在可再生自然资源的采收中，理想的情况是制定政策以允许最大程度地采收自然资源，同时又不使资源消耗到可持续水平以下。针对太平洋某一特定区域野生鲑鱼种群数量变化率，设计了一个简化的采收模型，公式如下：
$\frac{\text{d}P}{\text{d}t} = P(4 - P) - h,$
其中 $P$ 是时间为 $t$ 年时野生鲑鱼的种群数量（以百万为单位），$h$ 是恒定的采收率（以百万为单位）。

1. Express the given differential equation as a quadratic in completed square form to identify the turning point.

$\frac{\text{d}P}{\text{d}t} = 4P - P^2 - h$
$= -(P^2 - 4P) - h$
$= -[(P - 2)^2 - 4] - h$
$= -(P - 2)^2 + 4 - h$
Since the coefficient of $(P-2)^2$ is negative, the graph has a maximum turning point at $(2, 4 - h)$.

2. Identify the shape and $y$-intercept for the graph sketch.

The equation describes an inverted parabola.
When $P = 0$, $\frac{\text{d}P}{\text{d}t} = -h$.
The required sketch should display this parabolic shape passing through $(0, -h)$ on the vertical axis, reaching a peak at $(2, 4-h)$.

1. Use the maximum value of the rate of change to deduce the condition for sustainable harvest.

For the population to not face long-term depletion, there must be at least one value of $P$ where the population is not decreasing, meaning $\frac{\text{d}P}{\text{d}t} \ge 0$.
From part (i), the expression $\frac{\text{d}P}{\text{d}t} = -(P - 2)^2 + 4 - h$ has a maximum value of $4 - h$.
$4 - h \ge 0$
$h \le 4$
The largest sustainable harvest rate $h$ is therefore $4$.

1. Formulate the separable differential equation with the given constant harvest rate.

Substitute $h = 3$ into the given differential equation:
$\frac{\text{d}P}{\text{d}t} = P(4 - P) - 3$
$\frac{\text{d}P}{\text{d}t} = -P^2 + 4P - 3$
$\frac{\text{d}P}{\text{d}t} = -(P^2 - 4P + 3)$

2. Separate the variables and prepare for integration by completing the square.

Separating the variables yields:
$\int \frac{1}{P^2 - 4P + 3} \, \text{d}P = \int -1 \, \text{d}t$
Completing the square for the quadratic denominator:
$\int \frac{1}{(P - 2)^2 - 1^2} \, \text{d}P = \int -1 \, \text{d}t$

(Note: Using partial fractions $\frac{1}{P^2 - 4P + 3} = \frac{1/2}{P-3} - \frac{1/2}{P-1}$ is an equally valid method to prepare for integration).

3. Integrate both sides using the standard formula.

Using the standard integration result $\int \frac{1}{x^2 - a^2} \, \text{d}x = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|$:
$\frac{1}{2(1)} \ln \left| \frac{(P - 2) - 1}{(P - 2) + 1} \right| = -t + C$
$\frac{1}{2} \ln \left| \frac{P - 3}{P - 1} \right| = -t + C$

4. Apply the initial condition to determine the arbitrary constant $C$.

Let $t = 0$ represent the year 2015. We are given that $P = 3.2$ when $t = 0$:
$\frac{1}{2} \ln \left| \frac{3.2 - 3}{3.2 - 1} \right| = -(0) + C$
$C = \frac{1}{2} \ln \left( \frac{0.2}{2.2} \right) = \frac{1}{2} \ln \left( \frac{1}{11} \right)$

5. Rearrange to make $P$ the subject of the formula.

Substitute $C$ back into the integrated equation:
$\frac{1}{2} \ln \left| \frac{P - 3}{P - 1} \right| = -t + \frac{1}{2} \ln \left( \frac{1}{11} \right)$
Multiply by $2$:
$\ln \left| \frac{P - 3}{P - 1} \right| = -2t + \ln \left( \frac{1}{11} \right)$
Since $P = 3.2 > 3$, the fraction $\frac{P - 3}{P - 1} > 0$. We can thus drop the absolute value signs.
$\frac{P - 3}{P - 1} = e^{-2t + \ln(1/11)} = \frac{1}{11} e^{-2t}$
$11(P - 3) = (P - 1)e^{-2t}$
$11P - 33 = P e^{-2t} - e^{-2t}$
$P(11 - e^{-2t}) = 33 - e^{-2t}$
$P = \frac{33 - e^{-2t}}{11 - e^{-2t}}$
(Note: An equivalent and equally valid algebraic form is $P = \frac{33e^{2t} - 1}{11e^{2t} - 1}$)

6. Calculate the population of the wild salmon in the year 2016.

The year 2016 corresponds to $t = 1$.
$P = \frac{33 - e^{-2(1)}}{11 - e^{-2(1)}}$
$P = \frac{33 - e^{-2}}{11 - e^{-2}} \approx 3.0249$
The population in 2016 is approximately $3.02$ million.

1. State a valid assumption required for the differential equation model to hold

For the mathematical model $\frac{\text{d}P}{\text{d}t} = P(4 - P) - h$ to accurately predict the population, we assume that the population changes solely due to the natural birth/death rate modelled and the specified harvesting. This means there are no sudden external environmental changes, disease outbreaks, or migration.