Let's work through each part step-by-step: --- ### **Given:** - Initial volume: \( 100 \) liters of pure water - Brine inflow: \( 0.5 \) kg salt per liter at \( 3 \) L/min - Outflow: \( 2 \) L/min (well-mixed solution) - \( y(t) \): amount of salt (kg) in tank at time \( t \) (min) --- ## **(a) Differential Equation** #### **Salt Inflow Rate:** - Salt in per min = \( 3 \) L/min × \( 0.5 \) kg/L = **1.5 kg/min** #### **Salt Outflow Rate:** - Outflow concentration at time \( t \): \( \frac{y(t)}{V(t)} \) (kg/L) - Outflow rate: \( 2 \) L/min × \( \frac{y(t)}{V(t)} \) kg/L \( = \frac{2y(t)}{V(t)} \) kg/min #### **Volume at time \( t \):** - Initial: \( 100 \) L - Net inflow: \( 3 - 2 = 1 \) L/min - \( V(t) = 100 + t \) #### **Differential Equation:** \[ \frac{dy}{dt} = \text{rate in} - \text{rate out} \] \[ \frac{dy}{dt} = 1.5 - \frac{2y(t)}{100+t} \] --- ## **(b) Solve by Separation of Variables** Rewrite: \[ \frac{dy}{dt} + \frac{2}{100+t}y = 1.5 \] This is a linear ODE. Let's use integrating factor method. **Integrating Factor (\( \mu(t) \)):** \[ \mu(t) = \exp\left(\int \frac{2}{100 + t} dt\right) = (100 + t)^2 \] **Multiply both sides by integrating factor:** \[ (100 + t)^2 \frac{dy}{dt} + 2(100 + t)y = 1.5(100 + t)^2 \] \[ \frac{d}{dt} \left[ (100 + t)^2 y \right] = 1.5(100 + t)^2 \] Integrate both sides: \[ (100 + t)^2 y = 1.5 \int (100 + t)^2 dt + C \] \[ \int (100 + t)^2 dt = \frac{(100 + t)^3}{3} \] \[ (100 + t)^2 y = 1.5 \cdot \frac{(100 + t)^3}{3} + C \] \[ (100 + t)^2 y = 0.5 (100 + t)^3 + C \] \[ y(t) = 0.5(100 + t) + \frac{C}{(100 + t)^2} \] **Initial condition:** \( y(0) = 0 \) (pure water, no salt initially) \[ 0 = 0.5 \times 100 + \frac{C}{100^2} \] \[ 0 = 50 + \frac{C}{10000} \implies C = -500,000 \] **So the solution is:** \[ \boxed{ y(t) = 0.5(100 + t) - \frac{500,000}{(100 + t)^2} } \] --- ## **(c) Amount of Salt after 30 minutes** Plug \( t = 30 \): \[ y(30) = 0.5(100 + 30) - \frac{500,000}{(100 + 30)^2} \] \[ = 0.5 \times 130 - \frac{500,000}{16900} \] \[ = 65 - 29.585 \] \[ = 35.415 \text{ kg} \] --- ## **(d) Maximum Volume Before Overflow** - Initial volume: \( 100 \) L - Net inflow: \( 1 \) L/min **Tank will overflow when \( V(t) > \text{capacity} \):** Since the only info is that the initial volume is 100 L and inflow is 1 L/min: \[ \text{Maximum volume before overflow} = \infty \text{ (unless a max capacity is specified)} \] But usually, if asked, the tank will overflow after \( t = T \) minutes when \( V(T) = \text{max capacity} \). **If the question means before it exceeds initial volume:** \[ V(t) = 100 + t \] So the tank will overflow after infinite time, unless given a max capacity. --- ## **Summary of Answers** **(a) Differential equation:** \[ \frac{dy}{dt} = 1.5 - \frac{2y}{100+t} \] **(b) Solution:** \[ y(t) = 0.5(100 + t) - \frac{500,000}{(100 + t)^2} \] **(c) After 30 min:** \[ y(30) \approx 35.4 \, \text{kg} \] **(d) Maximum volume before overflow:** \[ \text{Maximum volume} = \infty \text{ (unless max capacity is specified)} \]

The amount of salt in the tank changes over time due to salt entering and leaving. Salt enters at a constant rate, while the salt leaves with the outflow, which depends on the current concentration. The volume of the liquid increases steadily because inflow exceeds outflow. By setting up a differential equation and solving it, we can find how salt amount varies with time and determine the salt quantity at any specific time.