5:00 PM is 17 hours after midnight. 9:00 AM is 9 hours after midnight. Part (a) The number of people who’ve entered the park can be found by integrating E(t)--the rate at which people enter the park. 5:00 PM is 17 hours after midnight. 9:00 AM is 9 hours after midnight. 15600 t2-24t+160 dt 9 17 = 6004.270 6004 people will enter the park between 9:00 AM and 5:00 PM.

We’ll multiply this result by the full-day price. ($15) Part (b) To find how much money the park has collected, we need to integrate E(t) two different times. We’ll multiply this result by the full-day price. ($15) 15600 t2-24t+160 dt 9 17 23 And this gets multiplied by the evening price. ($11)

Part (b) 15600 dt = (15)(6004.270) (15) t2-24t+160 (11) 9 17 23 (15) = (15)(6004.270) (11) = (11)(1271.283)

The park will take in $104,048 on a typical day. Part (b) (15)(6004.270) 104,048.165 (11)(1271.283) The park will take in $104,048 on a typical day.

Part (c) H(t) would represent the number of people in the park at any given time. Since H(17) = 3725, that’s how many people were in the park at 5:00 PM. H’(17) represents the rate at which people are entering/exiting the park at 5:00 PM.

Part (c) H’(t) = E(t) – L(t) H’(17) = E(17) – L(17) H’(17) = 380.488 – 760.769 H’(17) = -380.281 H’(17) represents the rate at which people are entering/exiting the park at 5:00 PM. This means that at 5:00 PM, the park’s attendance is dropping at a rate of 380 people/hour.

The number of people in the park, H(t), will be a max when H’(t) = 0. Part (d) The number of people in the park, H(t), will be a max when H’(t) = 0. H’(t) = E(t) – L(t) If H’(t) = 0, then E(t) = L(t). 15600 t2-24t+160 9890 t2-38t+370 =

Part (d) The algebra involved here would be absolutely insane to attempt. Instead, graph both functions and find where they intersect. t = 15.794 or 15.795 (3:48 PM) 15600 9890 = t2-24t+160 t2-38t+370