Answer :
To find the lowest temperature using the given function [tex]\( T(x) = 0.264x^2 - 4.752x + 81 \)[/tex] over the interval [tex]\( 0 \leq x \leq 18 \)[/tex], we'll follow these steps:
1. Understand the Function and Interval:
- The temperature function [tex]\( T(x) \)[/tex] is quadratic with coefficients [tex]\( a = 0.264 \)[/tex], [tex]\( b = -4.752 \)[/tex], and [tex]\( c = 81 \)[/tex].
- The interval for [tex]\( x \)[/tex] is from [tex]\( 0 \)[/tex] to [tex]\( 18 \)[/tex], representing hours after 5 PM.
2. Calculate the Vertex of the Quadratic Function:
- The vertex of a parabola given by [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula [tex]\( x = -\frac{b}{2a} \)[/tex].
- Plugging in the values from our function:
[tex]\[
x = -\frac{-4.752}{2 \cdot 0.264} = \frac{4.752}{0.528} \approx 9
\][/tex]
- So, the vertex occurs at [tex]\( x \approx 9 \)[/tex].
3. Evaluate the Temperature at Critical Points:
- We need to evaluate [tex]\( T(x) \)[/tex] at the vertex and at the endpoints of the interval [tex]\( x = 0 \)[/tex] and [tex]\( x = 18 \)[/tex].
- Calculate [tex]\( T(0) \)[/tex]:
[tex]\[
T(0) = 0.264 \cdot 0^2 - 4.752 \cdot 0 + 81 = 81
\][/tex]
- Calculate [tex]\( T(18) \)[/tex]:
[tex]\[
T(18) = 0.264 \cdot 18^2 - 4.752 \cdot 18 + 81 = 0.264 \cdot 324 - 85.536 + 81 \approx 85.536 - 85.536 + 81 = 81
\][/tex]
- Calculate [tex]\( T(9) \)[/tex], the vertex point:
[tex]\[
T(9) = 0.264 \cdot 9^2 - 4.752 \cdot 9 + 81 = 0.264 \cdot 81 - 42.768 + 81 = 21.384 - 42.768 + 81 = -21.384 + 81 \approx 59.616
\][/tex]
4. Find the Minimum Temperature:
- Compare the temperature values at [tex]\( x = 0 \)[/tex], [tex]\( x = 18 \)[/tex], and [tex]\( x = 9 \)[/tex]:
[tex]\[
T(0) = 81, \quad T(18) = 81, \quad T(9) \approx 59.616
\][/tex]
- The lowest temperature observed is 59.616 degrees Fahrenheit at [tex]\( x = 9 \)[/tex].
5. Round the Lowest Temperature to the Nearest Whole Degree:
- Rounding 59.616 to the nearest whole number gives us 60 degrees Fahrenheit.
Given the possible choices:
- a. 72 degrees
- b. 66 degrees
- c. 64 degrees
- d. 60 degrees
The correct answer is 60 degrees. Thus, the nearest whole degree for the lowest temperature reached is:
[tex]\[
\boxed{60}
\][/tex]
1. Understand the Function and Interval:
- The temperature function [tex]\( T(x) \)[/tex] is quadratic with coefficients [tex]\( a = 0.264 \)[/tex], [tex]\( b = -4.752 \)[/tex], and [tex]\( c = 81 \)[/tex].
- The interval for [tex]\( x \)[/tex] is from [tex]\( 0 \)[/tex] to [tex]\( 18 \)[/tex], representing hours after 5 PM.
2. Calculate the Vertex of the Quadratic Function:
- The vertex of a parabola given by [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula [tex]\( x = -\frac{b}{2a} \)[/tex].
- Plugging in the values from our function:
[tex]\[
x = -\frac{-4.752}{2 \cdot 0.264} = \frac{4.752}{0.528} \approx 9
\][/tex]
- So, the vertex occurs at [tex]\( x \approx 9 \)[/tex].
3. Evaluate the Temperature at Critical Points:
- We need to evaluate [tex]\( T(x) \)[/tex] at the vertex and at the endpoints of the interval [tex]\( x = 0 \)[/tex] and [tex]\( x = 18 \)[/tex].
- Calculate [tex]\( T(0) \)[/tex]:
[tex]\[
T(0) = 0.264 \cdot 0^2 - 4.752 \cdot 0 + 81 = 81
\][/tex]
- Calculate [tex]\( T(18) \)[/tex]:
[tex]\[
T(18) = 0.264 \cdot 18^2 - 4.752 \cdot 18 + 81 = 0.264 \cdot 324 - 85.536 + 81 \approx 85.536 - 85.536 + 81 = 81
\][/tex]
- Calculate [tex]\( T(9) \)[/tex], the vertex point:
[tex]\[
T(9) = 0.264 \cdot 9^2 - 4.752 \cdot 9 + 81 = 0.264 \cdot 81 - 42.768 + 81 = 21.384 - 42.768 + 81 = -21.384 + 81 \approx 59.616
\][/tex]
4. Find the Minimum Temperature:
- Compare the temperature values at [tex]\( x = 0 \)[/tex], [tex]\( x = 18 \)[/tex], and [tex]\( x = 9 \)[/tex]:
[tex]\[
T(0) = 81, \quad T(18) = 81, \quad T(9) \approx 59.616
\][/tex]
- The lowest temperature observed is 59.616 degrees Fahrenheit at [tex]\( x = 9 \)[/tex].
5. Round the Lowest Temperature to the Nearest Whole Degree:
- Rounding 59.616 to the nearest whole number gives us 60 degrees Fahrenheit.
Given the possible choices:
- a. 72 degrees
- b. 66 degrees
- c. 64 degrees
- d. 60 degrees
The correct answer is 60 degrees. Thus, the nearest whole degree for the lowest temperature reached is:
[tex]\[
\boxed{60}
\][/tex]
