The number N(t) of people in a community who are exposed to a particular advertisement is governed by the logistic equation. Initially, N(0) = 600, and it is observed that N(1) = 1200. Solve for N(t) if it is predicted that the limiting number of people in the community who will see the advertisement is 60,000. (Round all coefficients to four decimal places.)

Here N(0) ist the initial value and k is the charge capacity. Using the information of the question we have [tex]N(0) =600[/tex] and [tex]k=60000[/tex].

We need to find the a value, using the logístic equation:

[tex]F(1)=\frac{600*60000}{600+(60000-600)e^{-a} }=1200[/tex]

Isolating a, we have:

[tex]\frac{600*60000}{600+(60000-600)e^{-a} }=1200\\\frac{600*60000}{1200}=600+(60000-600)e^{-a}\\\frac{60000}{2}=600+(60000-600)e^{-a}\\30000=600+59400e^{-a}\\30000-600=59400e^{-a}\\29400=59400e^{-a}\\\frac{29400}{59400}=e^{-a}\\ln(\frac{29400}{59400})=-a\\-ln(\frac{29400}{59400})=a\\a=0.703[/tex]

This way the logistic equation to this problem is:

[tex]F(t)=\frac{600*60000}{600+59400e^{-0.703*t} }[/tex]

Lanuel Lanuel · Answer 2 · 2022-02-25T14:40:42+01:00

N(t) is given by this logistic equation [tex]F (t) = \frac{600 \times 60000 }{600\;+\;[60000\; - \;600]e^{-0.7032t}}[/tex]

Given the following data:

Initial value, N(0) = 600.

N(1) = 1200.

Carrying capacity, k = 60,000.

To solve for N(t) assuming the limiting number of people in the community is 60,000:

What is a logistic equation?

A logistic equation can be defined as a differential equation model of population growth that is typically used to relate the change in population ([tex]\frac{dP}{dt}[/tex]) to the total population at a given growth rate (r) and carrying capacity (k).

Mathematically, the logistic equation for this community is given by this equation:

[tex]F (t) = \frac{N(0) k}{N(0)\;+\;[k\; + \;N(0)]e^{-rt}}[/tex]

Substituting the given parameters into the formula, we have;

[tex]F (1) =1200 = \frac{600 \times 60000 }{600\;+\;[60000\; - \;600]e^{-r}}\\\\\frac{36000000}{1200} =600\;+\;[60000\; - \;600]e^{-r}\\\\30000=600+59400e^{-r}\\\\59400e^{-r}=30000-600\\\\59400e^{-r}=29400\\\\e^{-r}=\frac{29400}{59400} \\\\e^{-r}=0.4950\\\\r=ln(0.4950)\\\\[/tex]

r = 0.7032.

For N(t), we have:

[tex]F (t) = \frac{600 \times 60000 }{600\;+\;[60000\; - \;600]e^{-0.7032t}}[/tex]

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