Differential Equation Model Let's say we have a question like: 
As the salt KNO3 dissolves in methanol, the number x(t) of grams of the salt in a solution after t seconds satisfies the differential equation $ x'=0.8x-0.004x^2 $.
How does one find the maximum amount M of the salt that will ever dissolve in the methanol? 
 A: Observe that $x' = 0.8x−0.004x^2 = x(0.8 - 0.004x)$ has two roots at $x=0,200$. Hence, we know that $x'>0$ if $0<x<200$ and $x'<0$ if $x<0$ or $x>200$.
Now suppose that we start with an initial amount of $x(0)=x_0 \geq 0$. Then there are three cases to consider.
Case 1: If $x_0 = 0$ or $x_0 = 200$, then the amount will stay constant, so $M=x_0$.
Case 2: If $0<x_0<200$, then the amount will steadily increase towards the stable solution of $x=200$, so $x$ gets arbitrarily close to $200$.
Case 3: If $x_0 > 200$, then the amount will steadily decrease towards the stable solution of $x=200$, so $M = x_0$.
A: If one looks at $x'(t)$, one observes the following facts:
1.)  $x'(0 \, \text{grams}) = 0$;
2.) $x'(200 \,  \text{grams}) = 0$;
3.)  for $0 < x < 200 \,  \text{grams}$, $x'(t) > 0$; the salt is going into the alcohol;
4.)  for $x > 200 \,  \text{grams}$,  $x'(t) < 0$; the salt is coming out of solution.
Note that, by (2), if $0 \,  \text { grams} < x < 200 \,  \text {grams}$ initially, $x$ can never increase beyond $200 \,  \text{grams}$  since $x'(t) = 0$ there.  So assuming we start with $x$ as in (3), the salt will continue, on the net, to enter the solution, at an ever decreasing rate as $x$ gets close to $200 \,  \text{grams}$, so that $x(t) \to 200  \, \text{grams}$ as $t \to \infty$.  Eventually, the amount of salt in solution will be indistinguishable from $200 \,  \text{grams}$, but never more.  The model predicts a maximum observable amount $x = 200 \,  \text{grams}$, provided $x$ starts off as in (3).
Hope this helps.  Cheers, and
Fiat Lux!!!
