Calculation regarding to Lagrange Multiplier

In summary, the Lagrange multiplier method can be used to find the maximum or minimum of a function subject to a constraint. By setting up the Lagrange function and solving for the variables using the Lagrange equations, we can obtain the values of the variables in terms of the Lagrange multiplier, which can then be substituted into the constraint to find the maximum or minimum value. However, it is also possible to eliminate the Lagrange multiplier and solve for the variables directly by dividing the Lagrange equations and reducing the constraint to a single variable equation. By following this method, we can solve for the variables and obtain the maximum or minimum value of the function.
  • #1
linchien
3
0
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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  • #2
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
 
  • #3
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
 
  • #4
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
 
  • #5
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 [itex]\nabla F= \lambda\nabla G[/itex] so I would have written these as [itex]yz= -\lambda(4y+ 2z)[/itex], [itex]xz= -\lambda(4x+ 3z)[/itex] and [itex]xy= -\lambda(3y+ 2x)[/itex]. 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 [itex]\lambda[/itex] as part of the solution to this problem, we can immediately eliminate [itex]\lambda[/itex] by dividing one equation by another.
for example, dividing [itex]yz= -\lambda(4y+ 2z)[/itex] by [itex]xz= -\lambda(4x+ 3z)[/itex] we get
[tex]\frac{y}{x}= \frac{4y+ 2z}{4x+ 3z}[/tex]
or [itex] y(4x+ 3z)= x(4y+ 2z[/itex] or [itex]4xy+ 3yz= 4xy+ 2xz[/itex] which easily reducews to [itex]3y= 2x[/itex] or [itex]y= (2/3)x[/itex] or [itex]x= (3/2)y[/itex].

Similarly, we can divide two other equations, eliminating [itex]\lambda[/itex] and getting z as a function of either x or y. Then we can reduce the constraint itself [itex]4xy + 3yz + 2xz -72 =0[/itex] 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
 
  • #6
Thanks for the help from both of you, I have finally got it right this time!
 

Related to Calculation regarding to Lagrange Multiplier

What is the purpose of using Lagrange multipliers in calculations?

Lagrange multipliers are used in calculations to find the extreme values of a function subject to a constraint. This technique is commonly used in optimization problems in mathematics and physics.

How does Lagrange multiplier work?

Lagrange multiplier works by introducing an additional variable, called the Lagrange multiplier, into the equations representing the function and the constraint. This variable is then used to find the critical points of the function, which correspond to the extreme values.

What is the significance of the Lagrange multiplier in optimization problems?

The Lagrange multiplier provides a way to solve optimization problems with constraints by converting it into an unconstrained optimization problem. This allows for the use of techniques such as derivatives and gradients to find the optimal solution.

Can Lagrange multiplier be used for non-linear functions?

Yes, Lagrange multiplier can be used for non-linear functions. It is a general technique that can be applied to any function, regardless of its linearity.

What are some limitations of using Lagrange multipliers?

One limitation of using Lagrange multipliers is that it may not always provide the global optimal solution. Additionally, it can be computationally expensive for high-dimensional problems. Another limitation is that it requires the constraint to be differentiable, which may not always be the case.

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