Chapter 7.8, Problem 3GP

a.

To determine

To graph:

The given function $P\left(t\right)=\frac{16,500}{1+18e^{-0.085t}}$ for $0\le t\le 200$ .

Answer to Problem 3GP

The value k is $k=0.06931$ .

Explanation of Solution

Given information:

The given function is $P\left(t\right)=\frac{16,500}{1+18e^{-0.085t}}$ .

Graph:

  

Interpretation:

Make a table of values $\left(t\in \left[0,200\right]\right)$ .

$t$	$P\left(t\right)$
$0$	$868$
$20$	$3847$
$40$	$10,307$
$60$	$14,868$
$80$	$16,176$
$100$	$16,439$
$120$	$16,489$
$140$	$16,498$
$160$	$16,499.63$
$180$	$16,499.93$
$200$	$16,499.99$

b.

To determine

To Find:

The horizontal asymptote.

Answer to Problem 3GP

The horizontal asymptote will be $x=16,500$ .

Explanation of Solution

Given information:

The given function is $P\left(t\right)=\frac{16,500}{1+18e^{-0.085t}}$ .

When $t\to \infty \Rightarrow e^{-0.085t}\to 0$

  $\begin{array}{l}\Rightarrow 1+18e^{-0.085t}\to 1\\ \Rightarrow \frac{16,500}{1+18e^{-0.085t}}\to 16,500\end{array}$

Therefore, the horizontal asymptote will be $x=16,500$ .

c.

To determine

To find:

The maximum population.

Answer to Problem 3GP

The maximum population is $16,500$ .

Explanation of Solution

Given information:

The given function is $P\left(t\right)=\frac{16,500}{1+18e^{-0.085t}}$ .

Consider the given function $P\left(t\right)=\frac{16,500}{1+18e^{-0.085t}}$ .

The population gets close to its maximum when the denominator gets close to each minimum which is when $t\to \infty$ .

Hence, the maximum population is $16,500$ .

d.

To determine

To find:

When the population reaches $16,450$ .

Answer to Problem 3GP

The population reaches $16,450$ approximately by $102$ years.

Explanation of Solution

Given information:

The given function is $P\left(t\right)=\frac{16,500}{1+18e^{-0.085t}}$ .

Consider the given function $P\left(t\right)=\frac{16,500}{1+18e^{-0.085t}}$ .

Substitute $P\left(t\right)=16,450$

  $\begin{array}{l}16,450=\frac{16,500}{1+18e^{-0.085t}}\\ 16,450\left(1+18e^{-0.085t}\right)=16,500\\ 16,450+16,450\times 18e^{-0.085t}=16,500\\ 16,450\times 18e^{-0.085t}=16,500-16,450\\ 16,450\times 18e^{-0.085t}=50\\ e^{-0.085t}=\frac{50}{2,96,100}\\ e^{-0.085t}\approx 0.00016\\ \ln e^{-0.085t}\approx \ln 0.00016\\ -0.085t\approx -8.68643\\ t\approx \frac{-8.68643}{-0.085}\\ t\approx 102\end{array}$

Hence, the population reaches $16,450$ approximately by $102$ years.