Water flows in with rate [tex]R(t)[/tex] and out with rate [tex]D(t)[/tex]. The net rate - call it [tex]N(t)[/tex] - at which it flows through the pipe is [tex]N(t)=R(t)-D(t)[/tex].
a. This part is asking how much water flows into the pipe, regardless of how much flows out of it. To find this, we integrate [tex]R(t)[/tex]:
[tex]\displaystyle\int_0^8R(t)\,\mathrm dt\approx76.57\,\mathrm{ft}^3[/tex]
(I assume you know how to find this value with a graphing calculator?)
b. The amount of water in the pipe increases when [tex]N(t)>0[/tex]. For [tex]t=3[/tex], we have [tex]N(3)\approx-0.313<0[/tex], which means the amount of water is decreasing at this time.
c. You can use the first derivative test here. [tex]t_0[/tex] is a critical point if [tex]N(t_0)=0[/tex], and a minimum occurs at this [tex]t_0[/tex] if [tex]N(t)<0[/tex] for [tex]t<t_0[/tex], and [tex]N(t)>0[/tex] for [tex]t>t_0[/tex]. Refer to a plot - you'll see that this is the case for [tex]t_0[/tex] between [tex]t=2[/tex] and [tex]t=4[/tex]. Use your calculator to get a reasonable approximation (by solving [tex]N(t)=0[/tex] and picking the value between 2 and 4.)
d. The pipe starts with 30 ft^3 of water, so the amount of water in the pipe at any time [tex]w\ge0[/tex] is
[tex]30+\displaystyle\int_0^wN(t)\,\mathrm dt[/tex]
The pipe will overflow for the first time when this is equal to 50.
