Let t be time in seconds and let r(t) be the rate, in gallons per second, that water enters a reservoir:

r(t)=700−40t.

a) For 0≤t≤30, when does the reservoir have the most water?
b) For 0≤t≤30, when does the reservoir have the least water

Already answered. See your previous post.

im trying to look for this too

To find when the reservoir has the most and least water, we need to look at the behavior of the rate function, r(t).

a) To find when the reservoir has the most water, we need to find the maximum value of r(t) within the given time interval, 0 ≤ t ≤ 30.

The rate function is given by r(t) = 700 - 40t. To find the maximum value, we differentiate r(t) with respect to t and set it equal to zero:

r'(t) = -40

Setting -40 equal to zero gives us t = 0. Since this value is not within the given time interval, we exclude it.

So, to find the maximum value of r(t) within the interval 0 ≤ t ≤ 30, we evaluate r(t) at the endpoints of the interval:

r(0) = 700 - 40(0) = 700
r(30) = 700 - 40(30) = 700 - 1200 = -500

Therefore, the reservoir has the most water at t = 0 (the beginning of the interval), with a rate of 700 gallons per second.

b) To find when the reservoir has the least water, we need to find the minimum value of r(t) within the given time interval, 0 ≤ t ≤ 30.

Again, we differentiate r(t) with respect to t and set it equal to zero to find the minimum value:

r'(t) = -40

Setting -40 equal to zero gives us t = 0, but as mentioned before, this value is not within the given time interval.

So, we evaluate r(t) at the endpoints of the interval:

r(0) = 700 - 40(0) = 700
r(30) = 700 - 40(30) = 700 - 1200 = -500

The minimum value of r(t) within the interval 0 ≤ t ≤ 30 is -500 gallons per second, which occurs at t = 30 (the end of the interval).

Therefore, the reservoir has the least water at t = 30 with a rate of -500 gallons per second.