Question

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.

Answer 1

(a)

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


`\displaystyle \int_0^(300) r(t)\, dt =270`

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


`\displaystyle\int_0^(300) 44 \left( (t)/(100) \right)^3 \left(1 - (t)/(300) \right)^7`

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(b)


`\displaystyle 20 + \int_0^(300) \big[ r(t) - 0.7\big] dt = 80`

There are 80 people at time t = 300

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(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.


`80 + \int_0^t (0.7) \,dx = 0 \\ 80 + 0.7t = 0 \\ t \approx 114.286`

This is for t > 300, so

The first time t is approximately t = 414.286

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(d)

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

By graphing calculator,


`r(t) - 0.7 = 0 \implies t \approx 33.013, 166.575`

If
`P(t) = 20 + \int_0^t \left[ r(x) - 0.7 \right] dx` 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.