Q. Suppose that a tank initially contains 2000 gal of water and the rate of change of its volume after the tank drains for t min is`(t) = (0.5)t -30(in gallons per minute). How much water does the tank contain after it has been draining for 25 minutes?

A. $\displaystyle y(t) = .25t^2 -30t +c.$

$\displaystyle C = 2000.$ Substitute $\displaystyle t = 25.$

We get $\displaystyle y(25) = .25(25)^2 -30t + 2000 = 1406.25$ Gal remaining.