Langrange Multiplier

soyuz02 · Mar 29, 2007, 09:03 AM

Hi,

I am trying to solve this langrangean problem:

TC=5xz+4xy+4yz - MAIN FUNCTION

xyz=12 - CONSTRAINT FUNCTION

xyz-12=0

L=5xz+4xy+4yz+λ[xyz-12] - Langrangean Function

L=5xz+4xy+4yz+xyzλ-12λ

Partial Derivatives:

∂L/∂x=5z+4y+yzλ
∂L/∂y=4x+4z+xzλ
∂L/∂z=5x+4y+xyλ

However now I do not kow how to solve the partial derivatives, can anyone give me a hint or help?

Thank you very much in advance!

galactus · Mar 29, 2007, 12:26 PM

$f(x,y,z)=5xz+4xy+4yz$

$g(x,y,z)=xyz$

${\nabla}f=(4y+5z)i+(4x+4z)j+(5x+4y)k$

${\nabla}g=(yz)i+(xz)j+(xy)k$

${\nabla}g\neq{0}$ because any point on the surface xyz=12, since x,y,z are always non-zero on the surface. Therefore, ${\nabla}f={\lambda}{\nabla}g$

Hence, $(4y+5z)i+(4x+4z)j+(5x+4y)k={\lambda}(yzi+xzj+xyk)$

Equate:

1. $4y+5z=yz{\lambda}\rightarrow{\frac{4}{z}+\frac{5}{ y}={\lambda}$
2. $4x+4z=xz{\lambda}\rightarrow{\frac{4}{z}+\frac{4}{ x}={\lambda}$
3. $5x+4y=xy{\lambda}\rightarrow{\frac{5}{y}+\frac{4}{ x}={\lambda}$

From 1 and 2, $x=\frac{4y}{5}$
From 2 and 3, $z=\frac{4y}{5}$

Sub into constraint:

$(\frac{4y}{5})y(\frac{4y}{5})=12$

$y=\frac{6^{\frac{1}{3}}\cdot{5^{\frac{2}{3}}}}{2}= {2.66.....}$

x and z follow.

Check my work, please.

soyuz02 · Apr 1, 2007, 05:45 AM

Originally Posted by soyuz02

Hi,

I am trying to solve this langrangean problem:

TC=5xz+4xy+4yz - MAIN FUNCTION

xyz=12 - CONSTRAINT FUNCTION

xyz-12=0

L=5xz+4xy+4yz+λ[xyz-12] - Langrangean Function

L=5xz+4xy+4yz+xyzλ-12λ

Partial Derivatives:

∂L/∂x=5z+4y+yzλ
∂L/∂y=4x+4z+xzλ
∂L/∂z=5x+4y+xyλ

However now I do not kow how to solve the partial derivatives, can anyone give me a hint or help?

Thank you very much in advance!

Ah wonderful, makes sense.

Thank you very much again!