| the whole thing was easy. 1d use cylinders. u have height and radius. just 2(3.14)fnint ,(x)f(x),x,0,2 if i remember numbers right. for 3b you have the rate at which its entering and rate at which its leaving. because you have the RATE it implies it means it is the first derivative, so find the zeroes and use first derivative test to find maximum time. it is 25. for 3c you have the starting volume. and because until time 25 the amount entering exceeds the amount leaving, verified in part b, you will do 60,000 + (3.14)fnint, 2000-400(t)^.5, or something like that if i remember numbers right |