Project #1 Chapter 4.1 Q#11.

Contents

Group members: Raisa Ratri and Richard Li

A candy maker makes 500 pounds of candy per week, while his large family eats the candy at a rate equal to Q(t)/10 pounds per week, where Q(t) is the amount of candy present at the time.

1. Find Q(t) for t > 0 if the candy maker has 250 pounds of candy at t = 0.
2. Find limt–> ∞ Q(t).

Part A

Q’ = 500 – Q(t) / 10

= (5000 – Q(t)) / 10

= -dQ(t) / (5000 – Q(t)) = -dt / 10                    We find the derivative

= Ln(5000 – Q(t)) = -t / 10 + k                            K is a constant

= 5000 – Q(t) = Ce^(-t / 10)          C is any constant, but in this problem we need                                                                          to find  C

Plug in when t = 0 Q(0) = 250

250 = 500 – Ce^(0/10)

250 = 500 – C

C = 4750

Q(t) = 5000 – 4750e^(-t/10)

Part B

Limt–> ∞ Q(t) 5000 – 4750e^(-t/10) = 5000

It will go to 5000 because as t gets larger, the whole number gets smaller because it is a negative exponent, so 5000 – 0 is 5000.