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

We wish to optimize $$F(T)=\frac{1}{4}\underset{T}{\overset{T+4}{\int }}(1+e^{-(t-4{)}^2}+\cos (\pi t))\mathrm{d}t.$$ Ultimately, we wish to solve $F^{\prime }(T)=0$ 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 $F^{\prime }(T)$ 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: $$\underset{T}{\overset{T+4}{\int }}P(t)\mathrm{d}t=\underset{T}{\overset{c}{\int }}P(t)\mathrm{d}t+\underset{c}{\overset{T+4}{\int }}P(t)\mathrm{d}t=-\underset{c}{\overset{T}{\int }}P(t)\mathrm{d}t+\underset{c}{\overset{T+4}{\int }}P(t)\mathrm{d}t,$$ where $c$ is an real number in the domain of $P$. We see then that $$\begin{array}{rl}{F}^{\prime }(T)=P(T+4)-P(T)& =\frac{1}{4}(1+e^{-T^2}+\cos (\pi (T+4))-1-e^{-(T-4{)}^2}-\cos (\pi T))\\ & =\frac{1}{4}(e^{-T^2}-e^{-(T-4{)}^2}+\cos (\pi T+4\pi )-\cos (\pi T)).\end{array}$$ Since $\cos$ is a periodic function with period $2\pi$, $\cos (\pi T+4\pi )=\cos (\pi T)$, and $$F^{\prime }(T)=\frac{1}{4}(e^{-T^2}-e^{-(T-4{)}^2}).$$ We can now solve $F^{\prime }(T)=0$: $$\frac{1}{4}(e^{-T^2}-e^{-(T-4{)}^2})=0⟺e^{-T^2}=e^{-(T-4{)}^2}⟺T^2=(T-4{)}^2⟺T=2.$$ 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: $$F^{″}(T)=\frac{1}{4}(e^{-T^2}\cdot (-2T)-e^{-(T-4{)}^2}\cdot (-2(T-4))).$$ 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 $T=6$.