A particular species of orchid is being studied - Edexcel - A-Level Maths Pure - Question 3 - 2005 - Paper 5

We substitute a = 0.12 into the equation:

$p = \frac{2800 \cdot 0.12 \cdot e^{0.2t}}{1 + 0.12 \cdot e^{0.2t}}$

We set p equal to 1850:

$1850 = \frac{336 \cdot e^{0.2t}}{1 + 0.12 e^{0.2t}}$

Cross multiplying gives:

$1850(1 + 0.12 e^{0.2t}) = 336 e^{0.2t}$

This expands to:

$1850 + 222 e^{0.2t} = 336 e^{0.2t}$

Rearranging terms results in:

$1850 = 336 e^{0.2t} - 222 e^{0.2t} \Rightarrow 1850 = 114 e^{0.2t}$

Solving for e^{0.2t} gives:

$e^{0.2t} = \frac{1850}{114} \approx 16.23$

Taking the natural logarithm:

$0.2t = \ln(16.23)$

Therefore,

$t \approx \frac{\ln(16.23)}{0.2} \approx 14.09.$

So, it takes approximately 14 years for the population to reach 1850.

Using the established value of a, we want to simplify the original equation:

$p = \frac{2800 \cdot 0.12 \cdot e^{0.2t}}{1 + 0.12 \cdot e^{0.2t}}$

Calculating the numerator gives:

$2800 \cdot 0.12 = 336,$

So subbing this back into the equation yields:

$p = \frac{336 \cdot e^{0.2t}}{1 + 0.12 e^{0.2t}}$

This can be rewritten as:

$p = \frac{336}{\frac{1}{e^{0.2t}} + 0.12}$

Therefore, we arrive at:

$p = \frac{336}{0.12 + e^{0.2t}}.$

From the expression we derived:

$p = \frac{336}{0.12 + e^{0.2t}},$

we can analyze the behavior of the function as t approaches infinity. As t increases, the term $e^{0.2t}$ grows significantly. Thus,

$\lim_{t \to \infty} e^{0.2t} \to \infty,$

which implies:

$\lim_{t \to \infty} p = \frac{336}{0.12 + e^{0.2t}} \to \frac{336}{\infty} \to 0,$

and will always be finite. However, the maximum value of p would occur when $e^{0.2t}$ is at its minimal value (approaching 0), yielding:

$p \leq 336/(0.12) = 2800.$

Thus, the population cannot exceed 2800.