# Lagrange Multiplier Trouble

## Homework Statement

Find extrema for $f\left( x,y,z \right) ={ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }$
under the constraint $g\left( x,y,z \right) ={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=16$

## Homework Equations

(1) $\nabla f=\lambda \nabla g$
(2) $g\left( x,y,z \right) ={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=16$

## The Attempt at a Solution

(1)$\Rightarrow \left( 3{ x }^{ 2 },3{ y }^{ 2 },3{ z }^{ 2 } \right) =\lambda \left( 2x,2y,2z \right)$⇔(3) $x=y=z$
(3)→(2)$\Rightarrow x=y=z=\pm \frac { 4 }{ \sqrt { 3 } }$
But subbing in x=y=0, z=4 gives a greater value..
What am I doing wrong?

Last edited:

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Find extrema for $f\left( x,y,z \right) ={ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }$
under the constraint $g\left( x,y,z \right) ={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=16$

## Homework Equations

(1) $\nabla f=\lambda \nabla g$
(2) $g\left( x,y,z \right) ={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=16$

## The Attempt at a Solution

(1)$\Rightarrow \left( 3{ x }^{ 2 },3{ y }^{ 2 },3{ z }^{ 2 } \right) =\lambda \left( 2x,2y,2z \right)$⇔(3) $x=y=z$
(3)→(2)$\Rightarrow x=y=z=\pm \frac { 4 }{ \sqrt { 3 } }$
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.

All clear, thank you!