Question

a tank initially contains 200 gal of brine in which 50 lb of salt are dissolved. a brine containing 5 lb/gal of salt runs into the tank at the rate of 3 gal/min. the mixture is kept uniform by stirring and flows out of the tank at the rate of 2 gal/min. let y represent the amount of salt at time t. complete parts a through e.
d write down and solve the initial value problem describing the mixing process.
\\(\frac{dy}{dt} = 15 - \frac{2y}{200 + t}\\), \\(y(0) = 50\\).
what is the solution to the initial value problem?
\\(y = 5(200 + t) - \frac{38000000}{(200 + t)^2}\\)
e. find the concentration of salt in the tank 25 min after the process starts.
\\(square\\) lb/gal
(type an integer or decimal rounded to the nearest tenth as needed.)

Explanation:

Step1: Find the amount of salt at \(t = 25\)

We know the formula for \(y\) (the amount of salt) is \(y=5(200 + t)-\frac{38000000}{(200 + t)^{2}}\). Substitute \(t = 25\) into the formula.
First, calculate \(200+t\): \(200 + 25=225\)
Then, \(5(200 + t)=5\times225 = 1125\)
Next, \(\frac{38000000}{(200 + t)^{2}}=\frac{38000000}{225^{2}}=\frac{38000000}{50625}\approx750.6\)
So, \(y=1125-750.6 = 374.4\)

Step2: Find the volume of brine at \(t = 25\)

The initial volume is \(V_0=200\) gal. The inflow rate is \(r_{in}=3\) gal/min and the outflow rate is \(r_{out}=2\) gal/min.
The volume of brine at time \(t\) is \(V(t)=V_0+(r_{in}-r_{out})t\). Substitute \(t = 25\), \(V_0 = 200\), \(r_{in}=3\), \(r_{out}=2\)
\(V(25)=200+(3 - 2)\times25=200 + 25=225\) gal

Step3: Calculate the concentration

Concentration \(C=\frac{y}{V}\). Substitute \(y = 374.4\) and \(V = 225\)
\(C=\frac{374.4}{225}=1.664\approx1.7\) lb/gal

Answer:

\(1.7\) lb/gal