I misread the question and misinterpreted the amount of incoming salt.
There are 3 pounds per gallon of incoming brine, and 4 gallons being pumped in per minute, in other words, 12 pounds incoming.
If $A(t)$ is the function describing the total amount of salt as a function of time, then $$\frac{dA}{dt} = 12 - \frac{5A(t)}{100-t}.$$
Solving the differential equation gave me $$A(t) = [3(100-t)^{-4}+C](100-t)^5 $$ and then I solved for C by using $A(0) = 0$ as an initial condition (since the tank has no brine at the beginning). My final function for $$A(t)= 3(100-t)-\frac{3}{100^4}(100-t)^5$$ and $$A(30) = 159.57$$
:)
You have 100 gal of brine with 50 lbs of salt, so V0 = 100 gal, Q0 = 50 lbs.
We know that 5 gal is entering per second, and 3 gal is exiting, so to find the total volume in the tank:
$V=100+\left( V_{in}-V_{out} \right)t=100+\left( 3-2 \right)t$
where t is time in seconds, so at any time t:
$$V=100+2t$$
concentration = C = $\frac{Q}{V}$
$Q=\mbox{C}\cdot V$ and we can also take this as a rate:
$$\frac{dQ}{dt}=\mbox{C}\cdot \frac{dV}{dt}$$
So basically if $V_{in}=$ 5 gal flows in every second, with a concentration of 1 lb per gallon, then $Q_{in}=$ 5 lb flows in every second.
And if we also have a $V_{out}=$ 3 gal that flows out from the tank every second, with a concentration of Q/V (we don’t know the concentration here, but it’s the concentration of the tank), where Q is the total salt in the tank and V is the total volume (since we’re flushing out solution from the tank), then
$Q_{out}=\mbox{C}\cdot V_{out}=\frac{Q}{V}\cdot 3=\frac{3Q}{100+2t}$ lbs of salt flows out every second.
Now we subtract the two rates:
$$\frac{dQ}{dt}=5-\frac{3Q}{100+2t}$$
I think many people get confused with the Qout part, you need to know that Qout≠Q and Vout≠V (in context). Here we don't know the concentration as with Qin, but we do know that the concentration is that of the tank (since we ARE taking out volume from the tank), so we can take the concentration as Q/V, and then we multiply that by how much volume is going out (Vout) to get Qout every second.
Best Answer
Let $A(t)$ be the number of pounds of coffee in the tank at time $t$. We find an expression for $A'(t)$.
There is a standard pattern for setting up the appropriate differential equatiom. We look separately at the rate sugar is (i) entering the tank and (ii) leaving the tank.
Entering: Liquid is entering at $3$ gallons per minute, and each gallon has $\frac{1}{3}$ pound of sugar. Thus sugar is entering at the rate $\frac{1}{3}\cdot 3$, that is, $1$.
Leaving: If we set $t=0$ at the beginning, then the amount of liquid in the tank at time $t\ge 0$ is $100+2t$. The concentration of sugar at time $t$ is $\frac{A(t)}{100+2t}$. Since liquid is leaving at rate $1$, sugar is leaving at rate $\frac{A(t)}{100+2t}$.
A suitable differential equation for $A(t)$ is therefore $$A'(t)=1-\frac{A(t)}{100+2t}.$$ One needs to write down an appropriate initial condition. The differential equation now can be solved in any of the usual ways, for example by first considering the related homogeneous equation.