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

MATH101 B April 2024

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[hide] Question 18

A key statistic for describing how a population changes is the average population over a fixed time interval, such as a season, a year, or a generation. For any given function $f$, the average value from $t=a$ to $t=b$ is $${\frac {1}{b-a}}\int _{a}^{b}f(t)\mathrm {d} t.$$ Suppose the population (in thousands) of a species at time $t$ (in months) is given by the function $$P(t)=1+e^{-(t-4)^{2}}+\cos(\pi t).$$ For which four-month period (i.e., from $a=T$ to $b=T+4$) is the average value of $P$ the greatest? You need to justify your answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
This is an optimization problem. Start by writing down an expression for the objective function.

[show] Hint 2
We are tasked with finding the time $T$ for which the average population from $T$ to $T+4$ is as large as possible. This function is given by $$F(T)={\frac {1}{T+4-T}}\int _{T}^{T+4}P(t)\mathrm {d} t.$$ Can you proceed from here?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

[show] Solution

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 $T=6$.