Chapter 4: Problem 8
Show that for a rectangle of given perimeter \(K\) the one with maximum area is a square.
Short Answer
Expert verified
A rectangle with maximum area for a given perimeter is a square.
Step by step solution
01
Express the Perimeter Equation
We know that the perimeter of a rectangle is given by the equation:\[ P = 2(l + w) \]where \( l \) is the length and \( w \) is the width of the rectangle. Since the perimeter is given as \( K \), we write:\[ K = 2(l + w) \].
02
Solve for One Variable in Terms of the Other
From the perimeter equation, solve for \( w \) in terms of \( l \) (or vice versa):\[ l + w = \frac{K}{2} \]\[ w = \frac{K}{2} - l \].
03
Express the Area Function
The area \( A \) of the rectangle can be written as:\[ A = l imes w \]Substituting for \( w \) from Step 2, we get:\[ A = l \left(\frac{K}{2} - l\right) \]\[ A = \frac{K}{2}l - l^2 \].
04
Find the Vertex of the Quadratic
The equation \( A = \frac{K}{2}l - l^2 \) is a quadratic function in terms of \( l \). The maximum area occurs at the vertex of this parabola.To find the vertex, use the formula \( l = -\frac{b}{2a} \) for the quadratic equation \( A = al^2 + bl + c \), where \( a = -1 \) and \( b = \frac{K}{2} \):\[ l = -\frac{\frac{K}{2}}{2(-1)} = \frac{K}{4} \].
05
Conclude the Shape of the Rectangle
When \( l = \frac{K}{4} \), substitute back into the perimeter equation to find \( w \):\[ w = \frac{K}{2} - \frac{K}{4} = \frac{K}{4} \].Since both \( l \) and \( w \) are equal, the rectangle is a square, which confirms that the maximum area for a given perimeter is achieved when the rectangle is a square.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rectangle Perimeter
The perimeter of a rectangle offers a straightforward way to understand its boundary outline. Perimeter measures the total distance around a rectangle. For a standard rectangle, the perimeter equation is given by:
- \( P = 2(l + w) \)
- \( K = 2(l + w) \)
Quadratic Function
Quadratic functions come into play when dealing with equations related to maxima or minima, such as this area problem for a rectangle. These functions are of the form:
- \( f(x) = ax^2 + bx + c \)
- \( A = \frac{K}{2}l - l^2 \)
Area Optimization
Area optimization focuses on adjusting dimensions to maximize or minimize the area of a shape. With a fixed perimeter, our goal is to find the rectangle's dimensions that lead to the largest possible area.
Starting with:
Starting with:
- \( A = l \times w \)
- \( w = \frac{K}{2} - l \)
- \( A = l \left(\frac{K}{2} - l\right) \)
- \( A = \frac{K}{2}l - l^2 \)
Vertex of a Parabola
The vertex of a parabola is a crucial concept when working with quadratic functions in geometry, especially for maximizing areas. The vertex represents the highest or lowest point of a parabola, depending on its orientation.
For a downward-opening parabola such as the one in our area function \( A = \frac{K}{2}l - l^2 \), the vertex indicates a maximum. The formula to find the vertex of a quadratic function \( ax^2 + bx + c \) is given by:
For a downward-opening parabola such as the one in our area function \( A = \frac{K}{2}l - l^2 \), the vertex indicates a maximum. The formula to find the vertex of a quadratic function \( ax^2 + bx + c \) is given by:
- \( l = -\frac{b}{2a} \)
- \( a = -1 \) and \( b = \frac{K}{2} \)
- \( l = -\frac{\frac{K}{2}}{2(-1)} = \frac{K}{4} \)