在一个有 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。该公司确定，为了使业务盈利，它在任何时间点都必须拥有至少 100 万客户。 该公司从 0 个客户起步，在运营第三年结束时拥有 100 万客户。假设在任何时间点人口都保持不变。

1. Formulate the differential equation using the given rate of change.

Let the proportionality constant be $m$.
Since the total population is 5 million and $y$ is the number of customers (in millions), the number of people who are not customers is $5 - y$.
The rate of increase of customers is proportional to this number:
$\frac{\mathrm{d}y}{\mathrm{d}t} = m(5 - y)$

2. Separate the variables and integrate to find the general solution.

Separating the variables:
$\int \frac{1}{5-y} \,\mathrm{d}y = \int m \,\mathrm{d}t$
$-\ln(5 - y) = mt + C$
(Note: The modulus is dropped since $5 - y > 0$ as the number of customers cannot exceed the total population.)
$5 - y = e^{-mt - C} = A e^{-mt}$
where $A = e^{-C}$ is an arbitrary constant.

3. Apply the initial condition to find the constant of integration and show the given form.

The company starts with 0 customers, so $y = 0$ when $t = 0$:
$5 - 0 = A e^0 \implies A = 5$
Substituting $A = 5$ into the equation:
$5 - y = 5 e^{-mt}$
$y = 5(1 - e^{-mt})$
By letting $k = -m$, we can rewrite this in the required form:
$y = 5(1 - e^{kt})$

4. Apply the secondary condition to determine the exact value of $k$.

At the end of the third year ($t = 3$), the company has 1 million customers ($y = 1$):
$1 = 5(1 - e^{3k})$
$\frac{1}{5} = 1 - e^{3k}$
$e^{3k} = 1 - 0.2 = 0.8$
$3k = \ln 0.8$
$k = \frac{1}{3}\ln 0.8 \approx -0.0744 \text{ (to 3 s.f.)}$

1. Find the starting point (intercept) and the horizontal asymptote.

For $t = 0$:
$y = 5(1 - e^0) = 5(0) = 0$
The curve starts at the origin $(0,0)$.

From part (i), we know $k < 0$. As $t \to \infty$:
$e^{kt} \to 0$
$y \to 5(1 - 0) = 5$
So, the line $y = 5$ is a horizontal asymptote.

2. Sketch the graph for the practical domain ($t \ge 0$).

• Draw and label the horizontal axis as $t$ and the vertical axis as $y$.
• Draw a dashed line for the horizontal asymptote $y = 5$.
• Sketch a smooth curve starting from $(0,0)$, which is strictly increasing with a decreasing gradient (concave downwards), approaching the asymptote $y = 5$ from below.

1. Integrate the given differential equation to find the general form of $y$ in terms of $t$.

$\frac{\mathrm{d}^2 y}{\mathrm{d}t^2} = -0.1$

Integrating with respect to $t$:
$\frac{\mathrm{d}y}{\mathrm{d}t} = -0.1t + c$

Integrating again:
$y = -0.05t^2 + ct + d$

2. Substitute the given boundary conditions to find the constants of integration $c$ and $d$.

At the end of the 6th year ($t = 6$), $y = 1.8$:
$1.8 = -0.05(6)^2 + 6c + d$
$1.8 = -1.8 + 6c + d \implies 6c + d = 3.6 \quad \text{--- (1)}$

At the end of the following year, which is the 7th year ($t = 7$), $y = 1.65$:
$1.65 = -0.05(7)^2 + 7c + d$
$1.65 = -2.45 + 7c + d \implies 7c + d = 4.1 \quad \text{--- (2)}$

Subtracting (1) from (2):
$c = 0.5$

Substitute $c = 0.5$ into (1):
$6(0.5) + d = 3.6 \implies 3.0 + d = 3.6 \implies d = 0.6$

So, the number of customers in the new model is given by:
$y = -0.05t^2 + 0.5t + 0.6$

3. Set up and solve an inequality to find the time $t$ when the business becomes unprofitable.

The business becomes unprofitable when the customer base drops below 1 million ($y < 1$):
$-0.05t^2 + 0.5t + 0.6 < 1$
$-0.05t^2 + 0.5t - 0.4 < 0$

Dividing by $-0.05$ (which flips the inequality sign):
$t^2 - 10t + 8 > 0$

Solving $t^2 - 10t + 8 = 0$ using the quadratic formula:
$t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(8)}}{2(1)}$
$t = \frac{10 \pm \sqrt{68}}{2} = 5 \pm \sqrt{17}$

Since this model applies for $t \ge 6$, we take the relevant interval for the larger root:
$t > 5 + \sqrt{17} \approx 9.123$

Thus, the number of customers drops below 1 million for $t > 9.123$.

4. Determine the specific year of operation from the calculated value of $t$.

The $n$-th year of operation corresponds to the time interval $n-1 < t \le n$.
Since $9 < 9.123 < 10$, the time $t = 9.123$ occurs during the 10th year of operation. Consequently, the business becomes unprofitable in the 10th year.