Tank A contains 80 gallons of water in which 20 pounds of salt has been dissolved. Tank B contains 30 gallons of water in which 5 pounds of salt has been dissolved. A brine mixture with a concentration of 0.5 pounds of salt per gallon of water is pumped into tank A at the rate of 4 gallons per minute. The well-mixed solution is then pumped from tank A to tank B at the rate of 6 gallons per minute. The solution from tank B is also pumped through another pipe into tank A at the rate of 2 gallons per minute, and the solution from tank B is also pumped out of the system at the rate of 4 gallons per minute. The correct differential equations with initial conditions for the amounts, \( x(t) \) and \( y(t) \), of salt in tanks A and B, respectively, at time \( t \) are Select the correct answer. \( \frac{d x}{d t}=2-\frac{x}{40}+\frac{y}{5}, \frac{d y}{d t}=\frac{x}{40}-\frac{y}{3}, x(0)=20, y(0)=5 \) \( \frac{d x}{d t}=2-\frac{3 x}{40}+\frac{y}{15}, \frac{d y}{d t}=\frac{3 x}{40}-\frac{y}{5}, x(0)=20, y(0)=5 \) \( \frac{d x}{d t}=4-\frac{3 x}{40}+\frac{y}{15}, \frac{d y}{d t}=\frac{3 x}{40}-\frac{y}{5}, x(0)=20, y(0)=5 \) \( \frac{d x}{d t}=4-\frac{x}{40}+\frac{y}{5}, \frac{d y}{d t}=\frac{x}{40}-\frac{y}{3}, x(0)=20, y(0)=5 \) \( \frac{d x}{d t}=2-\frac{3 x}{40}+\frac{y}{5}, \frac{d y}{d t}=\frac{x}{40}-\frac{y}{5}, x(0)=20, y(0)=5 \) \( x \)

We begin by letting x(t) be the amount of salt (in pounds) in tank A at time t (in minutes) and y(t) be the amount of salt in tank B at time t. Step 1. Determine the volume in each tank. • Tank A: Initially 80 gallons. It receives 4 gal/min from an external brine (with 0.5 lb/gal) plus 2 gal/min from tank B, and it sends out 6 gal/min to tank B. Thus, its volume is constant:   Net flow = 4 + 2 – 6 = 0. • Tank B: Initially 30 gallons. It receives 6 gal/min from tank A, and it sends out 2 gal/min back to A and 4 gal/min out of the system. Its net flow is also 0. So, the volumes remain constant:  Tank A: 80 gallons  Tank B: 30 gallons Step 2. Write the salt balance for Tank A. Salt is added and removed as follows: • Inflow from external brine: 4 gal/min × 0.5 lb/gal = 2 lb/min. • Inflow from Tank B: The salt concentration in Tank B is y/30, and the inflow is 2 gal/min, so salt added = 2(y/30) = y/15. • Outflow to Tank B: The salt concentration in Tank A is x/80, and the outflow is 6 gal/min, so salt removed = 6(x/80) = (3x)/40. Thus, the differential equation for Tank A is:   dx/dt = (salt in) – (salt out)   dx/dt = 2 + (y/15) – (3x/40). Step 3. Write the salt balance for Tank B. Salt is added and removed as follows: • Inflow from Tank A: The salt concentration in Tank A is x/80 and the inflow is 6 gal/min, so salt added = 6(x/80) = (3x)/40. • Outflow from Tank B: There are two outlets:   – 2 gal/min to Tank A: Salt removed = 2(y/30) = y/15.   – 4 gal/min out of the system: Salt removed = 4(y/30) = 2y/15.   Total salt removed = y/15 + 2y/15 = y/5. Thus, the differential equation for Tank B is:   dy/dt = (salt in) – (salt out)   dy/dt = (3x/40) – (y/5). The initial amounts are given:  x(0) = 20 pounds  y(0) = 5 pounds Step 4. Match with the answer choices. We obtained:  dx/dt = 2 – (3x/40) + (y/15),  dy/dt = (3x/40) – (y/5), with initial conditions: x(0)=20, y(0)=5. This corresponds to the second option:  “dx/dt = 2 – (3x)/40 + (y)/15, dy/dt = (3x)/40 – (y)/5, x(0)=20, y(0)=5." Thus, the correct answer is:  dx/dt = 2 – (3x)/40 + (y)/15, dy/dt = (3x)/40 – (y)/5, with x(0)=20, y(0)=5.