Answer :
To determine when the patient's temperature reaches its maximum value and what that maximum temperature is, we need to analyze the given quadratic function for temperature:
[tex]\[ T(t) = -0.025 t^2 + 0.545 t + 98.2 \][/tex]
### Step-by-Step Solution:
Step 1: Identify the type of function
This is a quadratic function of the form [tex]\( T(t) = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -0.025 \)[/tex]
- [tex]\( b = 0.545 \)[/tex]
- [tex]\( c = 98.2 \)[/tex]
Since the coefficient [tex]\( a \)[/tex] is negative, this parabola opens downwards, meaning it has a maximum value.
Step 2: Find the vertex of the parabola
The maximum or minimum of a quadratic function [tex]\( T(t) = at^2 + bt + c \)[/tex] occurs at the vertex. The time [tex]\( t \)[/tex] at which the vertex occurs can be found using the formula:
[tex]\[ t_{max} = -\frac{b}{2a} \][/tex]
Step 3: Calculate [tex]\( t_{max} \)[/tex]
Given:
[tex]\[ a = -0.025 \][/tex]
[tex]\[ b = 0.545 \][/tex]
Substitute these values into the formula:
[tex]\[ t_{max} = -\frac{0.545}{2(-0.025)} \][/tex]
[tex]\[ t_{max} = -\frac{0.545}{-0.05} \][/tex]
[tex]\[ t_{max} = \frac{0.545}{0.05} \][/tex]
[tex]\[ t_{max} = 10.9 \][/tex]
So, the patient's temperature reaches its maximum value 10.9 hours after the illness begins.
Step 4: Find the maximum temperature
To find the maximum temperature, substitute [tex]\( t_{max} \)[/tex] back into the original function:
[tex]\[ T(10.9) = -0.025(10.9)^2 + 0.545(10.9) + 98.2 \][/tex]
First, calculate [tex]\( (10.9)^2 \)[/tex]:
[tex]\[ (10.9)^2 = 118.81 \][/tex]
Then, calculate each term:
[tex]\[ -0.025 \times 118.81 = -2.97025 \][/tex]
[tex]\[ 0.545 \times 10.9 = 5.9405 \][/tex]
Now sum these values with the constant term 98.2:
[tex]\[ T(10.9) = -2.97025 + 5.9405 + 98.2 \][/tex]
[tex]\[ T(10.9) = 101.17025 \][/tex]
Rounded to one decimal place:
[tex]\[ T(10.9) = 101.2 \][/tex]
### Answer:
- The patient's temperature reaches its maximum value after 10.9 hours.
- The patient's maximum temperature during the illness is 101.2°F.
[tex]\[ T(t) = -0.025 t^2 + 0.545 t + 98.2 \][/tex]
### Step-by-Step Solution:
Step 1: Identify the type of function
This is a quadratic function of the form [tex]\( T(t) = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -0.025 \)[/tex]
- [tex]\( b = 0.545 \)[/tex]
- [tex]\( c = 98.2 \)[/tex]
Since the coefficient [tex]\( a \)[/tex] is negative, this parabola opens downwards, meaning it has a maximum value.
Step 2: Find the vertex of the parabola
The maximum or minimum of a quadratic function [tex]\( T(t) = at^2 + bt + c \)[/tex] occurs at the vertex. The time [tex]\( t \)[/tex] at which the vertex occurs can be found using the formula:
[tex]\[ t_{max} = -\frac{b}{2a} \][/tex]
Step 3: Calculate [tex]\( t_{max} \)[/tex]
Given:
[tex]\[ a = -0.025 \][/tex]
[tex]\[ b = 0.545 \][/tex]
Substitute these values into the formula:
[tex]\[ t_{max} = -\frac{0.545}{2(-0.025)} \][/tex]
[tex]\[ t_{max} = -\frac{0.545}{-0.05} \][/tex]
[tex]\[ t_{max} = \frac{0.545}{0.05} \][/tex]
[tex]\[ t_{max} = 10.9 \][/tex]
So, the patient's temperature reaches its maximum value 10.9 hours after the illness begins.
Step 4: Find the maximum temperature
To find the maximum temperature, substitute [tex]\( t_{max} \)[/tex] back into the original function:
[tex]\[ T(10.9) = -0.025(10.9)^2 + 0.545(10.9) + 98.2 \][/tex]
First, calculate [tex]\( (10.9)^2 \)[/tex]:
[tex]\[ (10.9)^2 = 118.81 \][/tex]
Then, calculate each term:
[tex]\[ -0.025 \times 118.81 = -2.97025 \][/tex]
[tex]\[ 0.545 \times 10.9 = 5.9405 \][/tex]
Now sum these values with the constant term 98.2:
[tex]\[ T(10.9) = -2.97025 + 5.9405 + 98.2 \][/tex]
[tex]\[ T(10.9) = 101.17025 \][/tex]
Rounded to one decimal place:
[tex]\[ T(10.9) = 101.2 \][/tex]
### Answer:
- The patient's temperature reaches its maximum value after 10.9 hours.
- The patient's maximum temperature during the illness is 101.2°F.
