For each of the following, set up either a system of equations in two variables or a quadratic equation and solve. (Round to two decimal places if necessary.) 10. Landscaping The perimeter of a rectangular field is 750 ft. If the length is four times the width, find the dimensions of the field.
For each of the following, set up either a system of equations in two variables or a quadratic equation and solve. (Round to two decimal places if necessary.) 10. Landscaping The perimeter of a rectangular field is 750 ft. If the length is four times the width, find the dimensions of the field.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
For each of the following, set up either a system of equations in two variables or a quadratic equation and solve. (Round to two decimal places if necessary.)
**10. Landscaping:**
The perimeter of a rectangular field is 750 ft. If the length is four times the width, find the dimensions of the field.
---
**Solution Explanation:**
1. **Define Variables:**
- Let \( w \) be the width of the field.
- Let \( l \) be the length of the field.
2. **Equations Based on the Problem:**
- The formula for the perimeter \( P \) of a rectangle is given by:
\[
P = 2l + 2w
\]
- According to the problem, the length is four times the width:
\[
l = 4w
\]
- The perimeter is given as 750 ft:
\[
2l + 2w = 750
\]
3. **Substitute and Solve:**
- Substitute \( l = 4w \) into the perimeter equation:
\[
2(4w) + 2w = 750
\]
- Simplify and solve for \( w \):
\[
8w + 2w = 750
\]
\[
10w = 750
\]
\[
w = 75
\]
4. **Find Length:**
- Use the relationship \( l = 4w \) to find \( l \):
\[
l = 4(75) = 300
\]
5. **Conclusion:**
- The dimensions of the field are:
- Width = 75 ft
- Length = 300 ft
This structured approach allows you to solve for the dimensions of the field effectively using algebraic manipulation and substitution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8cd8c204-d1e8-4e49-9751-601d95aaf93f%2Fc5362a03-4467-4685-afb0-cfc06a494564%2Fgh4lly6_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
For each of the following, set up either a system of equations in two variables or a quadratic equation and solve. (Round to two decimal places if necessary.)
**10. Landscaping:**
The perimeter of a rectangular field is 750 ft. If the length is four times the width, find the dimensions of the field.
---
**Solution Explanation:**
1. **Define Variables:**
- Let \( w \) be the width of the field.
- Let \( l \) be the length of the field.
2. **Equations Based on the Problem:**
- The formula for the perimeter \( P \) of a rectangle is given by:
\[
P = 2l + 2w
\]
- According to the problem, the length is four times the width:
\[
l = 4w
\]
- The perimeter is given as 750 ft:
\[
2l + 2w = 750
\]
3. **Substitute and Solve:**
- Substitute \( l = 4w \) into the perimeter equation:
\[
2(4w) + 2w = 750
\]
- Simplify and solve for \( w \):
\[
8w + 2w = 750
\]
\[
10w = 750
\]
\[
w = 75
\]
4. **Find Length:**
- Use the relationship \( l = 4w \) to find \( l \):
\[
l = 4(75) = 300
\]
5. **Conclusion:**
- The dimensions of the field are:
- Width = 75 ft
- Length = 300 ft
This structured approach allows you to solve for the dimensions of the field effectively using algebraic manipulation and substitution.
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