Question

A candymaker 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 time t.A) Find Q(t) for t > 0 if the candymaker has 250 pounds of candy at t=0.B) Find lim Q(t) t-> infinity.

Answer 1

The function that describes the amount of candy Q(t) over time, given the production and consumption rates and the initial condition, is Q(t) = 5000 - 4750e^(-t/10). As time goes to infinity, the amount of candy approaches 5000 pounds.

Step-by-step explanation:

The student's question involves a differential equation that models a candymaker's candy production and consumption rates. Given that the candymaker makes 500 pounds of candy per week and the family eats at a rate of Q(t)/10 pounds per week, where Q(t) represents the amount of candy at time t, we need to establish a function Q(t) that reflects the candy amount over time.

To find Q(t), we set up the differential equation as follows:

dQ/dt = 500 - Q(t)/10

Since the initial condition is Q(0) = 250, we can solve this first-order linear differential equation using an integrating factor or separation of variables. The general solution will take the form:

Q(t) = Ce^(-t/10) + 5000

Using the initial condition to solve for C, we have:

250 = Ce^(0) + 5000

C = -4750

Therefore, the particular solution for Q(t) is:

Q(t) = 5000 - 4750e^(-t/10)

For part B, to find the limit of Q(t) as t approaches infinity, we evaluate:

lim Q(t) as t -> infinity = 5000 - 4750e^(-infinity)

As t approaches infinity, e^(-t/10) approaches 0, resulting in:

lim Q(t) as t -> infinity = 5000

This indicates that over time, the amount of candy will approach a steady state of 5000 pounds.