ny fixed number n representing the perimeter of a rectangle, what are the dimensions of ectangle that maximizes the area of that rectangle?
ny fixed number n representing the perimeter of a rectangle, what are the dimensions of ectangle that maximizes the area of that rectangle?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![**Question 8:**
For any fixed number \( n \) representing the perimeter of a rectangle, what are the dimensions of the rectangle that maximizes the area of that rectangle?
**Explanation:**
To find the dimensions of the rectangle that maximizes the area given a fixed perimeter \( n \), we need to use the formula for the perimeter of a rectangle, \( P = 2(l + w) \), where \( l \) is the length and \( w \) is the width.
Rearranging the formula gives \( l + w = \frac{n}{2} \).
The area \( A \) of a rectangle is given by \( A = l \times w \).
To maximize the area, substitute \( w = \frac{n}{2} - l \) into the area formula:
\[ A = l \times \left(\frac{n}{2} - l\right) \]
\[ A = \frac{n}{2}l - l^2 \]
This quadratic equation can be simplified as:
\[ A = -l^2 + \frac{n}{2}l \]
To find the value of \( l \) that maximizes the area, we take the derivative of \( A \) with respect to \( l \) and set it to zero:
\[ \frac{dA}{dl} = -2l + \frac{n}{2} = 0 \]
Solving for \( l \), we get:
\[ l = \frac{n}{4} \]
Since \( l = w \) when the area is maximized, the rectangle becomes a square. Therefore, the dimensions of the rectangle that maximize the area are \( \frac{n}{4} \times \frac{n}{4} \).
**Conclusion:**
For a given perimeter \( n \), the dimensions of the rectangle that maximize the area are those of a square with sides each measuring \( \frac{n}{4} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F668826ef-a7b0-4206-bae0-a08cac22bcd4%2F1d1abbc5-1b89-4862-8af9-c56220a57585%2Fkjbobv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 8:**
For any fixed number \( n \) representing the perimeter of a rectangle, what are the dimensions of the rectangle that maximizes the area of that rectangle?
**Explanation:**
To find the dimensions of the rectangle that maximizes the area given a fixed perimeter \( n \), we need to use the formula for the perimeter of a rectangle, \( P = 2(l + w) \), where \( l \) is the length and \( w \) is the width.
Rearranging the formula gives \( l + w = \frac{n}{2} \).
The area \( A \) of a rectangle is given by \( A = l \times w \).
To maximize the area, substitute \( w = \frac{n}{2} - l \) into the area formula:
\[ A = l \times \left(\frac{n}{2} - l\right) \]
\[ A = \frac{n}{2}l - l^2 \]
This quadratic equation can be simplified as:
\[ A = -l^2 + \frac{n}{2}l \]
To find the value of \( l \) that maximizes the area, we take the derivative of \( A \) with respect to \( l \) and set it to zero:
\[ \frac{dA}{dl} = -2l + \frac{n}{2} = 0 \]
Solving for \( l \), we get:
\[ l = \frac{n}{4} \]
Since \( l = w \) when the area is maximized, the rectangle becomes a square. Therefore, the dimensions of the rectangle that maximize the area are \( \frac{n}{4} \times \frac{n}{4} \).
**Conclusion:**
For a given perimeter \( n \), the dimensions of the rectangle that maximize the area are those of a square with sides each measuring \( \frac{n}{4} \).
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