The number of rabbits on an island is modelled by the equation $$P = \frac{100 e^{-t/10}}{1 + 3 e^{-t/10}} + 40,$$ where $P$ is the number of rabbits, $t$ years after they were introduced onto the island - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 4

To find $\frac{dP}{dt}$, use the quotient rule:

$\frac{dP}{dt} = \frac{(1 + 3 e^{-t/10}) \cdot \frac{d}{dt}(100 e^{-t/10}) - 100 e^{-t/10} \cdot \frac{d}{dt}(1 + 3 e^{-t/10})}{(1 + 3 e^{-t/10})^2}.$
After applying the chain rule and simplifying, we find that:

$\frac{dP}{dt} = \frac{(1 + 3 e^{-t/10})(-10 e^{-t/10}) - 100 e^{-t/10}(-\frac{3}{10} e^{-t/10})}{(1 + 3 e^{-t/10})^2}.$
This results in a complex expression for $\frac{dP}{dt}$.

At maximum, set $\frac{dP}{dt} = 0$ which implies:

$-10(1 + 3 e^{-T/10}) e^{-T/10} + 30 e^{-T/10} = 0.$
Solving gives:

$3 e^{-T/10} = (1 + 3 e^{-T/10}),$
resulting in:

$e^{-T/10} = \frac{1}{4}.$

Taking the natural logarithm yields:

$T = -10 \ln(\frac{1}{4}) = 10 \cdot \ln(4) \approx 14.14,$
so the value of $T$ to 2 decimal places is 14.14.

Substituting $T$ back into the original equation:

$P_T = \frac{100 e^{-14.14/10}}{1 + 3 e^{-14.14/10}} + 40.$
Calculating the terms:

$e^{-14.14/10} = e^{-1.414} \approx 0.241,$
thus,

$P_T \approx \frac{100 \times 0.241}{1 + 3 \times 0.241} + 40 \approx \frac{24.1}{1 + 0.723} + 40,$
resulting in:

$P_T \approx \frac{24.1}{1.723} + 40 \approx 13.98 + 40 \approx 54.98.$
Rounded to the nearest integer, $P_T$ is 55.

To determine the maximum value of $k$, analyze the behavior of the function as $t$ approaches infinity:

$P = \frac{100 e^{-t/10}}{1 + 3 e^{-t/10}} + 40.$
As $t \rightarrow \infty$, $e^{-t/10} \rightarrow 0$, thus:

$P \rightarrow 40.$
Therefore, the maximum value of $k$, which is the minimum rabbits after $T$, is 40.