The average rate of change of total area is calculated according to the usual formula. It does not follow the differentiation rules, which are about instantaneous rates of change. We let \(A\) represent the area of the city and \(t\) the time in years from now. The city currently has a total area of

\begin{equation*}
A(0) = 5 \times 3 = 15 \: \hbox{mi}^2\text{.}
\end{equation*}

After 10 years, the city will have a total area of

\begin{equation*}
A(10) = 8 \times 5 = 40 \: \hbox{mi}^2\text{.}
\end{equation*}

The change in area is \(\Delta A = A(10)-A(0) = 25 \: \hbox{mi}^2\) and the change in time is \(\Delta t = 10\) years. Consequently, the average rate of change of area is

\begin{equation*}
\frac{\Delta A}{\Delta t} \Big|_{0,10} = \frac{25}{10} = 2.5\: \hbox{mi}^2/\hbox{yr}\text{.}
\end{equation*}

To connect our intuition with functions and to prepare for the next calculations, let us introduce variables in addition to time \(t\) and total area \(A\text{.}\) The state of the city can be characterized more precisely with two more variables: the distance east-to-west, which we'll call the width \(W\) (mi), and the distance north-to-south, which we'll call the height \(H\) (mi). We think of \(W\text{,}\) \(H\) and \(A\) as being dependent variables as they are each a function of time \(t\text{.}\) They are related variables because the area always equals the product of \(W\) and \(H\text{:}\)

\begin{equation*}
A(t) = W(t) \cdot H(t).
\end{equation*}

To find the instantaneous rates of change, we need to know how fast the width and height measurements are changing in time. Because the problem stated that these changed at a constant rate, we can use the average rates of change to compute the instantaneous rates:

\begin{gather*}
\displaystyle \frac{dW}{dt} = \left. \frac{\Delta W}{\Delta t}\right|_{[0,10]}
= \frac{W(10)-W(0)}{10-0} = \frac{8-5}{10} = 0.3, \\
\displaystyle \frac{dH}{dt} = \left. \frac{\Delta H}{\Delta t}\right|_{[0,10]}
= \frac{H(10)-H(0)}{10-0} = \frac{5-3}{10} = 0.2.
\end{gather*}

Since the area \(A\) is the product of \(W\) and \(H\text{,}\) the product rule for derivatives will provide the instantaneous rate of change for area:

\begin{equation*}
\frac{dA}{dt} = \frac{d}{dt}[W \cdot H] = \frac{dW}{dt} \cdot H + W \cdot \frac{dH}{dt}.
\end{equation*}

This equation is the related rates equation.

When \(t=0\) we have \(W(0)=5\) and \(H(0)=3\) so that

\begin{align*}
\left. \frac{dA}{dt} \right|_{0}
&= \left.\frac{dW}{dt}\right|_{0} \cdot H(0) + W(0) \cdot \left.\frac{dH}{dt}\right|_{0}\\
&= 0.3(3) + 5(0.2) = 1.9.
\end{align*}

That is, at the beginning, the city is expanding at a rate of \(1.9 \: \hbox{mi}^2/\hbox{yr}\text{.}\) After 10 years, \(t=10\text{,}\) we have \(W(10)=8\) and \(H(10)=5\) so that

\begin{align*}
\left. \frac{dA}{dt} \right|_{10}
&= \left.\frac{dW}{dt}\right|_{10} \cdot H(10) + W(10) \cdot \left.\frac{dH}{dt}\right|_{10}\\
&= 0.3(5) + 8(0.2) = 3.1.
\end{align*}

At the end of the 10 years, the city is expanding at a rate of \(3.1 \: \hbox{mi}^2/\hbox{yr}\text{.}\)

The picture of expanding area helps provide some intuition for why the product rule is the appropriate technique. If we consider the city after 6 months (\(t=0.5\)), both the width and the height have changed by a small amount, as shown in the figure below. The total change in area has two primary contributions, corresponding to long, skinny rectangles with areas \(W(0) \cdot \Delta H\) and \(\Delta W \cdot H(0)\text{,}\) and a very small rectangle with area \(\Delta W \cdot \Delta H\text{.}\) The product rule corresponds to the rate of change coming from the two primary contributions while the small rectangle leads to a term that has a limit of zero in the calculation of the derivative.