Rejection sampling a Beta(3,2) with the tightest box

To simulate from $\text{Beta}(3,2)$, whose density is $f(x)=12x^2(1-x)$ on $(0,1)$, you draw points uniformly from the box $[0,1]\times[0,a]$ and keep a point only if it falls under the graph of $f$. Choose $a>0$ to minimise rejections. With this choice, what is the expected number of rejected points per accepted point?

$\tfrac79$. The box must cover the density, so the best $a$ is $\max f=f(\tfrac23)=\tfrac{16}9$. Each point is accepted with probability $\tfrac{1}{a}=\tfrac9{16}$, so rejections before an acceptance are geometric with mean $\tfrac{1-9/16}{9/16}=\tfrac79$.