Homework Help: Calculus

Posted by Jasmin on Wednesday, March 30, 2011 at 5:55am.

An epidemiologist determines that a particular epidemic spreads in such a way that t weeks after the outbreak, N hundred new cases will be reported where N(t)= 5t/(12+t*2). At what time is the epidemic at its worst? What is the maximum number of reported new cases?

Calculus - drwls, Wednesday, March 30, 2011 at 6:27am
The epidemic is a maximum when
dN/dt = 0

dN/dt = [-5t*(2t) + 5(12 + t^2)]/(12 + t^2) = 0
That will be zero when the numerator is zero, so
-10t^2 + 60 + 5 t^2 = 0
t^2 = 12
t = 3.46 weeks

N(3.46) = 17.3/23.97 = 0.748 hundred new cases (75 cases)

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