On the interval [0,1] the function $x^{25} (1–x)^{75}$ takes its maximum value at the point

1/4

Let $f(x) = x^{25} (1–x)^{75}$. The critical points of f are given by $f’(x) = 0$.

But $f’(x) = x^{24} (1–x)^{74} [25 – 25x –75x]$

$= 25x^{24} (1–x)^{74} (1–4x)$

Thus the critical points are 0,1,1/4

Since $f (0) = 0, f(a) = 0$ and $f(1/4) = (1/4)^{25} (3/4)^{75}$

so f takes its maximum value at $x = 1/4$