Science:Math Exam Resources/Courses/MATH101 B/April 2024/Question 18/Solution 1

We wish to optimize $$F(T)={\frac {1}{4}}\int _{T}^{T+4}\left(1+e^{-(t-4)^{2}}+\cos(\pi t)\right)\mathrm {d} t.$$ Ultimately, we wish to solve $F'(T)=0$ and then verify that the solution is indeed a local maximum. A first attempt might be to try integrating the above expression in order to obtain another expression for $F(T)$, but note the presence of $e^{-(t-4)^{2}}$ in the integrand. The function $x\mapsto e^{x^{2}}$ notoriously does not have a closed form antiderivative, so we should try to obtain $F'(T)$ without integrating. The fundamental theorem of calculus does precisely this, but we have to apply it carefully, since the independent variable $T$ appears in both integral bounds. A trick that can help here is to rewrite the integral as follows: $$\int _{T}^{T+4}P(t)\mathrm {d} t=\int _{T}^{c}P(t)\mathrm {d} t+\int _{c}^{T+4}P(t)\mathrm {d} t=-\int _{c}^{T}P(t)\mathrm {d} t+\int _{c}^{T+4}P(t)\mathrm {d} t,$$ where $c$ is an real number in the domain of $P$. We see then that $${\begin{aligned}F'(T)=P(T+4)-P(T)&={\frac {1}{4}}(1+e^{-T^{2}}+\cos(\pi (T+4))-1-e^{-(T-4)^{2}}-\cos(\pi T))\\&={\frac {1}{4}}(e^{-T^{2}}-e^{-(T-4)^{2}}+\cos(\pi T+4\pi )-\cos(\pi T)).\end{aligned}}$$ Since $\cos$ is a periodic function with period $2\pi$, $\cos(\pi T+4\pi )=\cos(\pi T)$, and $$F'(T)={\frac {1}{4}}(e^{-T^{2}}-e^{-(T-4)^{2}}).$$ We can now solve $F'(T)=0$: $${\frac {1}{4}}(e^{-T^{2}}-e^{-(T-4)^{2}})=0\iff e^{-T^{2}}=e^{-(T-4)^{2}}\iff T^{2}=(T-4)^{2}\iff T=2.$$ Therefore $T=2$ is a critical point of $F$. Now let us check that it is indeed a local maximum by computing $F''(2)$ with the chain rule: $$F''(T)={\frac {1}{4}}\left(e^{-T^{2}}\cdot (-2T)-e^{-(T-4)^{2}}\cdot (-2(T-4))\right).$$ We can compute $F(4)=-e^{-4}-e^{-4}<0$, which verifies that $T=4$ is indeed a maximum. Since the function $F$ is differentiable and has no other critical points, it follows that $T=4$ is a global maximum.

Thus the 4-month period that maximizes the average population is from $T=2$ to $T=6$.