Results 1 to 7 of 7

Math Help - Solving the equations arising from a Lagrange multipliers problem

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    68

    Unhappy Solving the equations arising from a Lagrange multipliers problem

    Find the maximum value of 2
    x+3y+7z, subject to the constraint x^4+y^4+z^4= 1.

    First I formed my Lagrangian for this problem:

    L(x,y,z) = 2x + 3y + 7z - λ(x^4 + y^4 + z^4 -1)


    I then took my partial derivatives:

    dL/dx = 2 - 4
    λx^3

    dL/dy = 3 - 4λy^3
    dL/dz = 7 - 4λz^3

    For each of these I made x,y and z the subject respectively, giving:

    x^3 = 1/2λ ; y^3 = 3/4λ ; z^3 = 7/4λ

    I now subbed these into my constraint. This gave a scary equation with the above λ expressions raised to the power of 4/3 being equal to 1. This is where I'm stuck, can anyone help me please?



    I found a solution, but not in surd form, 2.38534589 satisfies the λ equation being equal to 1. But my answer must be in the form of a surd
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by mitch_nufc View Post
    Find the maximum value of 2


    x+3y+7z, subject to the constraint x^4+y^4+z^4= 1.


    First I formed my Lagrangian for this problem:

    L(x,y,z) = 2x + 3y + 7z - λ(x^4 + y^4 + z^4 -1)

    I then took my partial derivatives:

    dL/dx = 2 - 4λx^3
    dL/dy = 3 - 4λy^3
    dL/dz = 7 - 4λz^3

    For each of these I made x,y and z the subject respectively, giving:

    x^3 = 1/2λ ; y^3 = 3/4λ ; z^3 = 7/4λ

    I now subbed these into my constraint. This gave a scary equation with the above λ expressions raised to the power of 4/3 being equal to 1. This is where I'm stuck, can anyone help me please?


    I found a solution, but not in surd form, 2.38534589 satisfies the λ equation being equal to 1. But my answer must be in the form of a surd
    i'm just wondering why you thought that this is an "abstract algebra" problem and definitely not a "calculus" problem!!??
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2008
    Posts
    68
    Because I dont have a problem with the calculus, I have a problem solving the equation that comes from the calculus, which is an algebra problem.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mitch_nufc View Post
    Find the maximum value of 2


    x+3y+7z, subject to the constraint x^4+y^4+z^4= 1.


    First I formed my Lagrangian for this problem:

    L(x,y,z) = 2x + 3y + 7z - λ(x^4 + y^4 + z^4 -1)

    I then took my partial derivatives:

    dL/dx = 2 - 4λx^3
    dL/dy = 3 - 4λy^3
    dL/dz = 7 - 4λz^3

    For each of these I made x,y and z the subject respectively, giving:

    x^3 = 1/2λ ; y^3 = 3/4λ ; z^3 = 7/4λ

    I now subbed these into my constraint. This gave a scary equation with the above λ expressions raised to the power of 4/3 being equal to 1. This is where I'm stuck, can anyone help me please?


    I found a solution, but not in surd form, 2.38534589 satisfies the λ equation being equal to 1. But my answer must be in the form of a surd
    Doing what you say you get:

     \left( \frac{1}{2} \lambda \right)^{4/3} + \left( \frac{3}{4} \lambda \right)^{4/3} + \left( \frac{7}{4} \lambda \right)^{4/3} = 1

    \Rightarrow \lambda^{4/3} \left( \left(\frac{1}{2}\right)^{4/3} + \left( \frac{3}{4} \right)^{4/3} + \left( \frac{7}{4}\right)^{4/3}\right) = 1

    Now make \lambda the subject. I don't see any simple surd form here but obviously there is a complicated surd form.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,957
    Thanks
    1631
    Quote Originally Posted by mitch_nufc View Post
    Find the maximum value of 2
    x+3y+7z, subject to the constraint x^4+y^4+z^4= 1.

    First I formed my Lagrangian for this problem:

    L(x,y,z) = 2x + 3y + 7z - λ(x^4 + y^4 + z^4 -1)


    I then took my partial derivatives:

    dL/dx = 2 - 4
    λx^3

    dL/dy = 3 - 4λy^3
    dL/dz = 7 - 4λz^3

    each of those is equal to 0, of course.

    For each of these I made x,y and z the subject respectively, giving:

    x^3 = 1/2λ ; y^3 = 3/4λ ; z^3 = 7/4λ

    I now subbed these into my constraint. This gave a scary equation with the above λ expressions raised to the power of 4/3 being equal to 1. This is where I'm stuck, can anyone help me please?



