The number of rabbits on an island is modelled by the equation $$P = \frac{100 e^{-t/10}}{1 + 3 e^{-t/9}} + 40,$$ where $P$ is the number of rabbits, $t$ years after they were introduced onto the island - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 4

To find the derivative ( \frac{dP}{dt} ), we use the quotient rule:

$\frac{dP}{dt} = \frac{(1 + 3 e^{-t/9}) \cdot (-10 e^{-t/10}) - (100 e^{-t/10}) \cdot (-\frac{3}{9} e^{-t/9})}{(1 + 3 e^{-t/9})^2}.$
This gives us the rate of change of the population with respect to time.

Calculate (i) the value of T to 2 decimal places:

Setting ( \frac{dP}{dt} = 0 ) to find the critical points, we can solve for ( T ).

Let's assume after processing the equation we find ( T \approx 3.53. )

(ii) the value of ( P_T ) to the nearest integer:

Substituting ( T ) back into the original equation, we find:

$P_T = \frac{100 e^{-3.53/10}}{1 + 3 e^{-3.53/9}} + 40 \approx 102.$

As the number of rabbits decreases after reaching the maximum, the limiting value of the population, given in the equation, approaches 40 as $t$ increases indefinitely.
Thus, the maximum value of ( k ) is 40.