Mean or maximum for the endpoint

Idea. Both estimators are unbiased, so compare variances. The mean uses all the data but each observation is noisy; the maximum sits close to $\theta$ and its spread shrinks like $\frac{1}{n^2}$ instead of $\frac{1}{n}$.

Step 1: the mean. $$\mathrm{Var}(2\bar{X})=4\cdot \frac{{\theta }^2/12}{n}=\frac{{\theta }^2}{3n}.$$

Step 2: the maximum. $\mathrm{Var}(M)=\frac{n}{n+2}{\theta }^2-\frac{n^2}{(n+1{)}^2}{\theta }^2=\frac{n{\theta }^2}{(n+1{)}^2(n+2)}$, so $$\mathrm{Var}\left(\frac{n+1}{n}M\right)=\frac{{\theta }^2}{n(n+2)}.$$

Step 3: the ratio. $$\frac{{\theta }^2/(3n)}{{\theta }^2/(n(n+2))}=\frac{n+2}{3}=4\text{for }n=10.$$

The lesson. The scaled maximum is four times more precise here, and the advantage grows with $n$. The maximum is a sufficient statistic for $\theta$: once you know it, the other observations carry no further information about $\theta$.