At a processing plant, over the course of an 8-hour day, workers move material i

Question

At a processing plant, over the course of an 8-hour day, workers move material into a pile, which is removed at a constant rate by a conveyor belt. The workers move material into the pile at rate
500e^(?0.5t)
units/hr while the conveyor belt removes the material at 125 units/hr.
(a) Find the net change in the size of the pile over the first 2 hours, and over the first 8 hours. Round your answers to the nearest hundredth.
2 hours _____


8 hours _____

units

(b) At what time is the amount of material in the pile the largest and the smallest?
largest t =

smallest t =

Explanation / Answer

Net rate of piling dP/dt = 500e^(-0.5t) - 125

a) Integrating P = 500e^(-0.5t)/(-0.5) - 125t + C = -1000e^(-0.5t) - 125t + C

P(2) - P(0) = -1000*e^(-0.5*2) - 125*2 - [-1000*e^(-0.5*0) - 125*0] = 382.12 units

p(8) - P(0) = -1000*e^(-0.5*8) - 125*8 - [-1000*e^(-0.5*0) - 125*0] = -18.32 units

b) For largest pile size, we put dP/dt = 0

500e^(-0.5t) - 125 = 0

t = (ln 4)/0.5 = 2.77 hours

For smallest pile size we compare P(0) and P(8). Since P(8) - P(0) is negative P(8) < P(0).

Hence, smallest pile size is at t = 8 hours

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.