The $A_{i}'s$ and $B_{i}'s$ are known. I seek the $p$ which maximizes $L(p)$. I thought it might be easier to maximize $\log L(p)$ instead $L(p)$, but I think it is a dead end
$\log L(p)=\sum_{i=1}^{20}\log \left( (1-p)A_{i}+p B_{i} \right)$
$\frac{d}{dp}\log L(p)=\sum_{i=1}^{20}\frac{B_{i}-A_{i}}{\left( (1-p)A_{i}+pB_{i}\right)}=0$
$\frac{d}{dp}\log L(p)=\sum_{i=1}^{20}\frac{B_{i}-A_{i}}{ A_{i}+p(B_{i}-A_{i})}=0$
$\sum_{i=1}^{20}\frac{p(B_{i}-A_{i})}{ A_{i}+p(B_{i}-A_{i}) }=0$
$\sum_{i=1}^{20} 1 - \frac{A_{i}}{ A_{i}+p(B_{i}-A_ {i}) }=0$
but I can't seem to isolate the p. Any ideas on how this can be solved?