Question

A patient has an illness that typically lasts about 24 hours. The temperature, T, in degrees Fahrenheit, of the patient t hours after the illness begins is given by:T(t) = -0.02t^2 +0.4968 +98.
a. When does the patient's temperature reach its maximum value
b. What is the patient's maximum temperature during illness

Answer 1

Step-by-step explanation:

Given that,

The temperature, T, in degrees Fahrenheit, of the patient t hours after the illness begins is given by:

$T(t)=-0.02t^2 +0.4968 t+98$ ...(1)

(a) We need to find when does the patient's temperature reach its maximum value.

For maximum value.

Put dT(t)/dt = 0

$\dfrac{d(T(t)}{dt}=\dfrac{d}{dt}(-0.02t^2 +0.4968t +98)\\\\=-0.04t+0.4968\\\\-0.04t+0.4968=0\\\\t=12.42\ h$

(b) Put t = 12.42 in equation (1) :

$T(12.42)=-0.02(12.42)^2 +0.4968 (12.42)+98\\\\=101.08^{\circ} F$

Hence, this is the required solution.