A Water Tank Filling Model
A garden water tank holds 600 litres when full. A hose fills it at a constant 25 litres per minute. The tank starts with 150 litres. A drain at the bottom, opened by mistake, removes water at 5 litres per minute. A gardener models the volume V (in litres) after t minutes of filling (with the drain open) as V(t) = 150 + 20t, valid only while the tank is neither empty nor overflowing. The gardener wants to know when the tank will be full and whether the model makes sense for large t. Later, the drain is closed, and the hose fills the remaining tank at the full 25 L/min.
1. Using V(t) = 150 + 20t, find the time until the tank is full and state the domain of the model.
Read the worked answer
Set V(t) = 600: 150 + 20t = 600 → 20t = 450 → t = 22.5 minutes. Domain: the tank must not overflow, so 0 ≤ t ≤ 22.5. (It never empties because V is increasing from 150.) So the model is valid only on [0, 22.5].
2. The drain is then closed and the hose runs at 25 L/min. If the tank is at 400 litres at that moment, how long until full?
Read the worked answer
Remaining volume = 600 − 400 = 200 L. Time = 200 / 25 = 8 minutes. So the tank fills 8 minutes after the drain is closed.
3. Explain why V(t) = 150 + 20t cannot be used for t = 40 minutes.
Read the worked answer
At t = 40, the formula gives V = 150 + 20·40 = 950 L, but the tank only holds 600 L. The model is a linear expression with no upper cap, so beyond t = 22.5 it predicts impossible volumes. For t = 40 the physical situation would have overflowed, so the formula is outside its valid domain.