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?
 A: The first problem is that you substituted $x=\frac12(1-a)$ into $f(x)$ incorrectly; the second (and more important) problem is that you need to define a new function whose independent variable is $a$. Specifically, let $g(a)$ be the maximum attained by the function $f(x)=ax(1-x-a)$; you want to find the value of $a$ that maximizes $g(a)$. Substituting $x=\frac12(1-a)$ into $f(x)$, we find that
$$\begin{align*}
g(a)&=a\left(\frac{1-a}2\right)\left(1-\frac{1-a}2-a\right)\\
&=\frac{a}4(1-a)\big(2-(1-a)-2a\big)\\
&=\frac{a}4(1-a)(1-a)\\
&=\frac14\left(a-2a^2+a^3\right)\;.
\end{align*}$$
Now $g'(a)=\frac14\left(1-4a+3a^2\right)$. Setting this equal to $0$, we have $3a^2-4a+1=0$. To solve for $a$ you can either use the quadratic formula or notice that $3a^2-4a+1=(3a-1)(a-1)$; either way, you find that $g'(a)=0$ for $a=1$ and $a=\frac13$. By analyzing the sign of $g'(a)$ or by using the second derivative test you can check that $g(a)$ has a local maximum (of $\frac1{27}$) at $a=\frac13$ and a local minimum (of $0$) at $a=1$.
However, a quick check of the graph of $g(a)$ will show you that it increases without bound as $a\to\infty$, and this is also clear algebraically: as $a\to\infty$, $1-a\to-\infty$, so $\frac14a(1-a)^2\to\infty$. Thus, you can make the maximum of $f(x)$ as large as you want by choosing $a$ large enough.
A: The maximum value of the function, which occurs at $x = \frac{1-a}{2}$, is:
$$\begin{align}f\big(\frac{1-a}{2}\big) & = a\big(\frac{1-a}{2}\big)(1-\frac{1-a}{2}-a)\\
&= a\big(\frac{1-a}{2}\big)\bigl(\frac{1-a}{2}\bigr)\\
&= a\big(\frac{1-a}{2}\big)^2.
\end{align}$$
As $a \rightarrow \infty$, this grows without bound.
A: Before we start differentiating, let's think about $f(x)$ a bit. Let $x=-100$, and let $a=102$. Then $f$ is large. It can be made arbitrarily large by choosing positive $a$ suitably. 
If one wants a maximum to exist, the problem needs to be modified.  For example, instead of $a\gt 0$, you could specify that $0\lt a\lt 1$, or equivalently that the maximum occurs at a positive value of $x$.   
A: Once you insert $x = \dfrac{1-a}{2}$, you can no longer differentiate it since $a$ is a constant and differentiating $f(a)$ will lead to $0$. At best, you can find out $f'(x)$ and substitute the value of x in terms of a to re-verify it is 0 (but then, that's how you got $x = \dfrac{1-a}{2}$ to begin with)
A: 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.
