What is the smallest value of $x^2+y^2$ when $x+y=6$? If $ x+y=6 $ then what is the smallest possible value for $x^2+y^2$?
Please show me the working to show where I am going wrong!
Cheers

Reorder $x+y = 6$ to
$y=6-x$
Substituting $y$ in $(x^2+y^2)$ yields
$x^2+(6-x)^2 = 2x^2-12x+36$
The minimum occurs where the derivative equals $0$
$4x -12 = 0$
Therefore at the minimum,
$x=3$
Hence the minimum is $2\cdot 3^2-12\cdot 3+36 = 18$
