P '(t) = 4t^3 - 16t

0 = 4t^3 - 16t

0 = 4t(t^2 - 4)

0 = 4t(t + 2)(t - 2)

t = 0 or t = -2, or t = 2

Discard t = -2 because a negative time makes no sense, leaving just

t = 0 and t = 2

P ''(t) = 12t^2 - 16

P ''(0) = -16

P ''(2) = 12(2^2) - 16 = 32 > 0 so a local minimum occurs at t = 2.

P(0) = 0^4 - 8(0^2) + 18 = 18

Check the other endpoint of the domain, namely t = 3.

P(3) = 3^4 - 8(3^2) + 18 = 81 - 72 + 18 = 27

A global maximum of 27 occurs at t = 3.

Hint: Take your time and read the question. It has nothing to do with calculus. 

Its a plain ol' highschool question. Once you realize that, you can then solve it on your own. E.g. What are the max & min of x² - 8x + 18, for x = 0 to 9? 

Do you see now that your question is a 'simple' quadratic/boundary/vertex queston? 

Moral: don't be intimidated by college/university math questions. 99% of the questions are actually highschool algebra questions! Just take your time to understand what they're actually asking! 

t=3, P(3)= 27                 

P(t)= t^4 - 8t^2 + 18

A virus is represented by a function of 

the number of people infected in a hotel after hours. 

You measure the infected population for the first 3 hour

P(3) = 99 - 72 = 27

At what hour is the most number of people infected 

and how many people are infected at that time?

4t^3 - 16t + 18