The annual sales of a new company are expected to grow at a rate proportional to the difference between the sales at time \(t\) and an upper limit of $\(5\) million. Suppose the sales are $0 initially and are $\(2\) million after \(4\) years of operation. Determine the annual sales at any time \(t\). How long will it take for the annual sales to reach $\(4\) million?

Solution Since there are no sales initially (\(R=0\)), we use equation (4). Then since the upper limit of sales is $\(5\) million, the sales at time \(t\) are \[ \begin{equation*} y(t)=5(1-e^{-kt})\qquad \color{#0066A7}{{M=5}} \end{equation*} \]

Since sales are $\(2\) million after \(4\) years, the boundary condition is \(y(4)=2.\) We use the boundary condition to find the constant \(k\): \[ \begin{eqnarray*} 2& =&5(1-e^{-4k})\qquad \color{#0066A7}{{y(4)=2}}\\ 5e^{-4k}& =&3 \\[2pt] e^{-4k}& =&0.6 \\[4pt] k& =&\frac{\ln 0.6}{-4} = -0.25\ln 0.6 \end{eqnarray*} \]

So, the function \[ \begin{equation*} y(t)=5\big(1-e^{( 0.25\ln 0.6) t}\big) \end{equation*} \]

models the company's sales in year \(t.\)

To find out how long it will take for sales to reach $\(4\) million, we solve the equation \(y( t) =4.\) Then \[ \begin{eqnarray*} 4& =&5\big( 1-e^{( 0.25\ln 0.6) t}\big) \\ 5e^{( 0.25\ln 0.6) t}& =&1 \\[1pt] (0.25\ln 0.6) t & =&\ln 0.2 \\[3pt] t& =&\frac{\ln 0.2}{0.25\ln 0.6}\approx 12.6 \hbox{years} \end{eqnarray*} \]

It will take approximately \(12.6\) years for annual sales to reach $\(4\) million.