Question

Help me in calculus
A water tank at Camp Newton holds 1200 gallons of water at time t = 0. During the time interval 0 <= t <= 18 hours, water is pumped into the tank at the rate \[\large w(t)=95\sqrt{5}\sin^2(\frac{ t }{ 6 })~~~gallons~per~hour.\]During the same time interval, water is removed from the tank at the rate\[\large R(t)=275\sin^2(\frac{ t }{ 3 })~~~gallons~per~hour.\]

Answer 1

At what time t, is the amount of water in the tank at an absolute minimum?

Answer 2

@sooobored @issimplcalcus

Answer 3

Do I do this and solve?
2000 + w(t) - R(t) = 0

Answer 4

ok, the functions given are related to rates

If we look at the function for total amount of water in the tank at time t

it might look something like
\[g(t)=1200 + \int\limits_{}^{} w(t)+\int\limits R(t) \]

Answer 5

assuming there is a minimum, this would mean that this is a critical point

also meaning at this point, g'(t)=0
or the first derivative of g(t) will be equal to zero and then you can solve for t

Answer 6

since the rates are already given as w(t) and R(t)
thus meaning you dont need to integrate then derive since you would just be given w(t) and R(t)

so you start by finding all the critical points when
0=w(t)+R(t)

or solving for t

Answer 7

once you have t, determine whether it is a minimum by substituting it back in or i believe using the 2nd derivative test

Answer 8

Why is it +, since water is removed, shouldn't you subtract the integral of R(t) ????

Answer 9

oh yes, it should be negative

Answer 10

so w(t)-R(t)

Answer 11

Ok, thanks for the help @sooobored you are the smartest guy I've ever met.

Answer 12

@sooobored So I got 2 solutions. I find the derivative of w(t)-R(t) correct?

Answer 13

Using the 2nd derivative, if the 2nd derivative is positive, it is a minimum correct?

Answer 14

i think so