Calculating maximum of function I want to determine the value of a constant $a > 0$ which causes the highest possible value of $f(x) = ax(1-x-a)$.
I have tried deriving the function to find a relation between $x$ and $a$ when $f'(x) = 0$, and found $x = \frac{1-a}{2}$. 
I then insert it into the original function: $f(a) = \frac{3a - 6a^2 - 5a^3}{8}$
I derived it to $f'(a) = \frac{-15a^2 + 12a - 3}{8}$
I thought deriving the function and equaling it to $0$ would lead to finding a maximum, but I can't find it. I can't go beyond $-15a^2 + 12a = 3$.
Where am I going wrong?

We can resort to some algebra along with the calculus you are using, to see what happens with this function:
$$f(x)=ax(1−x−a)=-ax^2+(a-a^2)x$$
Note that this is a parabola. Since the coefficient of the $x^2$ term is negative, it opens downward so that the maximum value is at the vertex. As you have already solved, the vertex has x-coordinate $x=\frac{1−a}{2}$. Additionally, Théophile showed that the vertex's y-coordinate is $f(\frac{1−a}{2})=a(\frac{1−a}{2})^2$ which is unbounded as $a$ increases.
