Lagrange Multiplier Trouble

  • #1

Homework Statement


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]


Homework 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]

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:

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


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]


Homework 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]

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?

##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.
 
  • #3
All clear, thank you!
 

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