Let
= amount of salt (in pounds) in the tank at time
(in minutes). Then
.
Salt flows in at a rate
![\left(0.6(\rm lb)/(\rm gal)\right) \left(3(\rm gal)/(\rm min)\right) = \frac95 (\rm lb)/(\rm min)](https://img.qammunity.org/2023/formulas/mathematics/high-school/79vl7tgyqadxv5bgsi3w2jg45owkdak0p9.png)
and flows out at a rate
![\left((A(t)\,\rm lb)/(75\,\rm gal + \left(3(\rm gal)/(\rm min) - 3.25(\rm gal)/(\rm min)\right)t)\right) \left(3.25(\rm gal)/(\rm min)\right) = (13A(t))/(300-t) (\rm lb)/(\rm min)](https://img.qammunity.org/2023/formulas/mathematics/high-school/bzeue499fn0ynopvi0r01piio0wgspsk8r.png)
where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.
Then the net rate of salt flow is given by the differential equation
![(dA)/(dt) = \frac95 - (13A)/(300-t)](https://img.qammunity.org/2023/formulas/mathematics/high-school/hvdp2io5cvor4ry2jl6nc3yc6frmdea47s.png)
which I'll solve with the integrating factor method.
![(dA)/(dt) + (13)/(300-t) A = \frac95](https://img.qammunity.org/2023/formulas/mathematics/high-school/q1xfpjwv0bxk6biikw6lzm21g51tsfg548.png)
![-\frac1{(300-t)^(13)} (dA)/(dt) - (13)/((300-t)^(14)) A = -\frac9{5(300-t)^(13)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/qrztg3wgsk0e5m6dms2m96iutqdfp35hhl.png)
![\frac d{dt} \left(-\frac1{(300-t)^(13)} A\right) = -\frac9{5(300-t)^(13)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/zodqvvvwzu5633ioaqc1a1h164mgv45g7k.png)
Integrate both sides. By the fundamental theorem of calculus,
![\displaystyle -\frac1{(300-t)^(13)} A = -\frac1{(300-t)^(13)} A\bigg|_(t=0) - \frac95 \int_0^t (du)/((300-u)^(13))](https://img.qammunity.org/2023/formulas/mathematics/high-school/vb6pumcjm1lxlqv1orv0afom9phas9vq2d.png)
![\displaystyle -\frac1{(300-t)^(13)} A = -(11)/(300^(13)) - \frac95 * \frac1{12} \left(\frac1{(300-t)^(12)} - \frac1{300^(12)}\right)](https://img.qammunity.org/2023/formulas/mathematics/high-school/do3nyl3pa2stwchs1s3kzjvxcao8e7nkxz.png)
![\displaystyle -\frac1{(300-t)^(13)} A = (34)/(300^(13)) - \frac3{20}\frac1{(300-t)^(12)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ftrb8gvc36fdsn7a0zkk7w2f133nhrwbdu.png)
![\displaystyle A = \frac3{20} (300-t) - (34)/(300^(13))(300-t)^(13)](https://img.qammunity.org/2023/formulas/mathematics/high-school/9f510whdggeyi2w2cr8zcz7w4ozx1xycbt.png)
![\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^(13)](https://img.qammunity.org/2023/formulas/mathematics/high-school/8vms6gxs3jxri4pe0799k6nlk8rktsbma5.png)
After 1 hour = 60 minutes, the tank will contain
![A(60) = 45 \left(1 - \frac {60}{300}\right) - 34 \left(1 - \frac {60}{300}\right)^(13) = 45\left(\frac45\right) - 34 \left(\frac45\right)^(13) \approx 34.131](https://img.qammunity.org/2023/formulas/mathematics/high-school/91avsa9bzobbh2z7wioowectci6k4bchqq.png)
pounds of salt.