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\\eq0`, 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\frac1{\sqrt3}`

which means there are two possible critical points,
`\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right)` and
`\left(-\frac1{\sqrt3},-\frac1{\sqrt3},-\frac1{\sqrt3}\right)`. It's clear enough that the first gives a maximum value of
`f\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right)=3\sqrt3`, and a minimum value of
`f\left(-\frac1{\sqrt3},-\frac1{\sqrt3},-\frac1{\sqrt3}\right)=-3\sqrt3`.