Math help?

Question

I am having trouble with this math problem in calculus:
The CDC has discovered a new highly contagious disease. They estimate that t days after the disease is first observed in a community, the percent of the population infected by the disease is approximated by P(t) = 20t^3 - t^4 / 1000 for 0 less than or equal to t less than or equal to 20. When will the population reach maximum infection? What percent of the population will be infected?

Answer 1

Take the first derivative of your function:

P(t) = (20t^3 - t^4) / 1000

P'(t) = (60t^2 - 4t^3) /1000

Now set this to zero to find the local maxima and minima:
0 = (60t^2 - 4t^3) /1000

Factor out a t²:
0 = t²(60 - 4t) / 1000

You can multiply both sides by 1000 to simplify the equation:
0 = t²(60 - 4t)

This will be zero when t² is zero, or when the expression in the parentheses is zero.

Case 1:
t² = 0
t = 0

Case 2:
60 - 4t = 0
60 = 4t
t = 15

On day t = 0 (0 days) the spread will be a minimum. (Confirm this by calculating P(0) which will be 0%).

On day t = 15 (15 days) the spread will be a maximum.
Just plug in t = 15 to get the percentage.

P(15) = (20(15)^3 - (15)^4) / 1000
P(15) = (67500 - 50625) / 1000
P(15) = 16875 / 1000
P(15) = 16.875%

After 15 days, the infection will reach a maximum of 16.875% of the population.

Answer 2

Well, there must be a mistake. The function
P=20*t^3 - t^4/1000
does have a maximum, but it is at t = 15000, much larger than 20...

However, the function
P=20*t^3 - t^4
has a maximum at t=15. The way to find it is to take the derivative of P with respect to t, to obtain
dP/dt=60*t^2 - 4*t^3
and set it to zero. Of course you are interested in t>0, so divide everything by t^2 to obtain
60 - 4*t = 0 --> t = 15 @ maximum.

Hope this helped.

--Darth Maria