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Calculation regarding to Lagrange Multiplier

  1. Aug 15, 2010 #1
    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
  2. jcsd
  3. Aug 15, 2010 #2


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    Homework Helper

    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.

  4. Aug 15, 2010 #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
  5. Aug 15, 2010 #4


    User Avatar
    Homework Helper

    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.

  6. Aug 16, 2010 #5


    User Avatar
    Science Advisor

    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 writen 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.

  7. Aug 16, 2010 #6
    Thanks for the help from both of you, I have finally got it right this time!
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