We can approach this problem by using a differential equation model. Here's a step-by-step outline:

Step 1: Define Variables

• Let $S(t)$ be the amount of salt (in pounds) in the tank at time $t$ (in minutes).
• The volume of solution in the tank decreases at a net rate of 4 gallons per minute (since 4 gal/min enters and 8 gal/min leaves). Therefore, the volume of solution at time $t$ is: $V(t) = 80 - 4t \quad \text{(gallons)}$ We are asked to find the amount of salt when $V(t) = 40$ gallons, i.e., when $t = 10$ minutes.

Step 2: Salt Input and Output Rates

• Input rate of salt: The inflow rate of salt is the concentration of the incoming brine solution (1 lb/gal) times the flow rate of the incoming brine (4 gal/min): $\text{Input rate} = 1 \cdot 4 = 4 \, \text{lb/min}$

• Output rate of salt: The outflow rate of salt is the concentration of the solution in the tank at time $t$ (which is $\frac{S(t)}{V(t)}$, the amount of salt per gallon) times the outflow rate (8 gal/min): $\text{Output rate} = \frac{S(t)}{V(t)} \cdot 8 = \frac{S(t)}{80 - 4t} \cdot 8$

Step 3: Set Up the Differential Equation

The rate of change of salt in the tank is given by: $\frac{dS(t)}{dt} = \text{Input rate} - \text{Output rate}$ Substitute the input and output expressions: $\frac{dS(t)}{dt} = 4 - \frac{8S(t)}{80 - 4t}$

Step 4: Solve the Differential Equation

This is a first-order linear differential equation. To solve it, we can use an integrating factor method or simply recognize the form. First, rearrange the equation: $\frac{dS(t)}{dt} + \frac{8}{80 - 4t}S(t) = 4$ The integrating factor $\mu(t)$ is: $\mu(t) = e^{\int \frac{8}{80 - 4t} dt} = e^{\int \frac{-2}{80 - 4t} dt} = (80 - 4t)^{-2}$ Multiply through by the integrating factor: $(80 - 4t)^{-2} \frac{dS(t)}{dt} + \frac{8}{80 - 4t} \cdot (80 - 4t)^{-2} S(t) = 4 \cdot (80 - 4t)^{-2}$ Simplify: $\frac{d}{dt} \left( S(t) (80 - 4t)^{-2} \right) = 4(80 - 4t)^{-2}$ Integrate both sides: $S(t)(80 - 4t)^{-2} = \int 4(80 - 4t)^{-2} dt$ The right-hand side integrates to: $S(t)(80 - 4t)^{-2} = \frac{1}{80 - 4t} + C$ Multiply through by $(80 - 4t)^2$: $S(t) = (80 - 4t) + C(80 - 4t)^2$

Step 5: Use Initial Condition

At $t = 0$, the initial amount of salt in the tank is: $S(0) = \frac{80}{8} = 10 \, \text{lb of salt}$ Substitute this into the solution: $10 = 80 + C(80)^2$ Solve for $C$: $C = \frac{10 - 80}{6400} = \frac{-70}{6400} = -\frac{7}{640}$

Step 6: Find $S(10)$

Now, substitute $t = 10$ into the solution to find the amount of salt when the volume is 40 gallons: $S(10) = (80 - 40) + \left(-\frac{7}{640}\right)(80 - 40)^2$ Simplify: $S(10) = 40 + \left(-\frac{7}{640}\right)(40)^2$ $S(10) = 40 - \frac{7 \cdot 1600}{640}$ $S(10) = 40 - 17.5 = 22.5 \, \text{lb}$

Thus, the amount of salt in the tank when the volume is 40 gallons is 22.5 pounds.

Do you want further details or have any questions? Here are some related questions:

1. How can we interpret the behavior of the salt concentration over time?
2. What happens if the inflow and outflow rates were equal?
3. Can we apply this method to solve for different initial conditions?
4. How would the answer change if the brine inflow concentration varied over time?
5. What are some common scenarios where similar differential equations arise?

Tip: Always pay attention to the inflow and outflow rates when modeling mixing problems. They directly influence the salt concentration.