Question

Find two nonnegative numbers whose sum is 1 such that the sum of their squares is a minimum.

Answer 1

x + y = 1
x^2 + y^2 = s
s = x^2 +(1-x)^2
s = x^2 + (1-2x+ x^2)
s = 2x^2 - 2x + 1
you now need to find minimum value of this trinomial - the value of x when its minimum then finding y is easy

Answer 2

find the coordinates of vertex of graph of the expression will give you your answer
x cood od of vertex = -b/2a for general form of the quadratic

Answer 3

Im a little confused. When I work it out I get the following:

x^2 - y^2 = s
s = x^2 + (1 - x)^2
ds = 2x + 2(1 - x)(-1) = 0
= 2x + 2 (-1 + x) = 0
= 2x - 2 + 2x = 0
= 4x - 2 = 0
x = 1/2

Am I wrong? Most likely yes, I just want to be sure I know where Im screwing up.