Question

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The populations P (in thousands) of a city from 2000 through 2020 can be modeled by

P = 2687/(1 + 0.089e^0.050t) where t represents the year, with t = 0 corresponding to 2000.

a. Use the model to find the population of the city (in thousands) in the years 2001, 2006, 2011, 2015, and 2020. (Round your answers to 3 decimal places.)

2001 P = ______________ thousand people
2006 P= ______________ thousand people
2011 P = ______________ thousand people
2015 P= ______________ thousand people
2020 P= _____________ thousand people

b. Use a graphing utility to graph the function to determine the year in which the population reached 2.2 million
The population will reach 2.2 million in __________.

c. Confirm your answer to part (b) algebraically.
The population will reach 2.2 million in __________.

Answer 1

Answer:

a)  2001:  P = 2457.106 thousand people

    2006:  P = 2398.813 thousand people

    2011:  P = 2327.899 thousand people

    2015:  P = 2260.998 thousand people

    2020:  P = 2163.573 thousand people

b)  2018

c)  2018

Step-by-step explanation:

Given function:

$P = \dfrac{2687}{1 + 0.089e^{0.050t}}$

where:

• P = population (in thousands)
• t = number of years after the year 2000

Part (a)

In 2001, t = 1:

$\begin{aligned}t = 1 \implies P &= \dfrac{2687}{1 + 0.089e^{0.050(1)}}\\\\&= \dfrac{2687}{1 + 0.089e^{0.050}}\\\\&=2457.106 \; \sf (3 \; d.p.)\end{aligned}$

In 2006, t = 6:

$\begin{aligned}t = 6 \implies P &= \dfrac{2687}{1 + 0.089e^{0.050(6)}}\\\\&= \dfrac{2687}{1 + 0.089e^{0.3}}\\\\&=2398.813 \; \sf (3 \; d.p.)\end{aligned}$

In 2011, t = 11:

$\begin{aligned}t = 11 \implies P &= \dfrac{2687}{1 + 0.089e^{0.050(11)}}\\\\&= \dfrac{2687}{1 + 0.089e^{0.55}}\\\\&=2327.899 \; \sf (3 \; d.p.)\end{aligned}$

In 2015, t = 15:

$\begin{aligned}t = 15 \implies P &= \dfrac{2687}{1 + 0.089e^{0.050(15)}}\\\\&= \dfrac{2687}{1 + 0.089e^{0.75}}\\\\&= 2260.998\; \sf (3 \; d.p.)\end{aligned}$

In 2020, t = 20:

$\begin{aligned}t = 20 \implies P &= \dfrac{2687}{1 + 0.089e^{0.050(20)}}\\\\&= \dfrac{2687}{1 + 0.089e^{1}}\\\\&=2163.573 \; \sf (3 \; d.p.)\end{aligned}$

Part (b)

See attached for the graph of the function.

2.2 million = 2,200,000 = 2200 thousand

Therefore, draw a line at y = 2200.

The point of intersection between P(t) and y = 2200 is (18.223, 2200).

Therefore, the population will reach 2.2 million during 2018.

$\implies \dfrac{2687}{1 + 0.089e^{0.050t}}=2200$

$\implies 2687=2200(1 + 0.089e^{0.050t})$

$\implies \dfrac{2687}{2200}=1 + 0.089e^{0.050t}$

$\implies \dfrac{2687}{2200}-1=0.089e^{0.050t}$

$\implies \dfrac{487}{2200}=0.089e^{0.050t}$

$\implies \ln \left(\dfrac{487}{2200}\right)=\ln \left(0.089e^{0.050t}\right)$

$\implies \ln \left(\dfrac{487}{2200}\right)=\ln \left(0.089 \right)+\ln \left(e^{0.050t}\right)$

$\implies \ln \left(\dfrac{487}{2200}\right)=\ln \left(0.089 \right)+0.050t$

$\implies \ln \left(\dfrac{487}{2200}\right)-\ln \left(0.089 \right)=0.050t$

$\implies \ln \left(\dfrac{2435}{979}\right)=0.050t$

$\implies t=20\ln \left(\dfrac{2435}{979}\right)$

$\implies t=18.223\; \sf (3 \; d.p.)$

The population will reach 2.2 million in 2018.