Set $B_1M=a, A_1M=b, AM=x, BM=y, \angle B_1AM=\alpha$. Then $$S_{CB_1A_1M} = ab=\frac{1}{2}xy\sin 2\alpha$$Note that $\frac{1}{2}\sin 2\alpha$ is constant. That is, we need to find the maximum $xy$ where $x+y=\text{constant} = c$. We have $xy=x(c-x)=xc-x^2=\frac{c^2}{4}-\left(x-\frac{c}{2}\right)^2 \leqslant \frac{c^2}{4}$, equality holds when $x=\frac{c}{2} \Rightarrow y = \frac{c}{2}$. $\text{Answer:}$ $\boxed{\text{When} \, M \, \text{is a midpoint of} \, AB}$