This is a very famous example of Differential Equation, and has been applied to numerous of real life problems as a model. It's originally a Population Model created by Verhulst, as studying the population's growth.
Let's let P(t) as the population's size in term of time t, and dP/dt represents the Population's growth.
Malthus' Exponential growth theory of population
Mr. Malthus first introduced the exponential growth theory for the population by using a fairly simple equation: Where P is the "Population Size", t is the "Time", r is the "Growth Rate".
Verhulst's Logistic growth theory of population
Mr. Verhulst enhanced the exponential growth theory of population, as saying that the population's growth is NOT ALWAYS growing, but there is always a certain LIMIT or a Carrying Capacity to the exponential growth. And combining the exponential growth with a limit, it's then called the Logistic Growth.
And the logistic growth got its equation:
Where P is the "Population Size" (N is often used instead), t is "Time", r is the "Growth Rate", K is the "Carrying Capacity". And the (1 - P/K) determines how close is the Population Size to the Limit K, which means as the population gets closer and closer to the limit, the growth gets slower and slower.
"It explains how density dependent limiting factors eventually decrease the growth rate until a population reaches a Carrying Capacity ( K )."
Carrying Capacity
Carrying Capacity means the "celling", the "limit", the "asymptote".
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Get the Original Population Function P(t)
It's gonna use the method Separable Equations, which introduced the initial condition as P₀ in this case.
We could directly solve the Logistic Equation as solving differential equation to get the antiderivative:
But we still have a constant C in the antiderivative, which required us to introduce an Initial Condition to get rid of C and get the specific function:
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Solving Logistic Model Problems
Example
Solve:
We know the Logistic Equation is dP/dt = r·P(1-P/K).
So twist the given derivative to the logistic form: dy/dt = 10·y(1-y/600).
Then we could see the K = 600, which is the limit, the Carrying capacity.
Example
Solve:
It's asking "growing fastest", means the Max value of Sale's function S(c).
For the max value of function, we let S'(c) = dS/dc = 0
And we get S = 0 or 20,000,000
So at two points they are getting fastest growing.
Yet we have to take the AVERAGE of the two points, which is (0+20,000,000)/2 = 10,000,000.
Where 
Where 
Solve:
Solve: