Question

Population growth of a species can be modelled as \[ \frac{dN(t)}{dt}=rN(t)\left(1-\frac{N(t)}{K}\right), \] where \(N(t)\) is the population size at time \(t\), \(r\) is the growth rate, and \(K\) is the carrying capacity of the environment. For \(K=9000\), \(\dfrac{dN(t)}{dt}\) is maximized at \(N=\) ______ (Answer in integer)

Hint:

Treat \(dN/dt\) as a function of \(N\) alone, differentiate it with respect to \(N\), and set that derivative to zero to find where it peaks.

Answer 1

Correct Answer: 4500

Step 1: Recognise the shape of the growth curve.
Write the growth rate as $g(N)=rN\left(1-\dfrac{N}{K}\right)$. Expand this out and it is $g(N)=rN-\dfrac{r}{K}N^2$, a downward opening parabola in $N$ because the coefficient of $N^2$ is negative.

Step 2: Find where the parabola touches zero.
Since $g(N)=rN\left(1-\dfrac{N}{K}\right)$ is a product of two factors, $g(N)=0$ exactly when $N=0$ or when $1-\dfrac{N}{K}=0$, that is $N=K$. So the parabola crosses the horizontal axis at $N=0$ and $N=K$.

Step 3: Use the symmetry of a parabola.
For any downward opening parabola, the highest point sits exactly midway between its two zeros, because the curve is a mirror image of itself around the vertical line through the vertex. Here the two zeros are $0$ and $K$, so the vertex, the peak of $g(N)$, sits at
\[ N=\frac{0+K}{2}=\frac{K}{2} \]

Step 4: Confirm this matches the calculus answer.
Differentiating $g(N)=rN-\dfrac{r}{K}N^2$ gives $g'(N)=r-\dfrac{2r}{K}N$, which is zero at $N=K/2$ as well, and the parabola opens downward so this is a maximum, not a minimum. Both routes agree.

Step 5: Plug in the numbers.
With $K=9000$,
\[ N=\frac{9000}{2}=4500 \] \[ \boxed{N=4500} \]