x+y=16
the sum of the squares is
x^2+y^2=sum
solve for y in first equation
x+y=16
y=16-x
subsitiute that for y in other equation
x^2+(16-x)^2=sum
x^2+x^2-32x+256=sum
2x^2-32x+256=sum
take derivitive to find the minimum value (or just find the vertex because the parabola opens up)
derivitive is
4x-32=derivitve of sum
the max/min is where the derivitive equals 0
4x-32=0
4x=32
x=8
at x=8
so then y=16-8=8
the smallest value then is 8^2+8^2=64+64=128
