Question

Find three real numbers​ x, y, and z whose sum is 6 and the sum of whose squares is as small as possible. g

Answer 1

We're minimizing $x^2+y^2+z^2$ subject to $x+y+z=6$. Using Lagrange multipliers, we have the Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x+y+z-6)$

with partial derivatives

$\begin{cases}L_x=2x+\lambda\\L_y=2y+\lambda\\L_z=2z+\lambda\\L_\lambda=x+y+z-6\end{cases}$

Set each partial derivative equal to 0:

$\begin{cases}2x+\lambda=0\\2y+\lambda=0\\2z+\lambda=0\\x+y+z=6\end{cases}$

Subtracting the second equation from the first, we find

$2x-2y=0\implies x=y$

Similarly, we can determine that $x=z$ and $y=z$ by taking any two of the first three equations. So if $x=y=z$ determines a critical point, then

$x+y+z=3x=6\implies x=y=z=2$

So the smallest value for the sum of squares is $2^2+2^2+2^2=12$ when $(x,y,z)=(2,2,2)$.