(a)
The amount of people that went on the escalator is given by the integral
[tex]\displaystyle
\int_0^{300} r(t)\, dt =270[/tex]
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
[tex]\displaystyle\int_0^{300} 44 \left( \frac{t}{100} \right)^3 \left(1 - \frac{t}{300} \right)^7[/tex]
_______________________
(b)
[tex]\displaystyle
20 + \int_0^{300} \big[ r(t) - 0.7\big] dt = 80[/tex]
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.
[tex]80 + \int_0^t (0.7) \,dx = 0 \\
80 + 0.7t = 0 \\
t \approx 114.286[/tex]
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,
[tex]r(t) - 0.7 = 0 \implies t \approx 33.013, 166.575[/tex]
If [tex] P(t) = 20 + \int_0^t \left[ r(x) - 0.7 \right] dx[/tex] 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.
