A cylindrical tank (with dimensions shown below) contains water that is 1-foot deep. If water is poured into the tank at a constant rate of 20 ft3/min for 20 minutes, will the tank overflow? Use 3.14 to estimate π.

Complete Question: The diagram referred to in the question is attached below

The tank will overflow

Step-by-step explanation:

From the diagram:

The height of the cylindrical tank, h = 12 ft

Radius of the cylindrical tank, r = 3 ft

Volume of the cylindrical tank, Vtank = πr²h

Where π = 3.14

Vtank = 3.14 * 3² * 12

Vtank = 3.14 * 9 * 12

Vtank = 339.12 ft³

Volume of the cylindrical tank = 339.12 ft³

Depth of water, h = 1 ft

The radius of water still takes the radius of the tank, r = 3 ft

Volume of water, Vwater = πr²h

Volume of water = 3.14 * 3² * 1

Volume of water = 28.26 ft³

The volume available in the tank = Vtank - Vwater

Volume available in the tank = 339.12 - 28.26

Volume available in the tank = 310.86 ft³

Rate at which water is poured, R = 20 ft³/min

Time taken to pour water, t = 20 mins

Volume of water poured into tank = Rate * time

Volume of water poured into tank = 20 * 20

Volume of water poured into tank = 400 ft³

Since the volume of water poured into tank ( 400 ft³) > Volume available in the tank (310.86 ft³), the tank will overflow

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