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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![**Optimization Problem: Maximizing the Product \( xy^2 \)**
**Problem Statement:**
The sum of two positive real numbers \( x \) and \( y \) is 12. Find \( x \) and \( y \) among these such that the product \( xy^2 \) is maximal.
**Input Fields:**
* \( x = \) ____________
* \( y = \) ____________
---
**Solution Outline:**
To solve this optimization problem, we need to:
1. **Express the Constraint:** Given that \( x + y = 12 \), we can express one variable in terms of the other.
2. **Formulate the Objective Function:** The objective function to be maximized is \( xy^2 \).
3. **Substitute and Simplify:** Substitute the expressed variable into the objective function.
4. **Differentiate and Find Critical Points:** Use calculus to find the maximum value.
---
**Example Solution Approach Using Calculus:**
1. **Express the constraint:**
\[ y = 12 - x \]
2. **Formulate the objective function:**
\[ P = xy^2 \]
Substitute \( y \) from the constraint:
\[ P = x(12 - x)^2 \]
3. **Expand and simplify:**
\[ P = x(144 - 24x + x^2) \]
\[ P = 144x - 24x^2 + x^3 \]
4. **Differentiate with respect to \( x \):**
\[ \frac{dP}{dx} = 144 - 48x + 3x^2 \]
5. **Find the critical points:**
Solve \( \frac{dP}{dx} = 0 \):
\[ 3x^2 - 48x + 144 = 0 \]
\[ x^2 - 16x + 48 = 0 \]
6. **Solve the quadratic equation:**
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we get:
\[ x = \frac{16 \pm \sqrt{256 - 192}}{2} \]
\[ x = \frac{16 \pm \sqrt{64}}{2} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc0c43ed0-a854-4f55-a30b-147647d0ed33%2F6c42adfb-09f3-4157-b8e7-bde6d8f023f8%2Fodqxy5k_processed.png&w=3840&q=75)
Transcribed Image Text:**Optimization Problem: Maximizing the Product \( xy^2 \)**
**Problem Statement:**
The sum of two positive real numbers \( x \) and \( y \) is 12. Find \( x \) and \( y \) among these such that the product \( xy^2 \) is maximal.
**Input Fields:**
* \( x = \) ____________
* \( y = \) ____________
---
**Solution Outline:**
To solve this optimization problem, we need to:
1. **Express the Constraint:** Given that \( x + y = 12 \), we can express one variable in terms of the other.
2. **Formulate the Objective Function:** The objective function to be maximized is \( xy^2 \).
3. **Substitute and Simplify:** Substitute the expressed variable into the objective function.
4. **Differentiate and Find Critical Points:** Use calculus to find the maximum value.
---
**Example Solution Approach Using Calculus:**
1. **Express the constraint:**
\[ y = 12 - x \]
2. **Formulate the objective function:**
\[ P = xy^2 \]
Substitute \( y \) from the constraint:
\[ P = x(12 - x)^2 \]
3. **Expand and simplify:**
\[ P = x(144 - 24x + x^2) \]
\[ P = 144x - 24x^2 + x^3 \]
4. **Differentiate with respect to \( x \):**
\[ \frac{dP}{dx} = 144 - 48x + 3x^2 \]
5. **Find the critical points:**
Solve \( \frac{dP}{dx} = 0 \):
\[ 3x^2 - 48x + 144 = 0 \]
\[ x^2 - 16x + 48 = 0 \]
6. **Solve the quadratic equation:**
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we get:
\[ x = \frac{16 \pm \sqrt{256 - 192}}{2} \]
\[ x = \frac{16 \pm \sqrt{64}}{2} \]
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