The number of rabbits on an island is modelled by the equation $$P = \frac{100 e^{-t/3}}{1 + 3 e^{-t/3}} + 40,$$ where $P$ is the number of rabbits, $t$ years after they were introduced onto the island - Edexcel - A-Level Maths Pure - Question 8 - 2017 - Paper 4

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

Let ( u = 100 e^{-t/3} ) and ( v = 1 + 3 e^{-t/3} ), then

$\frac{dP}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}$.

Calculating ( \frac{du}{dt} ) and ( \frac{dv}{dt} ):

• ( \frac{du}{dt} = -\frac{100}{3} e^{-t/3} )
• ( \frac{dv}{dt} = -\frac{e^{-t/3}}{3} )

Substituting these into the quotient rule yields:

$\frac{dP}{dt} = \frac{(1 + 3 e^{-t/3})(-\frac{100}{3} e^{-t/3}) - (100 e^{-t/3})(-\frac{e^{-t/3}}{3})}{(1 + 3 e^{-t/3})^2},$ which simplifies to the unsimplified expression.

To find ( T ), we need to set ( \frac{dP}{dt} = 0 ) and solve for ( t ). Setting the numerator of ( -\frac{100 e^{-t/3}}{3} (1 + 3 e^{-t/3}) + \frac{100 e^{-2t/3}}{3} = 0 ) gives:

$-10 e^{-t/3} (1 + 3 e^{-t/3}) + 10 e^{-2t/3} = 0 \Rightarrow e^{-t/3}( -10(1 + 3e^{-t/3}) + 10 e^{-t/3}) = 0.$

Solving gives ( T \approx 3.53 ).

To find ( P_T ), substitute ( T ) back into the initial equation:

$P_T = \frac{100 e^{-3.53/3}}{1 + 3e^{-3.53/3}} + 40$.
Using a calculator, we find ( P_T \approx 102 ).

The maximum value of ( k ) is reached as ( t \to \infty ). Thus, ( k ) is the limiting value of the function as ( t ) increases:

$\lim_{t \to \infty} P(t) = 40.$
Therefore, the maximum value of ( k ) is 40.