jmc

Note that maximizing $x(1-x{)}^5$ is equivalent to maximizing $5x(1-x{)}^5.$ Then by AM-GM, $$\frac{5x+(1-x)+(1-x)+(1-x)+(1-x)+(1-x)}{6}\ge \sqrt[6]{5x(1-x{)}^5}.$$This simplifies to $\sqrt[6]{5x(1-x{)}^5}\le \frac{5}{6}.$ (Note how the left hand-side simplifies to a constant, which is why we consider $5x(1-x{)}^5.$) Hence, $$x(1-x{)}^5\le \frac{1}{5}{\left(\frac{5}{6}\right)}^6=\frac{3125}{46656}.$$Equality occurs when $5x=1-x,$ or $x=\frac{1}{6},$ so the maximum value is $\boxed{\frac{3125}{46656}}.$