The problem presents a fish population $P(t)$ approximated by the equation:

$P(t) = 15e^{0.6t}$

where $t$ is in months. You're asked to:

1. Calculate $P(12)$.
2. Calculate $P'(12)$ (the derivative of the population function evaluated at $t = 12$).
3. Explain what these values mean with respect to the population, including units.

Let's break this down:

1. Finding $P(12)$:

We plug $t = 12$ into the function $P(t)$:

$P(12) = 15e^{0.6(12)}$

2. Finding $P'(t)$:

The derivative of $P(t) = 15e^{0.6t}$ with respect to $t$ is:

$P'(t) = 15 \times 0.6 \times e^{0.6t} = 9e^{0.6t}$

Now we can evaluate $P'(12)$:

$P'(12) = 9e^{0.6(12)}$

Let's compute both values.The results are as follows:

1. $P(12) \approx 20,091$: This represents the population of fish after 12 months. The units here are "number of fish."

2. $P'(12) \approx 12,055$: This represents the rate of change of the fish population at 12 months, indicating that the population is increasing by approximately 12,055 fish per month at $t = 12$.

Would you like more details on these calculations or further clarifications?

Here are five related questions that could expand your understanding:

1. How does the population change at earlier time points, such as $t = 6$?
2. What does the function's behavior suggest about long-term population growth?
3. How would you find when the population reaches a certain size, say 50,000 fish?
4. Can you find the time at which the growth rate $P'(t)$ reaches a specific value, like 15,000 fish per month?
5. How would changing the initial coefficient of the exponential function affect the population growth?

Tip: When dealing with exponential growth, the rate of change is proportional to the population size, which leads to accelerating growth over time.