Chapter 3.1, Problem 56E

Solution

a.

To find: the population of city NY in 1850.

1,798,234

Given information:

The population of city NY state can be modeled by

  $P\left(t\right)=\frac{19.875}{1+57.993e^{-0.03505t}}$

where P is the population in millions and t is the number of years since 1800.

Explanation:

Here, t corresponds to years after 1800, so year 1850 corresponds to $t=50$ .

Substitute $t=50$ in the given model,

  $\begin{array}{c}P\left(50\right)=\frac{19.875}{1+57.993e^{-0.03505\left(50\right)}}\\ \approx 1.798234\end{array}$

Which corresponds to the population of about 1,798,234

b.

To find: the population of city NY in 2015.

19,278,362

Given information:

The population of city NY state can be modeled by

  $P\left(t\right)=\frac{19.875}{1+57.993e^{-0.03505t}}$

where P is the population in millions and t is the number of years since 1800.

Explanation:

Here, t corresponds to years after 1800, so year 2015 corresponds to $t=215$ .

Substitute $t=215$ in the given model,

  $\begin{array}{c}P\left(215\right)=\frac{19.875}{1+57.993e^{-0.03505\left(215\right)}}\\ \approx 19.278362\end{array}$

Which corresponds to the population of about 19,278,362

c.

To find: city N’s maximum sustainable population.

19,875,000

Given information:

The population of city NY state can be modeled by

  $P\left(t\right)=\frac{19.875}{1+57.993e^{-0.03505t}}$

where P is the population in millions and t is the number of years since 1800.

Explanation:

The maximum sustainable population or the limit to growth of the given population model is found by finding its limit as t increases infinitely large.

Now as $t\to \infty$, $e^{-0.03505}\to 0$

So, the limit to growth is:

  $\begin{array}{c}\underset{t\to \infty }{\lim }P\left(t\right)=\frac{19.875}{1+57.993\left(0\right)}\\ =\frac{19.875}{1}\\ =19.85\end{array}$

This corresponds to a maximum population of 19,875,000.