将某种鱼引入池塘并每天记录其数量。经观察，$t$ 天后，其数量 $P$ 由 $P = 5000 + Ae^{kt}$ 给出，其中 $A$ 和 $k$ 是常数。最初，池塘里有 3000 条鱼。10 天后，还剩 3800 条鱼。

1. Substitute the initial conditions into the population model

Initially, $t = 0$ and $P = 3000$.
$3000 = 5000 + Ae^{k(0)}$

2. Solve for A

$3000 = 5000 + A(1)$
$A = 3000 - 5000$
$A = -2000$ (shown)

1. Substitute the given values into the population model

Given $P = 5000 - 2000e^{kt}$.
When $t = 10$, $P = 3800$:
$3800 = 5000 - 2000e^{10k}$

2. Isolate the exponential term

$2000e^{10k} = 5000 - 3800$
$2000e^{10k} = 1200$
$e^{10k} = 0.6$

3. Solve for $k$ using natural logarithms

$10k = \ln 0.6$
$k = \frac{\ln 0.6}{10}$
$k \approx -0.05108$
$k = -0.0511 \text{ (to 3 s.f.)}$

1. Substitute the value of A into the population model

From (i), $A = -2000$. The population model is $P = 5000 - 2000e^{kt}$.

2. Use the property of the exponential function to explain the upper bound

For any real value of $t$, $e^{kt} > 0$.
Therefore, $-2000e^{kt} < 0$.
Hence, $P = 5000 - 2000e^{kt} < 5000$.
This means the number of fishes will always be strictly less than 5000.