# Langrange Multiplier Equations

• Jul 26th 2009, 06:37 AM
Reskal
Langrange Multiplier Equations
I was given the two equations f=x^2y^2z^2 and x^2+y^2+z^2=1
(I'll use 'u' to be equal to lambda)
and then set up my equations to be x^2y^2z^2-ux^2-uy^2-uz^2+u

taking the partial derivatives I got

fx=2x*y^2*z^2-2ux+1=0
fy=2x^2*y*z^2-2uy+1=0
fz=2x^2*y^2*z-2uz+1=0
fu=-x^2-y^2-z^2+1=0

I cant seem to find a way to solve this system of equations, can anyone? Or tell me what I did wrong?

Thanks!
• Jul 26th 2009, 06:50 AM
Jester
Quote:

Originally Posted by Reskal
I was given the two equations f=x^2y^2z^2 and x^2+y^2+z^2=1
(I'll use 'u' to be equal to lambda)
and then set up my equations to be x^2y^2z^2-ux^2-uy^2-uz^2+u

taking the partial derivatives I got

fx=2x*y^2*z^2-2ux+1=0
fy=2x^2*y*z^2-2uy+1=0
fz=2x^2*y^2*z-2uz+1=0
fu=-x^2-y^2-z^2+1=0

I cant seem to find a way to solve this system of equations, can anyone? Or tell me what I did wrong?

Thanks!

First off, I believe you have a mistake in your derivatives.

$\displaystyle \frac{\partial}{\partial x} \left(x^2 y^2 z^2 - \lambda(x^2+y^2+z^2-1) \right) = 2x y^2 z^2 - 2 \lambda x = 2x\left(y^2 z^2 - \lambda\right)$

Similarly for the other two. Notice that I factored. This gives rise to

$\displaystyle 2x\left(y^2 z^2 - \lambda\right) = 0$

$\displaystyle 2y\left(x^2 z^2 - \lambda\right) = 0$

$\displaystyle 2z\left(x^2 y^2 - \lambda\right) = 0$

Now look at cases.

Let me ask, why use Lagrange multipliers when one of the variables can be easily eliminated?
• Jul 26th 2009, 07:13 AM
Reskal
Quote:

Originally Posted by Danny
First off, I believe you have a mistake in your derivatives.

$\displaystyle \frac{\partial}{\partial x} \left(x^2 y^2 z^2 - \lambda(x^2+y^2+z^2-1) \right) = 2x y^2 z^2 - 2 \lambda x = 2x\left(y^2 z^2 - \lambda\right)$

Similarly for the other two. Notice that I factored. This gives rise to

$\displaystyle 2x\left(y^2 z^2 - \lambda\right) = 0$

$\displaystyle 2y\left(x^2 z^2 - \lambda\right) = 0$

$\displaystyle 2z\left(x^2 y^2 - \lambda\right) = 0$

Now look at cases.

Let me ask, why use Lagrange multipliers when one of the variables can be easily eliminated?

The directions said to use Lagrange multipliers.

Useing the equations that you wrote does this imply that x,y,z are all equal to 0?

Thanks Danny for all your help.
• Jul 26th 2009, 07:34 AM
Jester
Quote:

Originally Posted by Reskal
The directions said to use Lagrange multipliers.

Useing the equations that you wrote does this imply that x,y,z are all equal to 0?

Thanks Danny for all your help.

The cases:

$\displaystyle x = 0$, any $\displaystyle y$ and $\displaystyle z$ satisfying $\displaystyle y^2+z^2=1$. Similarly for $\displaystyle y = 0$ and $\displaystyle z = 0$. The remaining case is when $\displaystyle x^2 y^2 = \lambda, \;\;x^2 z^2 = \lambda,\;\; y^2 z^2 = \lambda$ or

$\displaystyle y = \pm x,\;\; z = \pm x,\;\; z = \pm y$ all 8 cases. Sub these into $\displaystyle x^2+y^2+z^2 = 1$