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: Write the growth rate as a function of population size.
The logistic growth equation is
\[ \frac{dN}{dt}=rN\left(1-\frac{N}{K}\right)=rN-\frac{r}{K}N^2 \]
This is a quadratic function of \(N\) alone, since \(r\) and \(K\) are constants. Call this function \(g(N)=rN-\dfrac{r}{K}N^2\).

Step 2: Differentiate \(g(N)\) with respect to \(N\).
\[ \frac{dg}{dN}=r-\frac{2r}{K}N \]

Step 3: Set the derivative to zero to find the turning point.
\[ r-\frac{2r}{K}N=0 \]
Divide both sides by \(r\), since \(r\) is a positive growth rate and not zero:
\[ 1-\frac{2N}{K}=0 \] \[ N=\frac{K}{2} \]

Step 4: Confirm it is a maximum, not a minimum.
The second derivative of \(g(N)\) is
\[ \frac{d^2g}{dN^2}=-\frac{2r}{K} \]
Since \(r\) and \(K\) are both positive, this second derivative is negative everywhere, so the curve of \(g(N)\) is concave down. A concave down curve with a single turning point has a maximum there, not a minimum, so \(N=K/2\) is indeed where \(dN/dt\) is largest.

Step 5: Substitute the given carrying capacity.
\[ N=\frac{K}{2}=\frac{9000}{2}=4500 \]

Step 6: Final answer.
The growth rate \(dN/dt\) is maximized when the population reaches exactly half the carrying capacity.
\[ \boxed{N=4500} \]