I'm trying to find shortest distance from the point $(2,0-3)$ to $x+y+z=1$.
I found $d^2=(x-2)^2+y^2+(z+3)^2$
Substituting for $z$: $d^2=(x-2)^2+y^2+(-x-y+1+3)^2$
$f_x=2x-4-2(-x-y+4)$ and $f_y=2y-2(-x-y+4)$ Next would be finding zeros as the critical points and using them as the $xyz$ values in the formula for $d$ but I'm having issues finding those points.
Best Answer
We have $4x+2y=12$ and $2x+4y=8$. First equation reduces to $y=6-2x$. Plugging that into the second gives $2x+4(6-2x)=8$ which solves to $2x+24-8x=8$ or $x=8/3$. Plugging this answer into $y=6-2x$ yields $y=2/3$. This is one of the many ways to solve such a system. Anyway, with this info you can find $z$ and from here you can find the shortest distance.