the logistic growth function ( f(t)=\frac{110,000}{1 + 5500e^{-t}} ) describes the number of people, ( f(t)…

the logistic growth function ( f(t)=\frac{110,000}{1 + 5500e^{-t}} ) describes the number of people, ( f(t) ), who have become ill with influenza ( t ) weeks after its initial outbreak in a particular community.\n a. how many people became ill with the flu when the epidemic began?\n b. how many people were ill by the end of the fourth week?\n c. what is the limiting size of the population that becomes ill?\n a. the number of people initially infected is (square).\n (round to the nearest whole number as needed.)

the logistic growth function ( f(t)=\frac{110,000}{1 + 5500e^{-t}} ) describes the number of people, ( f(t) ), who have become ill with influenza ( t ) weeks after its initial outbreak in a particular community.\n a. how many people became ill with the flu when the epidemic began?\n b. how many people were ill by the end of the fourth week?\n c. what is the limiting size of the population that becomes ill?\n a. the number of people initially infected is (square).\n (round to the nearest whole number as needed.)

Answer

Explanation:

Step1: Find the initial number of infected people

When ( t = 0 ), substitute into the logistic - growth function ( f(t)=\frac{110000}{1 + 5500e^{-t}}). [ \begin{align*} f(0)&=\frac{110000}{1+5500e^{0}}\ &=\frac{110000}{1 + 5500\times1}\ &=\frac{110000}{5501}\ &\approx20 \end{align*} ]

Step2: Find the number of infected people at ( t = 4 )

When ( t = 4 ), substitute into the function ( f(t)=\frac{110000}{1 + 5500e^{-t}}). [ \begin{align*} f(4)&=\frac{110000}{1+5500e^{-4}}\ &=\frac{110000}{1+5500\times\frac{1}{e^{4}}}\ &\approx\frac{110000}{1+5500\times0.0183299}\ &=\frac{110000}{1 + 100.81445}\ &=\frac{110000}{101.81445}\ &\approx1080 \end{align*} ]

Step3: Find the limiting size of the population

For a logistic - growth function ( y=\frac{L}{1 + Ae^{-kt}}), the limiting value as ( t\rightarrow\infty) is ( L). In the function ( f(t)=\frac{110000}{1 + 5500e^{-t}}), as ( t\rightarrow\infty), (e^{-t}\rightarrow0). [ \begin{align*} \lim_{t\rightarrow\infty}f(t)&=\lim_{t\rightarrow\infty}\frac{110000}{1 + 5500e^{-t}}\ &=\frac{110000}{1+0}\ &=110000 \end{align*} ]

Answer:

a. (20) people; b. (1080) people; c. (110000) people.