Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 207 / 3 / a / i / Solution

Since $n_0=q{n}_k$ and the $k$ experimental arms each contain $n_k$ patients,ConsequentlyDifferentiation gives $V^{\prime }(q)=(1-k/q^2)/n_{\mathrm{tot}}$, so the stationary point is $q=\sqrt{k}$. Sincethis is the unique minimum. Thus a shared standard arm should be $\sqrt{k}$ times the size of each new-treatment arm.