Xét hiệu $C - \dfrac{1}{40} $
$ = \dfrac{x}{(x+10)^2} - \dfrac{1}{40}$
$ = \dfrac{40x- (x+10)^2}{40.(x+10)^2}$
$ = \dfrac{-x^2+20x - 100}{40.(x+10)^2}$
$ = \dfrac{-(x-10)^2}{40.(x+10)^2} ≤ 0 $
Do đó $C ≤ \dfrac{1}{40}$
Dấu "=" xảy ra $⇔x=10$
Vậy Max $C = \dfrac{1}{40}$ khi $x=10$