Free Responce

1. In order to find how much sand will be removed during a six hour period, one must find the integral between 0 to 6 hours of R(t)=2+5sin(4pi x/25)

Once you set that up, you solve and you end up getting about 31.815 cubic yds
2. We are given that when time is zero, there are 2500 cubic yards of sand. We are also given that the sand is added at a certain rate S(t)=15t/1+3t, and that it is removed at a rate of R(t)=s+5sin(4pi x/25). The formula would be the rate at which the sand is added minus the rate at which it is taken away plus the original amount. it would be fnInt(15t/1+3t)-(2+5sin(4pi x/25)+2500
3. in order to find the change at a certain time, all we have to do is replace the variable with the correct time, which in this casae would be 4 for t in fnInt(15t/1+3t)-(2sin(4pi x/25)+2500. It is S(4)-R(4) which ends up being -1.909 yards/hour
4. Besides the two end points, the only other critical point is at 5.118. When graphed, you can easily see that the end points are relative maximums rather than minimums. When inputs are plugged in before 5.118, the outputs of y' are negative, and the outputs of after 5.118 are positive, proving that this is certainly a minimum. now you plug in 5.118 instead of t. and you end up getting 2492.369 cubic yards