1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lagrange Multiplier Trouble

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Find extrema for [itex]f\left( x,y,z \right) ={ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }[/itex]
    under the constraint [itex]g\left( x,y,z \right) ={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=16[/itex]


    2. Relevant equations
    (1) [itex]\nabla f=\lambda \nabla g[/itex]
    (2) [itex]g\left( x,y,z \right) ={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=16[/itex]

    3. The attempt at a solution
    (1)[itex]\Rightarrow \left( 3{ x }^{ 2 },3{ y }^{ 2 },3{ z }^{ 2 } \right) =\lambda \left( 2x,2y,2z \right)[/itex]⇔(3) [itex]x=y=z[/itex]
    (3)→(2)[itex]\Rightarrow x=y=z=\pm \frac { 4 }{ \sqrt { 3 } }[/itex]
    But subbing in x=y=0, z=4 gives a greater value..
    What am I doing wrong?
     
    Last edited: Jan 22, 2013
  2. jcsd
  3. Jan 22, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    ##f_{x}=\lambda g_{x} \longrightarrow \: 3x^2 = 2\lambda x,## so either ##x = 0## or ##3x = 2 \lambda##. Similarly for y and z. There are many, many candidate solutions, some of which are maxima, some of which are minima and (perhaps) some of which are constrained saddle points. Besides solutions with x = y = z you also have solutions with, for example, x = 0 and y = z ≠ 0.
     
  4. Jan 22, 2013 #3
    All clear, thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Lagrange Multiplier Trouble
  1. Lagrange Multiplier (Replies: 2)

  2. Lagrange Multipliers (Replies: 7)

  3. Lagrange Multipliers (Replies: 8)

  4. Lagrange Multipliers (Replies: 8)

Loading...