A tank originally contains 200 liters of fresh water. Then, water containing
50g of salt per liter is poured into the tank at a rate of 1 liter/min. The well-stirred
mixture is allowed to leave at the same rate of 1 liter/min. After 12 minutes, the process
is stopped, and fresh water is poured into the tank at a rate of r liters/min, with the
well-stirred mixture again leaving at the same rate of r liters/min. Develop mathematical
models for both processes and find the rate r if the amount of salt in the tank after 24
minutes (12 minutes for each process) is 100g.


### Problem Overview
We have two processes happening in sequence:

1. **First Process (0 to 12 minutes)**: A saltwater solution containing 50 g of salt per liter enters the tank at a rate of 1 liter per minute, while a well-stirred mixture of water and salt leaves at the same rate (1 liter/min).
2. **Second Process (12 to 24 minutes)**: Fresh water enters at a rate \( r \) liters per minute, and a well-stirred mixture of water and salt leaves at the same rate \( r \) liters per minute.

### First Process: 0 to 12 minutes
Let \( x(t) \) represent the amount of salt (in grams) in the tank at time \( t \) (in minutes).

- **Inflow rate of salt**: Since the inflowing water contains 50 g of salt per liter and enters at 1 liter per minute, the rate at which salt enters the tank is \( 50 \, \text{g/min} \).
- **Outflow rate of salt**: The well-stirred mixture leaves at a rate of 1 liter per minute, and since the tank volume is constant at 200 liters, the concentration of salt in the tank at time \( t \) is \( \frac{x(t)}{200} \, \text{g/liter} \). So the outflow rate of salt is \( \frac{x(t)}{200} \, \text{g/min} \).

Thus, the differential equation describing the salt content in the tank during the first process is:
\[
\frac{dx}{dt} = 50 - \frac{x(t)}{200}
\]
This is a linear first-order differential equation.

#### Solution of the differential equation:
Rewriting the equation:
\[
\frac{dx}{dt} + \frac{1}{200}x = 50
\]
The integrating factor \( \mu(t) \) is:
\[
\mu(t) = e^{\int \frac{1}{200} dt} = e^{\frac{t}{200}}
\]
Multiplying both sides by \( e^{\frac{t}{200}} \):
\[
e^{\frac{t}{200}} \frac{dx}{dt} + \frac{1}{200} e^{\frac{t}{200}} x = 50 e^{\frac{t}{200}}
\]
Integrating both sides with respect to \( t \):
\[
e^{\frac{t}{200}} x = \int 50 e^{\frac{t}{200}} dt = 50 \cdot 200 e^{\frac{t}{200}} + C
\]
Simplifying:
\[
x(t) = 50 \cdot 200 + C e^{-\frac{t}{200}} = 10000 + C e^{-\frac{t}{200}}
\]
At \( t = 0 \), the tank contains only fresh water, so \( x(0) = 0 \):
\[
0 = 10000 + C \implies C = -10000
\]
Thus, the solution for \( x(t) \) during the first process is:
\[
x(t) = 10000 \left( 1 - e^{-\frac{t}{200}} \right)
\]
At \( t = 12 \) minutes, the amount of salt in the tank is:
\[
x(12) = 10000 \left(

A tank originally contains 200 liters of fresh water. Then, water containing 50g of salt per liter is poured into the tank at a rate of 1 liter/min. The well-stirred mixture is allowed to leave at the same rate of 1 liter/min. After 12 minutes, the process is stopped, and fresh water is poured into the tank at a rate of r liters/min, with the well-stirred mixture again leaving at the same rate of r liters/min. Develop mathematical models for both processes and find the rate r if the amount of salt in the tank after 24 minutes (12 minutes for each process) is 100g.

This detailed solution explores how to model a tank's salt concentration using differential equations. Learn how the inflow-outflow process works, find the rate at which fresh water must enter the tank, and understand how the amount of salt changes over time. Suitable for college-level mathematics students.

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