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

Use: r(t) = 700-40t

Determine whether the rate is ever zero within the domain, and if it is positive or negative prior and after that point. That is, are there local minima or maxima within the domain 0≤t≤30?

Use integration to determine the volume of the reservoir V(t) with respect to the initial volume V(0).

Find: V(t) = V(0) + ∫{0↔t} (700-40x) dx

Evaluate V(t) of the points of interest. At t=0, t=30, and of local minima or maxima you found within the domain.

The answers are the t of least and greatest V(t).

To find the time when the reservoir has the most and least water, we need to determine the maximum and minimum values of the function r(t) = 700 - 40t within the given time interval.

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

First, let's find the critical points of the function r(t) by taking its derivative:

r'(t) = -40

Setting r'(t) = 0, we can solve for t:

-40 = 0

This equation has no solution. Therefore, there are no critical points within the given interval.

Since we don't have any critical points to consider, we can evaluate the function at the endpoints of the interval to determine the maximum value.

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

Comparing the two values, we can see that the reservoir has the most water at the beginning of the interval, at t = 0.

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

As discussed in part a), since there are no critical points within the given interval, we can evaluate the function at the endpoints to determine the minimum value.

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

Comparing the two values, we can see that the reservoir has the least water towards the end of the interval, at t = 30.

To find when the reservoir has the most and least water, we can look at the critical points of the function. In this case, we need to find the points where the rate of water entering the reservoir changes.

The rate function is given as r(t) = 700 - 40t.

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

To do this, we can take the derivative of r(t) with respect to t and set it equal to zero to find critical points:

r'(t) = -40

Setting r'(t) = 0 and solving for t:

-40 = 0
t = 0

Since t = 0 is within the time range 0 ≤ t ≤ 30, we also need to check the endpoints of the interval (t = 0 and t = 30).

Now, we evaluate r(t) at all three points: t = 0, t = 30, and any critical points (if applicable).

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

From these calculations, we can see that at t = 0, the reservoir has the most water 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 range 0 ≤ t ≤ 30.

Since the function is linear and decreasing (negative slope), the minimum value of r(t) will occur at t = 30, which is the endpoint of the interval.

Therefore, at t = 30, the reservoir has the least water with a rate of -500 gallons per second. However, it is important to note that negative rate represents water leaving the reservoir rather than entering it.