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速率与非其客户的人数成正比。该公司认定，为了使其业务盈利，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年底时拥有 100 万客户，其初始客户基数为 0。假设在任何时间点人口总数都保持不变。

1. Formulate the differential equation from the problem situation.

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

2. Solve the differential equation by separating variables.

$\int \frac{1}{5 - y} \,\mathrm{d}y = \int k \,\mathrm{d}t$
$-\ln|5 - y| = kt + c$
Since $y < 5$, $|5 - y| = 5 - y$.
$5 - y = e^{-(kt + c)}$
$5 - y = A e^{-kt} \quad \text{where } A = e^{-c}$
$y = 5 - A e^{-kt}$

3. Use the initial condition to find $A$ and show the given equation.

The company starts with a customer base of 0, so when $t = 0$, $y = 0$.
$0 = 5 - A e^{0}$
$A = 5$
Substituting $A = 5$ back into the equation:
$y = 5 - 5e^{-kt}$
$y = 5(1 - e^{-kt}) \quad \text{(Shown)}$

4. Determine the value of the constant $k$ using the customer base at the end of year 3.

When $t = 3$, $y = 1$.
$1 = 5(1 - e^{-3k})$
$\frac{1}{5} = 1 - e^{-3k}$
$e^{-3k} = 1 - 0.2 = 0.8$
$-3k = \ln 0.8$
$k = -\frac{1}{3} \ln 0.8 = \frac{1}{3} \ln 1.25$
$k \approx 0.0744 \text{ (to 3 s.f.)}$

1. Determine the starting point and long-term behaviour (asymptote) of the graph.

Given $y = 5(1 - e^{-kt})$ from part (i) with $k > 0$.
At the start of the operation, $t = 0$:
$y = 5(1 - e^0) = 0$
So the graph starts at the origin $(0,0)$.

As $t \to \infty$, the term $e^{-kt} \to 0$.
$y \to 5(1 - 0) = 5$
This indicates a horizontal asymptote at $y = 5$.

2. Identify the shape and sketch the required features.

The function represents an exponential growth model restricted to $t \ge 0$.
Since $\frac{dy}{dt} > 0$ and $\frac{d^2y}{dt^2} < 0$, the curve is strictly increasing but at a decreasing rate (concave down).
The student should draw axes labelled $t$ and $y$, mark the origin, draw the dashed asymptote $y=5$, and sketch the concave-down curve rising from $(0,0)$ towards $y=5$. The point $(3,1)$ given in the context can also be plotted for completeness.

1. Integrate the given second-order differential equation to find the general equation for $y$.

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

Integrating with respect to $t$:
$\frac{dy}{dt} = -0.1t + c$

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

2. Use the given initial conditions at $t=6$ and $t=7$ to determine the constants $c$ 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 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):
$c = 4.1 - 3.6 = 0.5$

Substituting $c = 0.5$ into (1):
$6(0.5) + d = 3.6 \implies 3 + d = 3.6 \implies d = 0.6$

So, the model for the number of customers (for $t \ge 6$) is:
$y = -0.05t^2 + 0.5t + 0.6$

3. Formulate an inequality to find when the business becomes unprofitable.

The business is profitable as long as $y \ge 1$. It becomes unprofitable when $y < 1$.
$-0.05t^2 + 0.5t + 0.6 < 1$
$-0.05t^2 + 0.5t - 0.4 < 0$

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

4. Solve the quadratic inequality to find the critical value of $t$.

First, find the roots of $t^2 - 10t + 8 = 0$ using the quadratic formula:
$t = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(8)}}{2} = \frac{10 \pm \sqrt{68}}{2} = 5 \pm \sqrt{17}$
$t \approx 0.877 \quad \text{or} \quad t \approx 9.123$

Since the new model starts at $t = 6$, we take the root $t = 5 + \sqrt{17} \approx 9.123$.
The inequality $t^2 - 10t + 8 > 0$ for $t \ge 6$ gives:
$t > 9.123$

5. Interpret the value of $t$ to state the specific year.

Since $t$ represents the number of years from the start, the 1st year is $0 < t \le 1$, the 2nd year is $1 < t \le 2$, and so on.
The time $t = 9.123$ falls in the interval $9 < t < 10$.

Therefore, the business will become unprofitable during the 10th year of operation.