Find two numbers whose product is 16 and whose sum of squares is minimum

Answer 1

let the two numbers be x and y
xy = 16
y = 16/x

Let S be the sum of their squares
S = x^2 + y^2
= x^2 + 256/x^2 = x^2 + 256x^-2
dS/dx = 2x - 512x^-3
= 2x - 512/x^3
= 0 for a min of S
2x = 512/x^3
2x^4 = 512
x^4 = 256
x = ±4
then 16/±4 = ±4

the two numbers are 4 and 4 , (you probably anticipated that answer)
or -4 and -4