A particular species of orchid is being studied - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 6

Using the equation with $a = 0.12$, we have:

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

Setting $p = 1850$, we get:

$1850 = \frac{2800 \cdot 0.12 e^{0.2t}}{1 + 0.12 e^{0.2t}}$

This simplifies to:

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

Expanding the left side:

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

Now, rearranging this gives:

$1850 = (336 - 222)e^{0.2t}$ $1850 = 114 e^{0.2t}$

Dividing both sides by 114:

$e^{0.2t} = \frac{1850}{114} \approx 16.2$

Taking the natural logarithm:

$0.2t = \ln(16.2)$ $t = \frac{\ln(16.2)}{0.2} \approx 14$

Starting with the equation:

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

Substituting $0.12$ for $a$, we simplify:

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

Factoring out 336 gives:

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

To have it in the desired form:

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

From the equation derived:

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

As $t \to \infty$, $e^{0.2t} \to \infty$, which implies:

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

Therefore, the population $p$ is always limited by the numerator 336, and as $e^{0.2t}$ increases, the population will never exceed the limiting value:

$p < 2800$