Question

Find the maximum and minimum values of f(x,y,z)=3x+3y+3zf(x,y,z)=3x+3y+3z on the sphere x2+y2+z2=1x2+y2+z2=1.

Answer 1

Lagrangian:

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

Set partial derivative equal to 0 and look for critical points:

$L_x=3+2\lambda x=0$
$L_y=3+2\lambda y=0$
$L_z=3+2\lambda z=0$
$L_\lambda=x^2+y^2+z^2-1=0$

$xL_x+yL_y+zL_z=3(x+y+z)+2\lambda(x^2+y^2+z^2)=0$
$3(x+y+z)+2\lambda=0$

We require $\lambda\neq0$, so we find that

$L_x-L_y=2\lambda(x-y)=0\implies x=y$
$L_y-L_z=2\lambda(y-z)=0\implies y=z$
$L_x-L_z=2\lambda(x-z)=0\implies x=z$

$3(3x)+2\lambda=9x+2\lambda=0$

$\begin{cases}9x+2\lambda=0\\3+2\lambda x=0\end{cases}\implies2\lambda=-9x$
$\implies 3+(-9x)x=3-9x^2=0\implies x=\pm\dfrac1{\sqrt3}$

which means there are two possible critical points, $\left(\dfrac1{\sqrt3},\dfrac1{\sqrt3},\dfrac1{\sqrt3}\right)$ and $\left(-\dfrac1{\sqrt3},-\dfrac1{\sqrt3},-\dfrac1{\sqrt3}\right)$. It's clear enough that the first gives a maximum value of $f\left(\dfrac1{\sqrt3},\dfrac1{\sqrt3},\dfrac1{\sqrt3}\right)=3\sqrt3$, and a minimum value of $f\left(-\dfrac1{\sqrt3},-\dfrac1{\sqrt3},-\dfrac1{\sqrt3}\right)=-3\sqrt3$.