在一个拥有 500 万人口的城市中，一家新公司的客户数量增长率与非其客户的人数成正比。该公司认定，为了使业务盈利，在任何时间点，它必须拥有至少 100 万客户。 该公司在运营第三年结束时拥有 100 万客户，其初始客户数量为 0。假设在任何时间点人口都保持不变。

1. Formulate the differential equation based on the text.

The rate of change of the number of customers, $y$, is proportional to the number of people who are not its customers, which is $(5 - y)$. Thus, the differential equation is:
$\frac{dy}{dt} = k(5 - y)$
where $k$ is a positive constant of proportionality.

2. Separate variables and integrate both sides.

$\int \frac{1}{5 - y} \, dy = \int k \, dt$
$-\ln|5 - y| = kt + C$

3. Apply the initial condition to find the constant of integration $C$.

Starting from a customer base of $0$ means $y = 0$ when $t = 0$. Since $5 - y > 0$, we can drop the absolute value:
$-\ln(5 - 0) = k(0) + C$
$C = -\ln 5$

4. Make $y$ the subject to show the required equation.

Substitute $C$ back into the equation:
$-\ln(5 - y) = kt - \ln 5$
$\ln(5 - y) = -kt + \ln 5$
$5 - y = e^{-kt + \ln 5}$
$5 - y = e^{\ln 5} e^{-kt}$
$5 - y = 5e^{-kt}$
$y = 5 - 5e^{-kt} = 5(1 - e^{-kt})$
(Shown)

5. Determine the value of the constant $k$.

The company has 1 million customers at the end of the third year, meaning $y = 1$ when $t = 3$. Substitute these values into the equation:
$1 = 5(1 - e^{-3k})$
$1 - e^{-3k} = 0.2$
$e^{-3k} = 0.8$
$-3k = \ln 0.8$
$k = -\frac{1}{3}\ln 0.8 \approx 0.0744 \text{ (to 3 s.f.)}$

1. Determine the starting point and domain of the function.

For $t = 0$, $y = 5(1 - e^0) = 0$. The context of the problem dictates that time $t \ge 0$, so the graph must start from the origin $(0,0)$.

2. Find the long-term behaviour to determine any asymptotes.

As $t \to \infty$, the term $e^{-kt} \to 0$ (since $k$ is a positive constant). Therefore, $y \to 5(1 - 0) = 5$. The graph has a horizontal asymptote at $y = 5$. The curve approaches this line from below with a decreasing gradient.

1. Find the general solution for $y$ by integrating twice.

$\frac{d^2 y}{dt^2} = -0.1$
$\frac{dy}{dt} = \int -0.1 \,dt = -0.1t + c_1$
$y = \int (-0.1t + c_1) \,dt = -0.05t^2 + c_1t + c_2$

2. Set up simultaneous equations using the values of $y$ at $t=6$ and $t=7$.

At the end of the 6th year, $t = 6$ and $y = 1.8$:
$1.8 = -0.05(6)^2 + 6c_1 + c_2$
$6c_1 + c_2 = 3.6 \quad \text{--- (1)}$
At the end of the 7th year, $t = 7$ and $y = 1.65$:
$1.65 = -0.05(7)^2 + 7c_1 + c_2$
$7c_1 + c_2 = 4.1 \quad \text{--- (2)}$

3. Solve the simultaneous equations to find the specific equation for $y$.

Subtracting (1) from (2):
$c_1 = 4.1 - 3.6 = 0.5$
Substitute $c_1 = 0.5$ into (1):
$6(0.5) + c_2 = 3.6 \implies c_2 = 0.6$
Therefore, the model is:
$y = -0.05t^2 + 0.5t + 0.6$

4. Set $y < 1$ (the profitability threshold) to find the critical value of $t$.

The business is unprofitable when $y < 1$:
$-0.05t^2 + 0.5t + 0.6 < 1$
$-0.05t^2 + 0.5t - 0.4 < 0$
Divide by $-0.05$ (reversing the inequality):
$t^2 - 10t + 8 > 0$
Solving the corresponding quadratic equation $t^2 - 10t + 8 = 0$:
$t = \frac{10 \pm \sqrt{100 - 32}}{2} = 5 \pm \sqrt{17}$
$t \approx 9.123 \quad \text{or} \quad t \approx 0.877$

5. Conclude the correct year based on the condition $t \ge 6$.

Since the model is valid from $t = 6$, we reject $t \approx 0.877$ and require $t > 9.123$.
The number of customers drops below 1 million when $t > 9.123$.
Because $9 < 9.123 \le 10$, this occurs during the 10th year of operation.