Application of max. value of quadratic functions(20m)Urgent!

Let x and (12 - x) be the 2 numbers.
Consider their product, y = x(12 - x)
y = -x^2 + 12x = -(x^2 - 12x) = -(x^2 - 12x + 36 - 36)
= -(x^2 - 2(x)(6) + 6^2) + 36
= -(x - 6)^2 + 36
y is at maximum when x = 6
Therefore, the 2 numbers are 6 and 6, and the product is 36.

let one of number be y..another one in (12-y)
product is at the maximum:
y(12-y)
=-(y^2-12y)
=-(y^2-(2)(6)(y)+6^2-6^2)
=-(y-6)^2+6^2
=-(y-6)^2+36

so the maximum product is 36

y(12-y)=36
-y^2+12y=36
y^2-12y+36=0
(y-6)(y-6)=0
y=6 (repeated)
so the number is 6 and 6