Chapter 3.5, Problem 41E

a.

To determine

Find the number of cell sites in the given years.

Answer to Problem 41E

The number of cell sites in $1998$ , $2003$ and $2006$ are $55557$ , $147644$ and $203023$ respectively.

Explanation of Solution

Given:

The cell sites from $1985$ through $2011$ are modeled by $y=\frac{269,573}{1+985e^{-0.308t}}$ .

Calculation:

  $y=\frac{269,573}{1+985e^{-0.308t}}$

The number of cell sites in $1998$ .

  $t=1998-\left(1985-5\right)=18$

Substitute the value of $t=18$ in the equation.

  $\begin{array}{l}\Rightarrow y=\frac{269,573}{1+985e^{-0.308\cdot 18}}\\ =\frac{269,573}{1+3.852189}\\ =\frac{269,573}{4.852189}\\ =55556.99\\ \approx 55557\end{array}$

The number of cell sites in $2003$ .

  $t=2003-\left(1985-5\right)=23$

Substitute the value of $t=23$ in the equation.

  $\begin{array}{l}\Rightarrow y=\frac{269,573}{1+985e^{-0.308\cdot 23}}\\ =\frac{269,573}{1+0.825836589}\\ =\frac{269,573}{1.825836589}\\ =147643.55\\ \approx 147644\end{array}$

The number of cell sites in $2006$ .

  $t=2006-\left(1985-5\right)=26$

Substitute the value of $t=26$ in the equation.

  $\begin{array}{l}\Rightarrow y=\frac{269,573}{1+985e^{-0.308\cdot 26}}\\ =\frac{269,573}{1+0.3277977886}\\ =\frac{269,573}{1.3277977886}\\ =203022.6\\ \approx 203023\end{array}$

Hence the number of cell sites in $1998$ , $2003$ and $2006$ are $55557$ , $147644$ and $203023$ respectively.

b.

To determine

Draw a graph of the given function using graphing utility.

Explanation of Solution

Given:

The cell sites from $1985$ through $2011$ are modeled by $y=\frac{269,573}{1+985e^{-0.308t}}$ .

.Calculation:

The given function is $y=\frac{269,573}{1+985e^{-0.308t}}$ .

By the use of graphing utility, the graph of the function is given below.

  

c.

To determine

Draw a graph of the given function using graphing utility.

Answer to Problem 41E

The number of cell sites reached $250,000$ in the year $2011$ .

Explanation of Solution

Given:

The cell sites from $1985$ through $2011$ are modeled by $y=\frac{269,573}{1+985e^{-0.308t}}$ .

.Calculation:

The given function is $y=\frac{269,573}{1+985e^{-0.308t}}$ .

By the use of graphing utility, the graph of the function is given below.

  

From the graph, the number of cell sites reached $250,000$ in $30.649\approx 31$ years.

  $1980+31=2011$

Hence the number of cell sites reached $250,000$ in the year $2011$ .

d.

To determine

Find the year in which the cell sites will be reach at the given numbers.

Answer to Problem 41E

The number of cell sites reached $250,000$ in the year $2011$ .

Explanation of Solution

Given:

The cell sites from $1985$ through $2011$ are modeled by $y=\frac{269,573}{1+985e^{-0.308t}}$ .

.Calculation:

The given function is $y=\frac{269,573}{1+985e^{-0.308t}}$ .

Substitute the value of $y=250000$ in the function.

  $250000=\frac{269,573}{1+985e^{-0.308t}}$

Multiply both sides by $1+985e^{-0.308t}$ .

  $\begin{array}{l}\Rightarrow 250000\left(1+985e^{-0.308t}\right)=\frac{269,573}{\left(1+985e^{-0.308t}\right)}\cdot \left(1+985e^{-0.308t}\right)\\ \Rightarrow 250000\left(1+985e^{-0.308t}\right)=269,573\end{array}$

Substitute both sides by $250000$ .

  $\begin{array}{l}\Rightarrow \frac{250000}{250000}\left(1+985e^{-0.308t}\right)=\frac{269,573}{250000}\\ \Rightarrow 1+985e^{-0.308t}=\frac{269,573}{250000}\end{array}$

Subtract $1$ from both sides.

  $\begin{array}{l}\Rightarrow 1+985e^{-0.308t}-1=\frac{269,573}{250000}-1\\ \Rightarrow 985e^{-0.308t}=\frac{269,573-250000}{250000}\\ \Rightarrow 985e^{-0.308t}=\frac{19573}{250000}\end{array}$

Divide both sides by $985$ .

  $\begin{array}{l}\Rightarrow \frac{985}{985}{e}^{-0.308t}=\frac{19573}{250000\cdot 985}\\ \Rightarrow e^{-0.308t}=\frac{19573}{246250000}\end{array}$

Taking natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln e^{-0.308t}=\ln \frac{19573}{246250000}\\ \Rightarrow -0.308t=-9.43995149\\ \Rightarrow 0.308t=9.43995149\end{array}$

Divided both sides by $0.308$ .

  $\begin{array}{l}\Rightarrow \frac{0.308}{0.308}t=\frac{9.43995149}{0.308}\\ \Rightarrow t=\frac{9.43995149}{0.308}\\ \Rightarrow t=30.649\\ \Rightarrow t\approx 31\end{array}$

Now,

  $1980+31=2011$

Hence the number of cell sites reached $250,000$ in the year $2011$ .