Question

A 200 gallon tank initially contains 100 gallons of water with 20 pounds of salt. A salt solution with 1/4 pound of salt per gallon is added to the tank at 4 gal/min, and the resulting mixture is drained out at 2 gal/min. Find the quantity of salt in the tank as it's about to overflow.

Answer 1

this is a easy problem but takes a long time to solve, with that said i'm not gonna help you here sorry :(

Answer 2

@SithsAndGiggles

Answer 3

The first sentence gives you a clue about the initial condition. If \(Q(t)\) denotes the amount of salt (in lbs) in the tank at time \(t\), then \(Q(0)=20\).

The rate at which the amount of salt changes, \(\dfrac{dQ(t)}{dt}\), is given as the following relation:
\[\begin{align*}\frac{dQ(t)}{dt}=&\text{(rate in)(concentration in)}-\text{(rate out)(concentration out)}\\\\
&=\left(4\frac{\text{gal}}{\text{min}}\right)\left(0.25\frac{\text{lb}}{\text{gal}}\right)-\left(2\frac{\text{gal}}{\text{min}}\right)\frac{Q(t)}{100+(4-2)t}\\\\
Q'+\frac{1}{50+t}Q&=1
\end{align*}\]
which is linear, and I'm sure you're capable of solving it :)

Lastly, we want to know quantity of salt when the tank is about to overflow. To do this you need to know the time when the tank overflows. The volume of water in the tank is
\[V=100+(4-2)t=100+2t\]
Solve for \(t\) when \(V=200\).

Answer 4

Thank you so much!