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Question

the temperature, T of a person during an illness is given by T(t) = -.1t^2 + 1.2t + 98; for 0 is less than or equal to t which is less than or equal to 12 where T is the temp, (F) at time t, in days, find the maximum value of the tempand when i occurs

i have no idea where to begin

Answer 1

dT / dt = -0.2t + 1.2.
d^2T / dt^2 = -0.2
The maximum or minimum occurs when dT/dt = 0.
-0.2t + 1.2 = 0
t = 1.2 / 0.2 = 6 days.
As d^2T / dt^2 < 0, the temperature at time t = 6 days is a maximum.
Putting t = 6 in the original formula, the temperature at that time is:
-0.1*36 + 1.2*6 + 98
= -3.6 + 7.2 + 98
= 101.6degF.

Answer 2

find the initial value first, that is at time 0 or t=0 little t is time and big T is temp.
T(0) = -.1(0)^2+1.2(0)+98= 98 so the ill person at the beginning (time=0) was at a healthy 98 deg F.
so plug in for t the number of days the illness has lasted i.e on the 4th day the patient's temp will be
T(4)= -.1(4)^2+1.2(4)+98= -1.6 + 4.8 + 98= 101.2 deg F
and so on.
after the 12th day there is no need to calculate T at t=13 because the person will have expired.