= Solution
For $N_{b0}=0$, differentiate the distinct-<nuclear decay constant> result:
$$
\dot N_b=\frac{\lambda_aN_{a0}}{\lambda_b-\lambda_a}
(-\lambda_ae^{-\lambda_at}+\lambda_be^{-\lambda_bt}).
$$
Equating the production and loss terms gives $e^{(\lambda_b-\lambda_a)t}=\lambda_b/\lambda_a$, hence
$$
\boxed{t_{\max}=\frac{\ln(\lambda_b/\lambda_a)}{\lambda_b-\lambda_a}.}
$$
For positive <nuclear decay constants> and $N_{a0}>0$, this is positive regardless of their ordering. Initially $\dot N_b>0$, while $N_b\to0$ at late times; the stationary point is the unique maximum. At that time $\lambda_aN_a=\lambda_bN_b$. If the <nuclear decay constants> are equal, differentiating $\lambda N_{a0}te^{-\lambda t}$ gives $\boxed{t_{\max}=1/\lambda}$, also the continuous limit of the expression above.
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