Calculate maximum duration nuclear power plant can fulfill demand

Let $P(t)$ be the electrical power generated by the nuclear power plant at time $t$, and let $P_0 = P(0)$ be the initial power output at $t = 0$.

Since the power generated by the fuel is directly proportional to its rate of radioactive decay (activity), $P(t)$ decreases exponentially over time with half-life $T$: $P(t) = P_0 \left(\frac{1}{2}\right)^{\frac{t}{T}}$

Let $P_{\text{req}}$ be the constant power requirement of the community. It is given that at $t = 0$, the required power is $12.5\%$ of the plant's output: $P_{\text{req}} = 12.5\% \text{ of } P_0 = \frac{12.5}{100} P_0 = \frac{1}{8} P_0$

The power plant can fulfill the community's demand as long as the generated power $P(t)$ is greater than or equal to the required power $P_{\text{req}}$: $P(t) \ge P_{\text{req}}$

The maximum duration $t = nT$ occurs when the generated power decreases to match the required power exactly: $P(nT) = P_{\text{req}}$

Substituting the expressions for $P(t)$ and $P_{\text{req}}$: $P_0 \left(\frac{1}{2}\right)^{\frac{nT}{T}} = \frac{1}{8} P_0$

Dividing both sides by $P_0$: $\left(\frac{1}{2}\right)^n = \frac{1}{8}$

Since $\frac{1}{8} = \left(\frac{1}{2}\right)^3$: $\left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^3 \implies n = 3$

Thus, the value of $n$ is $3$.