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 1 answerLast reply by: Bharat PatelMon Mar 25, 2013 2:29 PMPost by Raudel Ulloa on August 6, 2012example 4, i got 152.9 pounds and the same k

### Logistic Growth

• Memorize:
• ,
• Solving logistic growth problems:
• Apply formula with given variables
• Solve unknown variables using given conditions
• Complete growth model with solved variables
• Evaluate function (growth model)

### Logistic Growth

The population of Sedreville was 60 million in 2002 and 75 million in 2017.
Sedreville is speculated to have at most, 90 million denizens.
What is Sedreville's growth model?
• Apply known values for the Logistic Growth model
• K = 90
• P(0) = 60
• P(15) = 75
• With P and K in terms of millions and t in terms of years
• Solve for A
• A = [(K − P0)/(P0)]
• = [(90 − 60)/60]
• = [1/3]
• Find the k constant with P(15)
• P(t) = [K/(Ae − kt + 1)]
• 75 = [90/([1/3]e − 15k + 1)]
• [1/3]e − 15k + 1 = [90/75]
• [1/3]e − 15k = [6/5] − 1
• [1/3]e − 15k = [1/5]
• e − 15k = [3/5]
• lne − 15k = ln[3/5]
• − 15k = ln[3/5]
• k = [(ln[3/5])/( − 15)]
• k = 0.034
Definite the logistic growth model
P(t) = [90/([1/3]e − 0.034t + 1)]
What will the population be in 2020?
• Find t in the context of 2020
• 2020 − 2002 = 8
• t = 8
• Find P( 8 )
• P(8) = [90/([1/3]e − 0.0348 + 1)]
≈ 71.8 million
What is the rate of the population growth for Sedreville at 2020?
[dP/dt] = kP( 1 − [P/K] ) = (0.034)(71.8)( 1 − [71.8/90] ) = 0.49
How long would it take, since 2002, for Sedreville to have 68 million denizens?
• Find P(t) = 68 to solve for t
• 68 = [90/([1/3]e − 0.034t + 1)]
• [1/3]e − 0.034t + 1 = [90/68]
• [1/3]e − 0.034t = [90/65] − 1
• [1/3]e − 0.034t = [11/34]
• e − 0.034t = [33/34]
• lne − 0.034t = ln[33/34]
• − 0.034t = ln[33/34]
• t = [(ln[33/34])/( − 0.034)]
• t = 0.878
• 0.878*365 = 320.47
It will take about 320.47 days later for Sedreville to reach 68 million
Sedreville has a plan to restructure their city when their population reaches 85 million.
When do they restructure?
• Find P(t) = 85 to solve for t
• 85 = [90/([1/3]e − 0.034t + 1)]
• [1/3]e − 0.034t + 1 = [90/85]
• [1/3]e − 0.034t = [90/85] − 1
• [1/3]e − 0.034t = [1/17]
• e − 0.034t = [3/17]
• lne − 0.034t = ln[3/17]
• − 0.034t = ln[3/17]
• t = [(ln[3/17])/( − 0.034)]
• t = 51.02
• 2002 + 51.02 ≈ 2053
Sedreville should restructure in 2053
The peaceful town of Cloudydale was in chaos when news brought forth that an infected kitten was carrying a lethal virus. On July 10th, a couple was found dead from the virus.
45 days later on August 24th, Cloudydale has an infected total of 2403 out of the 13578 townsfolk.
What is the growth model for the virus' carnage on Cloudydale?
• Apply known values for the Logistic Growth model
• K = 13578
• P(0) = 2
• P(45) = 2403
• With t in terms of days
• Solve for A
• A = [(K − P0)/(P0)]
• = [(13578 − 2)/2]
• = 6788
• Find the k constant with P(45)
• 45 = [13578/(6788e − 45k + 1)]
• 6788e − 45k + 1 = [13578/45]
• 6788e − 45k = [13578/45] − [45/45]
• 6788e − 45k = 300.73
• e − 45k = 0.044
− 45k = − 3.11k = 0.07P(t) = [13578/(6788e − 0.07t + 1)]
Quarantine occur when 50% of the population is infected. When is the quarantine?
• Determine 50% of the population
• 50% of 13578 = 6789
• Solve for P(t) = 6789
• 6789 = [13578/(6788e − 0.07t + 1)]
• 6788e − 0.07t + 1 = [13578/6789]
• 6788e − 0.07t = 2 − 1
• 6788e − 0.07t = 1
• e − 0.07t = [1/6788]
• lne − 0.07t = ln[1/6788]
• − 0.07t = − 8.82
• t = 126.04
The quarantine will occur about 126 days from July 10th
How much of the quarantine until 99% of Cloudydale is infected?
• Determine 99% of the population
• 99% of 13578 ≈ 13577
• Solve for P(t) = 13577
• 13577 = [13578/(6788e − 0.07t + 1)]
• 6788e − 0.07t + 1 = [13578/13577]
• 6788e − 0.07t = [13578/13577] − [13577/13577]
• 6788e − 0.07t = [1/13577]
• e − 0.07t = [1/13577]( [1/6788] )
• lne − 0.07t = ln( [1/13577*6788] )
• − 0.07t = − 18.34
• t ≈ 262
• 262 − 126 = 136
Cloudydale will become 99% infected 136 days after they're quarantined
Jerry just benched 155 pounds today. However, his personal best was 195 last year.
When he started working out 2 years ago, he could only bench 55 pounds.
What is Jerry's mathematical model for his progress?
• Apply known values for the Logistic Growth model
• K = 195
• P(0) = 55
• P(2) = 155
• With P and K in terms of pounds, and t in terms of years.
• Solve for A
• A = [(K − P0)/(P0)]
• = [(195 − 55)/55]
• = [28/11]
• Find the k constant with P(2)
• 155 = [195/([28/11]e − 2k + 1)]
• [28/11]e − 2k + 1 = [195/155]
• [28/11]e − 2k = [195/155] − [155/155]
• [28/11]e − 2k = [40/155]
• e − 2k = [40/155]( [11/28] )
• lne − 2k = ln( [40*11/155*28] )
• − 2k = − 2.29
• k = 1.145
P(t) = [195/([28/11]e − 1.15t + 1)]
How much longer until Jerry can bench 194 pounds?
• Solve P(t) for 194
• 194 = [195/([28/11]e − 1.15t + 1)]
• [28/11]e − 1.15t + 1 = [195/194]
• [28/11]e − 1.15t = [195/194] − [194/194]
• [28/11]e − 1.15t = [1/194]
• e − 1.15t = [1/194]( [11/11] )
• lne − 1.15t = ln( [11/194*28] )
• − 1.15t = − 6.20
• t = [( − 6.20)/( − 1.15)]
• ≈ 5.4
• 5.4 − 3 = 2.4
It would take Jerry 2.4 more years to reach 194

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Logistic Growth

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Logistic Growth Function 0:07
• Defining Variables
• Equation Parts
• Logistic Growth 2:04
• Example 1 2:59
• Example 2 7:13
• Example 3 11:29
• Example 4 15:21