    I found a solution, but not in surd form, 2.38534589 satisfies the λ equation being equal to 1. But my answer must be in the form of a surd
    It would have made more sense if you had made \lambda the subject of each equation, the set them all equal: \lambda= \frac{1}{2x^3}= \frac{3}{4y^3}= \frac{7}{4y^3}. Or, you can eliminate \lambda by dividing one equation by another. From 4\lambda x^3= 2 and 4\lambda y^4= 3, we get \frac{4\lambda x^3}{4\lambda y^3}= \frac{x^3}{y^3}= \frac{2}{3} so that x^3= \frac{2}{3}y^3 and from 4\lambda y^3= 3 and 4\lambda z^3= 7 we get \frac{4\lambda z^3}{4\lambda y^3}= \frac{z^3}{y^3}= \frac{7}{3} so z^3= \frac{7}{3}y^3. Now x= \sqrt[3]{2/3}y so x^4= (2/3)^{4/3}y^4 and z= \sqrt[3]{7/3}y so z^4= (7/3)^{4/3}y^4. You can put those into the constraint equation to get a single equation for y, and then get x and z.
    Last edited by HallsofIvy; May 7th 2009 at 03:38 AM. Reason: Fixed quote and some latex
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member chella182's Avatar
    Joined
    Jan 2008
    Posts
    267
    Really need to start checking what you've posted before I post Matthew looks like what I've done thus far is right. I just thought it was a bit too complicated. It's far more complicated than the one on the examples sheet we were given, because the constraint there was only x^2+y^2+z^2=1 :\
    Last edited by chella182; May 6th 2009 at 05:39 PM. Reason: Bad grammar
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,802
    Thanks
    692
    Hello, mitch_nufc!

    Find the maximum value of 2x+3y +7z, subject to the constraint x^4 + y^4 + z^4 \:=\: 1

    First I formed my Lagrangian for this problem:

    L(x,y,z,\lambda) \:= \:2x + 3y + 7z - \lambda(x^4 + y^4 + z^4 -1)

    I then took my partial derivatives:

    . . \begin{array}{ccccc}\dfrac{dL}{dx} &=&  2 - 4\lambda x^3 &=& 0 \\ \\[-3mm]<br />
\dfrac{dL}{dy} &=&  3 - 4\lambda y^3 &=& 0 \\ \\[-3mm]<br />
\dfrac{dL}{dz} &=& 7 - 4\lambda z^3 &=& 0 \end{array}
    I would solve for \lambda\!:\quad \lambda \;=\;\frac{1}{2x^3} \;=\;\frac{3}{4y^3} \;=\;\frac{7}{4z^3}

    We have: . 14x^3 \:=\:4z^3 \quad\Rightarrow\quad x \:=\:\left(\tfrac{2}{7}z^3\right)^{\frac{1}{3}} \quad\Rightarrow\quad x^4 \:=\:\left(\tfrac{2}{7}\right)^{\frac{4}{3}}\!z^4 .[1]

    We have: . 28y^3 \:=\:12z^3 \quad\Rightarrow\quad y\:=\:\left(\tfrac{3}{7}z^3\right)^{\frac{1}{3}} \quad\Rightarrow\quad y^4 \:=\:\left(\tfrac{3}{7}\right)^{\frac{4}{3}}\!z^4 .[2]


    Substitute [1] and [2] into the constraint: . \left(\tfrac{2}{7}\right)^{\frac{4}{3}}\!z^4 + \left(\tfrac{3}{7}\right)^{\frac{4}{3}}\!z^4 + z^4 \:=\:1

    Factor: . \bigg[\left(\tfrac{2}{7}\right)^{\frac{4}{3}} + \left(\tfrac{3}{7}\right)^{\frac{4}{3}} + 1\bigg]\,z^4 \;=\;1  \quad\Rightarrow\quad\bigg[\frac{2^{\frac{4}{3}} + 3^{\frac{4}{3}} + 7^{\frac{4}{3}}}{7^{\frac{4}{3}}}\bigg]\,z^4 \;=\;1

    . . z^4 \;=\;\frac{7^{\frac{4}{3}}}{2^{\frac{4}{3}} + 3^{\frac{4}{3}} + 7^{\frac{4}{3}}} \quad\Rightarrow\quad z \;=\;\frac{7^{\frac{1}{3}}} {\left(2^{\frac{4}{3}} + 3^{\frac{4}{3}} + 7^{\frac{4}{3}}\right)^{\frac{1}{4}}}

    And we can back-substitute to determine x and y.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 27th 2011, 09:27 PM
  2. Standard optimization problem using Lagrange Multipliers
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: October 23rd 2011, 09:21 PM
  3. Lagrange multipliers problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 4th 2009, 02:03 PM
  4. Lagrange Multipliers :D
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 13th 2008, 02:29 PM
  5. Lagrange multipliers
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 28th 2007, 08:38 PM

Search Tags


/mathhelpforum @mathhelpforum