Science:Math Exam Resources/Courses/MATH102/December 2015/Question 18 (b)

MATH102 December 2015

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Question 18 (b)

A dangerous infectious disease spreads through Vancouver as described by the differential equation $I'=\beta I(N-I)-\mu I=\beta I\left(N-{\frac {\mu }{\beta }}-I\right)$ where $\beta >0$ is the transmission rate constant, $N$ is the total population size (constant), $\mu >0$ is the recovery rate and $I(t)$ is the number of infected individuals. Assume that $N>{\frac {\mu }{\beta }}$. (b) At what value of $I$ is the infection rate $(dI/dt)$ largest?

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Find the critical point(s) of the function $\beta I\left(N-{\frac {\mu }{\beta }}-I\right)$ and see the concavity.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since ${\frac {dI}{dt}}$ is given as ${\frac {dI}{dt}}=\beta I\left(N-{\frac {\mu }{\beta }}-I\right)$, the question can be rewritten as follows; find $I$ at which the absolute maximum of the function $f(I)=\beta I\left(N-{\frac {\mu }{\beta }}-I\right)$ is achieved. For this purpose, we first find the critical points. Using the product rule, we get the derivative of $f$; $f'(I)=\beta \left(N-{\frac {\mu }{\beta }}-I\right)-\beta I=\beta \left(N-{\frac {\mu }{\beta }}-2I-\right)$. Then, $I={\frac {1}{2}}\left(N-{\frac {\mu }{\beta }}\right)$ is the only critical point. Since the second derivative of $f$ is always negative $f''(I)=-2\beta <0$, the function $f$ is concave down. Therefore, at the critical point ${\frac {1}{2}}\left(N-{\frac {\mu }{\beta }}\right)$, we obtain its maximum. Answer: $\color {blue}{\frac {1}{2}}\left(N-{\frac {\mu }{\beta }}\right)$