To solve the problem, we will set up a differential equation based on the information provided about the tank, the inflow of brine, and the outflow of the mixture.
### Step-by-step Solution:
1. **Define Variables**:
Let \( u(t) \) be the amount of salt in pounds at time \( t \) (in minutes). Initially, the tank contains 50 gallons of fresh water, and we want to find how long it takes for the salt content to increase from 40 to 80 pounds.
2. **Rate of Salt Inflow**:
The brine flows into the tank at a rate of 2 gallons per minute, with a concentration of 2 pounds of salt per gallon. Therefore, the rate of salt inflow is:
\[
\text{Inflow} = 2 \, \text{gallons/min} \times 2 \, \text{pounds/gallon} = 4 \, \text{pounds/min}
\]
3. **Rate of Salt Outflow**:
The mixture is well-stirred, so the concentration of salt in the tank at time \( t \) is \( \frac{u(t)}{50} \) pounds per gallon. The outflow rate is also 2 gallons per minute, so the rate of salt outflow is:
\[
\text{Outflow} = 2 \, \text{gallons/min} \times \frac{u(t)}{50} \, \text{pounds/gallon} = \frac{2u(t)}{50} = \frac{u(t)}{25} \, \text{pounds/min}
\]
4. **Set Up the Differential Equation**:
The rate of change of salt in the tank can be expressed as:
\[
\frac{du}{dt} = \text{Inflow} - \text{Outflow} = 4 - \frac{u(t)}{25}
\]
5. **Rearranging the Equation**:
Rearranging gives:
\[
\frac{du}{dt} = 4 - \frac{u}{25}
\]
6. **Separate Variables**:
We can separate the variables to integrate:
\[
\frac{du}{4 - \frac{u}{25}} = dt
\]
7. **Integrate Both Sides**:
Integrating the left side:
\[
\int \frac{du}{4 - \frac{u}{25}} = \int dt
\]
The left side can be simplified:
\[
\int \frac{25 \, du}{100 - u} = 25 \ln |100 - u| + C
\]
The right side integrates to:
\[
t + C
\]
8. **Combine and Solve for \( u \)**:
Setting the integrated expressions equal gives:
\[
25 \ln |100 - u| = t + C
\]
Solving for \( u \):
\[
|100 - u| = e^{\frac{t + C}{25}}
\]
Thus,
\[
u = 100 - e^{\frac{t + C}{25}}
\]
9. **Initial Condition**:
At \( t = 0 \), \( u(0) = 0 \):
\[
0 = 100 - e^{\frac{C}{25}} \implies e^{\frac{C}{25}} = 100 \implies C = 25 \ln(100)
\]
10. **Final Equation**:
Substitute \( C \) back into the equation:
\[
u(t) = 100 - 100 e^{-\frac{t}{25}}
\]
11. **Find Time for \( u(t) = 80 \) and \( u(t) = 40 \)**:
- For \( u(t_1) = 40 \):
\[
40 = 100 - 100 e^{-\frac{t_1}{25}} \implies e^{-\frac{t_1}{25}} = 0.6 \implies t_1 = -25 \ln(0.6)
\]
- For \( u(t_2) = 80 \):
\[
80 = 100 - 100 e^{-\frac{t_2}{25}} \implies e^{-\frac{t_2}{25}} = 0.2 \implies t_2 = -25 \ln(0.2)
\]
12. **Calculate Time Difference**:
The time taken for the salt content to increase from 40 to 80 pounds is:
\[
\Delta t = t_2 - t_1 = -25 \ln(0.2) + 25 \ln(0.6) = 25 (\ln(0.6) - \ln(0.2)) = 25 \ln\left(\frac{0.6}{0.2}\right) = 25 \ln(3)
\]
13. **Convert to Seconds**:
Since \( \Delta t \) is in minutes, convert to seconds:
\[
\Delta t = 25 \ln(3) \times 60 \text{ seconds}
\]
14. **Equate to Given Expression**:
We know \( \Delta t = 206\lambda \):
\[
25 \ln(3) \times 60 = 206\lambda
\]
15. **Solve for \( \lambda \)**:
\[
\lambda = \frac{25 \ln(3) \times 60}{206}
\]
### Final Calculation:
Using \( \ln(3) \approx 1.0986 \):
\[
\lambda \approx \frac{25 \times 1.0986 \times 60}{206} \approx 8
\]
Thus, the value of \( \lambda \) is \( 8 \).
