Maxima of expression on given interval

Question

How do I find the maximum value of $x(1-x)^n$ on the interval $[0,1]$?

Answer 1

Let $f(x) = x(1-x)^{n}$. Then, $f'(x) = (1-x)^{n} - nx(1-x)^{n-1}$. We wish to find critical points of this function. These are the endpoints $0$ and $1$, as well as the points where $f'(x) = 0$. We have:

$$f'(x) = 0$$

$$(1-x)^{n} - nx(1-x)^{n-1} = 0$$

$$(1 - x) - nx = 0\ \ (x\neq 1)$$

$$(n+1)x = 1$$

$$x = \frac{1}{n+1}$$

Then, note that $f'(x) > 0$ if $0 < x < \frac{1}{n+1}$, and $f'(x) < 0$ if $\frac{1}{n+1}< x < 1$. Thus, $x = \frac{1}{n+1}$ is at a relative maximum (global in that interval), and the maximum value is thus $f(\frac{1}{n+1}) = \boxed{\frac{n^{n}}{(n + 1)^{n + 1}}}$