People enter a line for an escalator at a rate modeled by the function r given by (see image) where r(t) is measured in people per second and t is measured in seconds. as people get on the escalator, they exit the line at a constant rate of 0.7 person per second. there are 20 people in line at time t = 0. (a) how many people enter the line for the escalator during the time interval 0 ≤ t ≤ 300? (b) during the time interval 0 ≤ t ≤ 300, there are always people in line for the escalator. how many people are in line at time t = 300 ? (c) for t > 300, what is the first time t that there are no people in line for the escalator? (d) for 0 ≤ t ≤ 300, at what time t is the number of people in line a minimum? to the nearest whole number, find the number of people in line at this time. justify your answer.

(a)

The amount of people that went on the escalator is given by the integral



270 people enter the elevator during the time interval 0 ≤ t ≤ 300

You can save time by just writing that and getting an answer from your calculator. You are not expected to write out the entire integrand. Since this is for 0 ≤ t ≤ 300, you would be typing this integral into your calculator



_______________________

(b)



There are 80 people at time t = 300

_______________________

(c)

Since there are 80 people at time t = 300 and r(t) = 0 for t > 300, the rate of people in line is only determined constant exiting rate of 0.7 person per second. The amount of people in line is linear for t > 300.



This is for t > 300, so

The first time t is approximately t = 414.286

_______________________

(d)

The absolute minimum will occur at a critical point where r(t) - 0.7 = 0 or at an endpoint.

By graphing calculator,



If represents the amount of people in line for 0 ≤ t ≤ 300, then

P(0) = 20 people (given)
P(33.013) ≈ 3.803
P(166.575) ≈ 166.575
P(300) = 80

Therefore, at t = 33.013, the number of people in line is a minimum with 4 people.