In a city of 5 million people, the number of customers of a new company increases at a rate proportional to the number of people who are not its customers. The company determines that in order for its business to be profitable, it must have at least 1 million customers at any point in time. The company has 1 million customers at the end of the third year of operation, starting from a customer base of 0. Assume that the population remains unchanged at any point in time.

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.