Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that for a rectangle of given perimeter \(K\) the one with maximum area is a square.

Step-by-Step Solution

Answer

A rectangle with maximum area for a given perimeter is a square.

Step 1: 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:\[ K = 2(l + w) \].

Step 2: Solve for One Variable in Terms of the Other

From the perimeter equation, solve for \( w \) in terms of \( l \) (or vice versa):\[ l + w = \frac{K}{2} \]\[ w = \frac{K}{2} - l \].

Step 3: 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:\[ A = l \left(\frac{K}{2} - l\right) \]\[ A = \frac{K}{2}l - l^2 \].

Step 4: 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 \( b = \frac{K}{2} \):\[ l = -\frac{\frac{K}{2}}{2(-1)} = \frac{K}{4} \].

Step 5: 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.

Key Concepts

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:

• \( P = 2(l + w) \)

where \( l \) is the length and \( w \) is the width of the rectangle. This formula is essential because it helps set the starting condition for maximizing or minimizing rectangle properties, like area. When the perimeter is constant, such as in this problem, we establish that the outer limits of the shape remain unchanged while we consider potential changes in its interior dimensions. In the case of a given perimeter \( K \), we express that fixed boundary as:

• \( K = 2(l + w) \)

This lets us isolate and connect variables within the rectangle, paving the way for further calculation and analysis.

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:

• \( f(x) = ax^2 + bx + c \)

Their graphs are parabolas, which are U-shaped curves. A critical aspect of quadratic functions is their ability to tell us about extreme values — in this case, maximum area. In our rectangle scenario, we derive a quadratic area expression:

• \( A = \frac{K}{2}l - l^2 \)

Here, \( A \) represents the area of the rectangle, expressed as a function of its length \( l \). By transforming the problem into this quadratic form, we enable the use of techniques from calculus and algebra to find desired solutions, such as finding the vertex for optimal values.

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:

• \( A = l \times w \)

we substitute the expression for \( w \) from the perimeter equation:

• \( w = \frac{K}{2} - l \)

This substitution results in:

• \( A = l \left(\frac{K}{2} - l\right) \)

Simplified to:

• \( A = \frac{K}{2}l - l^2 \)

This form means we can apply quadratic optimization methods to find the maximum area. By focusing our efforts on these rearrangements, area optimization uncovers the most efficient use of space, guiding us towards the ideal size and shape our parameters will allow.

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:

• \( l = -\frac{b}{2a} \)

From the function \( A = -l^2 + \frac{K}{2}l \), we see:

• \( a = -1 \) and \( b = \frac{K}{2} \)

Plug into the vertex formula:

• \( l = -\frac{\frac{K}{2}}{2(-1)} = \frac{K}{4} \)

This value of \( l \) is substituted back into our perimeter equation to find \( w \), ensuring both \( l \) and \( w \) are \( \frac{K}{4} \) — making it a square. Thus, the vertex aids in confirming the shape and size that result in the maximum area, solving our problem efficiently.