Storm sewer backup causes your basement to flood at the steady tate of 1 in. of depth per hour. The basement floor area is \(1500 \mathrm{ft}^{2}\) What capacity (gal/min) pump would you rent to (a) kecp the water eccumulated in your basement at a constant level urtil the storm sewer is blocked off. and (b) reduce the water accumulation in ycur basement at a rate of 3 in fhr even while the backup problem exiss?

First, calculate the pump capacity to keep the water at a constant level. Since the water level increases at the rate of 1 inch per hour, the volume of water that enters the basement per minute is given by: Volume = Area x Height. The area is given as \(1500 \, \text{ft}^2\), and the height (rate of water increase) is \(1 \, \text{in/hr}\). Convert everything to consistent units before calculation. \n As 1 foot = 12 inches, 1 hr = 60 mins, and 1 cubic foot = 7.48 gallons, convert the area to square inches, the time to minutes, and the volume to gallons. We get thus the volume per minute in gallons: \( Volume = (1500 \, \text{ft}^2 * 144 \, \text{in}^2/\text{ft}^2 * 1 \, \text{in/hr}) / 60 \, \text{min/hr} * 7.48 \, \text{gal}/\text{ft}^3\)

Calculate the above expression to determine the pump capacity needed to maintain a constant water level: \(Volume = 37.4 \, \text{gal/min}\).

Next, calculate the pump capacity to reduce the water level at a rate of 3 inches per hour. The total volume of water to be removed includes both the water entering the basement (1 inch per hour, calculated in Step 1), and the additional water to lower the level (3 inches per hour). The total volume of water removed per minute is then: \( Volume_{total} = Area * Rate\_total = 1500 \, \text{ft}^2 * 144 \, \text{in}^2/\text{ft}^2 * 4 \, \text{in/hr} / 60 \, \text{min/hr} * 7.48 \, \text{gal}/\text{ft}^3 \)

Calculate the above expression to get the pump capacity required to reduce the water level: \(Volume_{total} = 149.5 \, \text{gal/min}\).