Calculation regarding to Lagrange Multiplier

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Homework Help Overview

The discussion revolves around a problem involving Lagrange multipliers in the context of optimization. The original poster presents a function f = xyz subject to a constraint involving a combination of the variables x, y, and z. The goal is to find the values of x, y, and z that optimize the function while satisfying the constraint.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting up the Lagrange function and differentiating it with respect to the variables and the multiplier. There are attempts to solve the resulting equations, with some participants expressing confusion about the process and the nature of the solutions.

Discussion Status

Several participants have provided insights and suggestions on how to approach the problem, including methods to eliminate variables and express them in terms of the Lagrange multiplier. The original poster has indicated difficulty in progressing, and while some guidance has been offered, there is no explicit consensus on the solution.

Contextual Notes

Participants note that the variables x, y, and z must be positive lengths, and there is a mention of a trivial solution that does not satisfy the constraint. The discussion reflects a mix of interpretations and methods being explored without a definitive resolution.

linchien
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Hi, Dear Math forum users,

I was practicing with my optimization course problem and encountered one type of
Lagrange multiplier question which I have trouble with. I am wondering if anyone could
enlighten me for the following Lagrange problem.

function f = x*y*z subject to 4xy+3yz+2xz = 72
since x y z are lengths, they are all positive.
I set up the Lagrange function:

L = xyz + u*(4xy + 3yz + 2xz - 72)

Differentiate w.r.t x, y, z and u gives:

Lx = yz + u*(4y + 2z)
Ly = xz + u*(4x + 3z)
Lz = xy + u*(3y + 2x)
Lu = 4xy + 3yz + 2xz - 72

Set Lx = Ly = Lz = 0 and solve for x, y, z, and u.
I know that x = y = z = 0 is a trivial solution.
I would like to find x, y, z each in terms of u respectively.

I have tried it on paper many times but still cannot solve it.
Any form of help is appreciated.

Thank you very much
 
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linchien said:
function f = x*y*z subject to 4xy+3yz+2xz = 72
since x y z are lengths, they are all positive.
I set up the Lagrange function:

L = xyz + u*(4xy + 3yz + 2xz - 72)

Differentiate w.r.t x, y, z and u gives:

Lx = yz + u*(4y + 2z)
Ly = xz + u*(4x + 3z)
Lz = xy + u*(3y + 2x)
Lu = 4xy + 3yz + 2xz - 72

Set Lx = Ly = Lz = 0 and solve for x, y, z, and u.
I know that x = y = z = 0 is a trivial solution.

x=y=z does not fulfil the condition 4xy+3yz+2xz = 72.

Solving this system of equations is rather easy. Show what you did so far.

ehild
 
Hello, thank you for the reply.
Here is the steps that I have gone through.

Lx = yz + u(4y + 2z) =0
Ly = xz + u(4x + 3z) =0
Lz = xy + u(3y + 2x) =0
Lu = 4xy + 3yz + 2xz -72 =0,

x*Lx = xyz + u(4xy + 2xz) =0
y*Ly = xyz + u(4xy + 3yz) =0
z*Lz = xyz + u(3yz + 2xz) = 0

since 4xy + 3yz + 2xz -72 = 0, we can rearrange x*Lx, y*Ly and z*Lz to be
x*Lx = xyz + u(4xy + 2xz) = xyz + u( 72 - 3yz) = 0
y*Ly = xyz + u(4xy + 3yz) = xyz + u(72 - 2xz) = 0
z*Lz = xyz + u(3yz + 2xz) = xyz + u(72 - 4xy) = 0, and

u(72 - 3yz) = u(72 - 2xz) = u(72 - 4xy),

this is the last step I have got.
The answer says that
y = -4u
x = -6u
z = -8u,

I have try to use different methods but still cannot get the result.
I am not sure where I got stuck with.
Any help is appreciated.
Thank you
 
You made it more complicated as it was originally. You need to cancel the variables x,y,z, till you get an equation for all of them in terms of u. Express one coordinate from the first equation, say y. y(z+4u)=-2uz--->y=-2uz/(z+4u).
Do the same with x. Use the third equation to get z in terms of u.
I hope that gives you a start.

ehild
 
linchien said:
Hello, thank you for the reply.
Here is the steps that I have gone through.

Lx = yz + u(4y + 2z) =0
Ly = xz + u(4x + 3z) =0
Lz = xy + u(3y + 2x) =0
Lu = 4xy + 3yz + 2xz -72 =0,
I always think of "Lagrange multipliers" as saying that the max or min of F(x,y,z) subject to the constraint G(x,y,z)= constant are where \nabla F= \lambda\nabla G so I would have written these as yz= -\lambda(4y+ 2z), xz= -\lambda(4x+ 3z) and xy= -\lambda(3y+ 2x). Of course, the equations you give can be easily rewritten in that form.

My point is that, since we don't really need to find \lambda as part of the solution to this problem, we can immediately eliminate \lambda by dividing one equation by another.
for example, dividing yz= -\lambda(4y+ 2z) by xz= -\lambda(4x+ 3z) we get
\frac{y}{x}= \frac{4y+ 2z}{4x+ 3z}
or y(4x+ 3z)= x(4y+ 2z or 4xy+ 3yz= 4xy+ 2xz which easily reducews to 3y= 2x or y= (2/3)x or x= (3/2)y.

Similarly, we can divide two other equations, eliminating \lambda and getting z as a function of either x or y. Then we can reduce the constraint itself 4xy + 3yz + 2xz -72 =0 to an equation in the single variable, x or y.


x*Lx = xyz + u(4xy + 2xz) =0
y*Ly = xyz + u(4xy + 3yz) =0
z*Lz = xyz + u(3yz + 2xz) = 0

since 4xy + 3yz + 2xz -72 = 0, we can rearrange x*Lx, y*Ly and z*Lz to be
x*Lx = xyz + u(4xy + 2xz) = xyz + u( 72 - 3yz) = 0
y*Ly = xyz + u(4xy + 3yz) = xyz + u(72 - 2xz) = 0
z*Lz = xyz + u(3yz + 2xz) = xyz + u(72 - 4xy) = 0, and

u(72 - 3yz) = u(72 - 2xz) = u(72 - 4xy),

this is the last step I have got.
The answer says that
y = -4u
x = -6u
z = -8u,

I have try to use different methods but still cannot get the result.
I am not sure where I got stuck with.
Any help is appreciated.
Thank you
 
Thanks for the help from both of you, I have finally got it right this time!
 

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