Problem 75 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that, among all rectangles with given perimeter \(p\), the square has the largest area. Hint: If \(a\) and \(b\) denote the lengths of adjacent sides of a rectangle of perimeter \(p\), then the area is \(a b\), and for the square the area is \(a^{2}=[(a+b) / 2]^{2} .\) Now see Problem 74 .

Step-by-Step Solution

Answer

The rectangle with the largest area for a given perimeter is a square.

Step 1: Set up the Problem

Given the perimeter of a rectangle is \( p \), let \( a \) and \( b \) be the side lengths of the rectangle. Hence, \( 2(a+b) = p \). To find the area \( A \) of the rectangle, use the formula \( A = a \times b \).

Step 2: Express One Variable in Terms of the Other

From the perimeter equation \( 2(a+b) = p \), solve for \( b \) in terms of \( a \): \( b = \frac{p}{2} - a \). This allows us to express the area \( A \) in terms of \( a \) alone: \( A = a \left( \frac{p}{2} - a \right) \).

Step 3: Formulate the Area Function

Substitute \( b \) into the area formula: \( A(a) = a \left( \frac{p}{2} - a \right) = \frac{p}{2}a - a^2 \). This is the quadratic function we need to maximize.

Step 4: Determine the Maximum Area

The function \( A(a) = \frac{p}{2}a - a^2 \) is a quadratic equation that describes an upside-down parabola (since the coefficient of \( a^2 \) is negative). The maximum value for this quadratic function will occur at its vertex.

Step 5: Compute the Vertex of the Quadratic

To find the vertex of the quadratic equation, use the vertex formula \( a = -\frac{b}{2c} \). Here, \( b = \frac{p}{2} \) and \( c = -1 \). So, \( a = -\frac{\frac{p}{2}}{2 \times (-1)} = \frac{p}{4} \).

Step 6: Validate the Solution for a Square

When \( a = \frac{p}{4} \), substituting into the expression for \( b \) gives \( b = \frac{p}{2} - \frac{p}{4} = \frac{p}{4} \). Thus, \( a = b = \frac{p}{4} \), meaning the rectangle is a square.

Key Concepts

Rectangle Perimeter and Area

Understanding the concept of a rectangle's perimeter and area is essential for solving many optimization problems in calculus.
The perimeter of a rectangle is the total length of all its sides. For a rectangle with length and width denoted as \( a \) and \( b \) respectively, the perimeter \( p \) is given by the formula:

• \( p = 2(a + b) \)

The area of the rectangle, which describes how much space is inside the figure, is calculated as:

• \( A = a \times b \)

These simple formulas allow us to solve for other variables when the perimeter or area is given.
For instance, if the perimeter \( p \) is known, we can rearrange the formula to express one side in terms of the other. This rearrangement aids in discussing different configurations of a rectangle while keeping a constant perimeter.
This is fundamental when trying to find the maximum area of a rectangle with a certain perimeter, leading us towards more complex calculus-based optimization methods.

Quadratic Functions

Quadratic functions are pivotal when working with optimization problems, especially those involving maximum or minimum values for areas or other characteristics.
A quadratic function is typically in the form \( f(x) = ax^2 + bx + c \). These functions graph as parabolas, which can either be U-shaped or inverted, depending on the sign of the leading coefficient (\( a \)).
In our rectangle area problem, we express one variable in terms of the other, resulting in a quadratic function that describes how the area changes:

• \( A(a) = \frac{p}{2}a - a^2 \)

Here, the function is "upside-down" because \( -a^2 \) means \( a \) is negative, indicating an inverted parabola. Understanding this helps us identify if we are looking for a maximum or a minimum area.
By getting the quadratic function into its standard form, we can effectively use calculus techniques or vertex calculations to find the optimal solutions, such as the maximum area in this exercise.

Vertex of a Parabola

The vertex of a parabola is a critical concept when maximizing or minimizing values described by quadratic functions.
The vertex is the point where the parabola changes direction, which means it represents either a maximum or minimum point on the graph depending on its orientation.
For a quadratic equation in standard form \( ax^2 + bx + c \), the vertex can be calculated using the formula:

• \( x = -\frac{b}{2a} \)

In our problem, we have the quadratic \( A(a) = \frac{p}{2}a - a^2 \). To find the vertex, we substitute \( b = \frac{p}{2} \) and \( a = -1 \) into the formula, yielding:

• \( a = \frac{p}{4} \)

This result gives the side length of the rectangle that maximizes the area. When the length and width are equal, the rectangle is a square, confirming the optimal shape for maximizing area with a fixed perimeter. The vertex, therefore, not only gives us a numerical solution but also a geometric understanding of optimizing shapes.