x,y,z can not be all different.

Two of them can be equal and in this case, you would use μ=-1 to get λ and the possible values of x,y,z.

ehild