Results 1 to 4 of 4

Math Help - Lagrange multipliers

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    94

    Lagrange multipliers

    Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint(s).


    9. f (x,y,z) = xyz; x^2 + 2y^2 + 3z^2 = 6


    I keep going in circles with this question. I get the gradient of f(x) and g(x) and make them equal each other while the gradient of g(x) is multiplied by lambda (the Lagrange multiplier), and then I multiply every equation on the left and right by a variable so that each equation on the left is xyz. I then try to do some algebraic manipulation but since I can't cancel out the sufficient number of variables I can't solve the problem.

    To get the gradients you have to take the partial derivatives with respect to x, y and z for both f(x, y, z) = xyz and g(x, y, z) = x^2 + 2y^2 + 3z^2 = 6.

    Then you have to multiply the gradient of g(x, y, z) by lambda and make the gradient of f equal the gradient of g (when g is multiplied by lambda).
    Last edited by Undefdisfigure; November 29th 2007 at 11:18 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Undefdisfigure View Post
    Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint(s).


    9. f (x,y,z) = xyz; x^2 + 2y^2 + 3z^2 = 6


    I keep going in circles with this question. I get the gradient of f(x) and g(x) and make them equal each other while the gradient of g(x) is multiplied by lambda (the Lagrange multiplier), and then I multiply every equation on the left and right by a variable so that each equation on the left is xyz. I then try to do some algebraic manipulation but since I can't cancel out the sufficient number of variables I can't solve the problem.

    To get the gradients you have to take the partial derivatives with respect to x, y and z for both f(x, y, z) = xyz and g(x, y, z) = x^2 + 2y^2 + 3z^2 = 6.

    Then you have to multiply the gradient of g(x, y, z) by lambda and make the gradient of f equal the gradient of g (when g is multiplied by lambda).
    You form the Lagrangian:

    L(x,y,z,\lambda)=xyz + \lambda (x^2+2y^2+3z^2-6)

    Then your system of equations is:

     <br />
\frac{\partial}{\partial x}L(x,y,x,\lambda)=0<br />

     <br />
\frac{\partial}{\partial y}L(x,y,x,\lambda)=0<br />

     <br />
\frac{\partial}{\partial z}L(x,y,x,\lambda)=0<br />

     <br />
\frac{\partial}{\partial \lambda}L(x,y,x,\lambda)=0<br />

    (the last of these is just the constraint equation)

    Is that what you have tried to do?

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2007
    Posts
    94
    Quote Originally Posted by CaptainBlack View Post
    You form the Lagrangian:

    L(x,y,z,\lambda)=xyz + \lambda (x^2+2y^2+3z^2-6)

    Then your system of equations is:

     <br />
\frac{\partial}{\partial x}L(x,y,z,\lambda)=0<br />

     <br />
\frac{\partial}{\partial y}L(x,y,z,\lambda)=0<br />

     <br />
\frac{\partial}{\partial z}L(x,y,z,\lambda)=0<br />

     <br />
\frac{\partial}{\partial \lambda}L(x,y,z,\lambda)=0<br />

    (the last of these is just the constraint equation)

    Is that what you have tried to do?

    RonL
    Yeah, but why did you make the third variable in L, i.e. L(x ,y, x, lambda) x?
    Shouldn't it be z?

    And I got as far as you've outlined. Now I have to solve for a variable and get a numerical value for a variable so I can solve the remaining ones. How do I do this?

    Help plz, thanks.
    Last edited by CaptainBlack; November 29th 2007 at 09:14 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Undefdisfigure View Post
    Yeah, but why did you make the third variable in L, i.e. L(x ,y, x, lambda) x?
    Shouldn't it be z?

    And I got as far as you've outlined. Now I have to solve for a variable and get a numerical value for a variable so I can solve the remaining ones. How do I do this?

    Help plz, thanks.
    Sorry should be z - typo and copy and paste there-of.

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. lagrange multipliers
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 10th 2009, 03:24 AM
  2. Lagrange Multipliers
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 1st 2009, 12:49 PM
  3. Lagrange Multipliers in 4D
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 23rd 2009, 10:37 AM
  4. Lagrange multipliers
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 22nd 2009, 02:17 PM
  5. Lagrange Multipliers
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 2nd 2009, 10:26 PM

Search Tags


/mathhelpforum @mathhelpforum