A radioactive nucleus A of half-life 1.0 hour decays to a radioactive nucleus B of half-life 4.0 hours, which decays to a stable nucleus. At t=0 only…

Let the decay constants be $\lambda_A=\ln2/1\ \mathrm{h^{-1}}=0.6931\ \mathrm{h^{-1}}$ and $\lambda_B=\ln2/4=0.1733\ \mathrm{h^{-1}}$. The rate equation for $B$ is $\dfrac{dN_B}{dt}=\lambda_AN_A-\lambda_BN_B$ with $N_A=N_0e^{-\lambda_At}$. The number of $B$ nuclei is maximum when $dN_B/dt=0$, i.e. when the activities are equal, $\lambda_AN_A=\lambda_BN_B$. Solving the equation gives $N_B=\dfrac{\lambda_AN_0}{\lambda_B-\lambda_A}\left(e^{-\lambda_At}-e^{-\lambda_Bt}\right)$. Setting its derivative to zero gives $t_{max}=\dfrac{\ln(\lambda_A/\lambda_B)}{\lambda_A-\lambda_B}=\dfrac{\ln4}{0.5199}=\dfrac{1.3863}{0.5199}=2.67$ hours. This is the classic transient-equilibrium situation: before $t_{max}$ the production of $B$ exceeds its decay and $N_B$ rises; afterwards the shrinking supply of $A$ makes $N_B$ fall. Note that the answer does not depend on $N_0$, and it is not simply the mean of the two half-lives or the time when $N_A$ equals $N_B$.