At an interior maximum, $\dot E(t)=0$, equivalently
$$
q_{EI}e^{-q_{EI}t}=q_{SE}e^{-q_{SE}t}.
$$
Thus
$$
t_{\max}=\frac{\log(q_{EI}/q_{SE})}{q_{EI}-q_{SE}}.
$$
This is positive for distinct positive rates. Its continuous limit when $q_{EI}=q_{SE}=q$ is $1/q$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 207 / 2 / d / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-207/2/d/iv/solution