Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 3: Problem 8
Show that for a rectangle of given perimeter \(K\) the one with maximum area is a square.
Short Answer
Expert verified
Among rectangles with a fixed perimeter, a square has the maximum area.
Step by step solution
01
Understand the Problem
We need to show that, among all rectangles with a given perimeter \( K \), the rectangle with the maximum area is a square.
02
Set up the Relations
Let the length and width of the rectangle be \( l \) and \( w \), respectively. The perimeter is given by the formula \( P = 2l + 2w = K \). Therefore, \( w = \frac{K}{2} - l \).
03
Express the Area
The area \( A \) of the rectangle is given by \( A = l \times w \). Substitute for \( w \) from the previous step: \[ A = l \left( \frac{K}{2} - l \right) = \frac{K}{2}l - l^2. \]
04
Find the Maximum Area
We need to maximize \( A = \frac{K}{2}l - l^2 \). This can be done by finding the derivative \( \frac{dA}{dl} = \frac{K}{2} - 2l \). Set \( \frac{dA}{dl} = 0 \) to find the critical points.Thus, \( \frac{K}{2} - 2l = 0 \) implies \( l = \frac{K}{4} \).
05
Confirm it's a Maximum with the Second Derivative Test
Calculate the second derivative \( \frac{d^2A}{dl^2} = -2 \). Since \( \frac{d^2A}{dl^2} < 0 \), the function has a maximum at \( l = \frac{K}{4} \).
06
Conclude the Shape
Since \( l = \frac{K}{4} \) and \( w = \frac{K}{2} - l = \frac{K}{4} \), the rectangle is a square with side \( \frac{K}{4} \). Thus, a square has the maximum area for a given perimeter.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Perimeter
In geometry, the perimeter of a shape is the total distance around it. For a rectangle, the perimeter is calculated by the formula \( P = 2l + 2w \), where \( l \) is the length and \( w \) is the width. This formula is crucial when determining the relationship between the sides of a rectangle, especially when dealing with a fixed perimeter.
When you fix a perimeter \( K \), you can express one dimension in terms of the other, like we did with the width \( w = \frac{K}{2} - l \). This allows us to substitute into other equations, such as the area, to explore different properties and optimize certain quantities.
When you fix a perimeter \( K \), you can express one dimension in terms of the other, like we did with the width \( w = \frac{K}{2} - l \). This allows us to substitute into other equations, such as the area, to explore different properties and optimize certain quantities.
- Keeping the perimeter constant while changing other dimensions helps in examining how different shapes, like rectangles, translate into maximum or minimum areas.
- This property is widely utilized in optimization problems, allowing us to find extreme values 鈥� in this case, maximum area.
Critical Points
Critical points are values that give potential maximum or minimum values of a function. In optimization, finding critical points is the first step in determining where these extreme values occur.
For the rectangle's area function \( A = \frac{K}{2}l - l^2 \), we took the derivative \( \frac{dA}{dl} \), which resulted in \( \frac{dA}{dl} = \frac{K}{2} - 2l \). Setting this derivative to zero helps us find the critical points of the function, which are the values of \( l \) where the behavior of the function changes.
For the rectangle's area function \( A = \frac{K}{2}l - l^2 \), we took the derivative \( \frac{dA}{dl} \), which resulted in \( \frac{dA}{dl} = \frac{K}{2} - 2l \). Setting this derivative to zero helps us find the critical points of the function, which are the values of \( l \) where the behavior of the function changes.
- In this exercise, solving \( \frac{K}{2} - 2l = 0 \) allowed us to find the critical point \( l = \frac{K}{4} \).
- Finding this critical point is vital because it pinpoints where we should check for maximum or minimum areas of rectangles with a given perimeter.
Second Derivative Test
The second derivative test is used to verify if a critical point found is a maximum, minimum, or a point of inflection. By looking at the second derivative of a function, we can determine the concavity at a point.
For the area function of the rectangle \( A = \frac{K}{2}l - l^2 \), the second derivative is \( \frac{d^2A}{dl^2} = -2 \). The fact that this second derivative is negative indicates that the function is concave down at the critical point \( l = \frac{K}{4} \). This means the area has a maximum at this length.
For the area function of the rectangle \( A = \frac{K}{2}l - l^2 \), the second derivative is \( \frac{d^2A}{dl^2} = -2 \). The fact that this second derivative is negative indicates that the function is concave down at the critical point \( l = \frac{K}{4} \). This means the area has a maximum at this length.
- A negative second derivative implies a local maximum, which means our critical point is indeed providing the rectangle with the largest possible area given the perimeter \( K \).
- The second derivative test helps confirm that the rectangle indeed takes the form of a square, proving the shape for maximum area with fixed perimeter.
