The height of a certain amusement park ride follows the function h(t) = −³+ 11t² — 32t + 28, where his measured in meters and t is measured in minutes. The ride lasts 7 minutes, so 0 ≤ t ≤ 7. Using Calculus, find the maximum height of the ride as well as what time it reaches that height. Give your answers in exact form, and make sure to check all possible times that the maximum could occur.
The height of a certain amusement park ride follows the function h(t) = −³+ 11t² — 32t + 28, where his measured in meters and t is measured in minutes. The ride lasts 7 minutes, so 0 ≤ t ≤ 7. Using Calculus, find the maximum height of the ride as well as what time it reaches that height. Give your answers in exact form, and make sure to check all possible times that the maximum could occur.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Mathematics Education Resource**
**Topic: Calculus - Finding Maximum Heights**
**Problem Statement:**
The height of a certain amusement park ride follows the function \( h(t) = -t^3 + 11t^2 - 32t + 28 \), where \( h \) is measured in meters and \( t \) is measured in minutes. The ride lasts 7 minutes, so \( 0 \leq t \leq 7 \).
Using Calculus, find the maximum height of the ride as well as the time it reaches that height. Give your answers in **exact form**, and make sure to check **all possible times that the maximum could occur**.
**Solution Approach:**
1. **Function Analysis:**
The given function is \( h(t) = -t^3 + 11t^2 - 32t + 28 \).
2. **First Derivative:**
To find the critical points, we first need the derivative of the function:
\[
h'(t) = \frac{d}{dt}(-t^3 + 11t^2 - 32t + 28)
= -3t^2 + 22t - 32
\]
3. **Finding Critical Points:**
Set the first derivative equal to zero to find the critical points:
\[
-3t^2 + 22t - 32 = 0
\]
4. **Solving for \( t \):**
Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
t = \frac{-22 \pm \sqrt{(22)^2 - 4(-3)(-32)}}{2(-3)}
\]
5. **Second Derivative Test:**
To verify whether these critical points give a maximum or minimum, use the second derivative:
\[
h''(t) = \frac{d}{dt}(-3t^2 + 22t - 32)
= -6t + 22
\]
6. **Boundary Analysis:**
Evaluate \( h(t) \) at the endpoints \( t = 0 \) and \( t = 7 \), as the maximum could occur at these points.
**Note:**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F56488219-115f-4b7a-9fca-52b6982bead0%2Fae2e4233-a6c0-495b-8fae-be158d3bf643%2Fp0dne0si_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Mathematics Education Resource**
**Topic: Calculus - Finding Maximum Heights**
**Problem Statement:**
The height of a certain amusement park ride follows the function \( h(t) = -t^3 + 11t^2 - 32t + 28 \), where \( h \) is measured in meters and \( t \) is measured in minutes. The ride lasts 7 minutes, so \( 0 \leq t \leq 7 \).
Using Calculus, find the maximum height of the ride as well as the time it reaches that height. Give your answers in **exact form**, and make sure to check **all possible times that the maximum could occur**.
**Solution Approach:**
1. **Function Analysis:**
The given function is \( h(t) = -t^3 + 11t^2 - 32t + 28 \).
2. **First Derivative:**
To find the critical points, we first need the derivative of the function:
\[
h'(t) = \frac{d}{dt}(-t^3 + 11t^2 - 32t + 28)
= -3t^2 + 22t - 32
\]
3. **Finding Critical Points:**
Set the first derivative equal to zero to find the critical points:
\[
-3t^2 + 22t - 32 = 0
\]
4. **Solving for \( t \):**
Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
t = \frac{-22 \pm \sqrt{(22)^2 - 4(-3)(-32)}}{2(-3)}
\]
5. **Second Derivative Test:**
To verify whether these critical points give a maximum or minimum, use the second derivative:
\[
h''(t) = \frac{d}{dt}(-3t^2 + 22t - 32)
= -6t + 22
\]
6. **Boundary Analysis:**
Evaluate \( h(t) \) at the endpoints \( t = 0 \) and \( t = 7 \), as the maximum could occur at these points.
**Note:**
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