Problem 43 - Intermediate Algebra

Question

Use quadratic functions. Two hundred forty meters of fencing is available to enclose a rectangular playground. What should be the dimensions of the playground to maximize the area?

Step-by-Step Solution

Answer

The dimensions to maximize the area are 60 meters by 60 meters.

Step 1: Define Variables

Let the length of the playground be \( L \) meters and the width be \( W \) meters. The perimeter of the rectangle is \( 2L + 2W = 240 \) meters.

Step 2: Simplify the Perimeter Equation

From the perimeter equation \( 2L + 2W = 240 \), simplify to find one variable in terms of the other:\[ L + W = 120 \]Solving for \( L \), we get:\[ L = 120 - W \]

Step 3: Express the Area as a Function of One Variable

The area \( A \) of the rectangle is given by \( A = L \times W \). Substitute \( L = 120 - W \) into this equation:\[ A = (120 - W) \times W = 120W - W^2 \].

Step 4: Identify the Function Type and Vertex

The equation \( A = 120W - W^2 \) is a quadratic function in the standard form \(-W^2 + 120W\), which opens downwards. The maximum area is at the vertex of this parabola.

Step 5: Find the Vertex of the Parabola

The vertex of a quadratic function \( ax^2 + bx + c \) is given by \( x = \frac{-b}{2a} \). Here, \( a = -1 \) and \( b = 120 \), so:\[ W = \frac{-120}{2(-1)} = 60 \].

Step 6: Solve for Corresponding Length

Substitute \( W = 60 \) back into the equation \( L = 120 - W \) to find the length:\[ L = 120 - 60 = 60 \].

Step 7: Conclusion on Dimensions

The dimensions of the playground that will maximize the area are \( 60 \) meters for both length and width, forming a square.

Key Concepts

Rectangular Area Optimization

When it comes to maximizing the area of a rectangle with a fixed perimeter, understanding the balance between length and width is key. Imagine you have 240 meters of fencing. Your goal is to enclose a rectangular area using this fence. The challenge is to figure out the dimensions that maximize the area of your rectangle.

First, define the problem with some variables. If we let the length of the rectangle be denoted as \( L \) and the width as \( W \), the perimeter of the rectangle is represented by the equation: \( 2L + 2W = 240 \). This equation captures the idea that the sum of all sides of the rectangle should equal the total available fencing.

After simplifying this equation to \( L + W = 120 \), you can express one variable in terms of the other, making it easier to optimize the area. Here, it turns out that a square (equal length and width) often gives the maximum area for a given perimeter. This process of using mathematical equations to find optimal dimensions is known as **rectangular area optimization**.

Vertex of a Parabola

The quadratic equation derived from this problem, \( A = 120W - W^2 \), is in a specific form that allows us to identify its highest point, known as the vertex. Because this is a parabola that opens downward, the vertex will give us the maximum value of the function, which corresponds to the maximum area of our rectangle.

To find the vertex of a parabola described by the equation \( ax^2 + bx + c \), the formula \( x = \frac{-b}{2a} \) is used. Here, \( a = -1 \) and \( b = 120 \), so we substitute these values into the vertex formula:

• \( x = \frac{-120}{2(-1)} = 60 \)

This calculation shows that the width \( W \) that maximizes the area is 60 meters. The concept of the vertex is crucial in determining points of maximum or minimum values in quadratic functions.

Quadratic Function Application

Quadratic functions are particularly useful in real-world applications involving optimization, such as maximizing or minimizing areas. In this problem, the relationship \( A = 120W - W^2 \) is a **quadratic function** because it features a squared term \( W^2 \).

Quadratics are prevalent because they can model various phenomena, such as projectiles' trajectories or areas related to perimeters, like the one here. They are identifiable by their parabolic shape, which either opens upwards or downwards. The orientation of the parabola (either opening up or down) depends on the leading coefficient (the sign of \( a \)).

• If \( a > 0 \), the parabola opens upward, and the function has a minimum.
• If \( a < 0 \), the parabola opens downward, and the function has a maximum.

In this situation, since \( a = -1 \), the parabola opens downward, indicating that our function will reach a maximum, allowing us to find the optimal dimensions for the highest possible area of a fenced rectangle.

Perimeter of a Rectangle

The **perimeter of a rectangle** is fundamentally the total distance surrounding it. Understanding the perimeter is essential, especially when you're tasked with finding dimensions that maximize or minimize another attribute, like area.

In our exercise, the perimeter formula is given as \( 2L + 2W = 240 \). By simplifying this to \( L + W = 120 \), we can express either \( L \) or \( W \) in terms of each other, providing flexibility in solving for the best configuration.

Perimeter calculations are not just about solving for length. They allow you to manipulate and substitute values, in turn revealing relationships between different dimensions of geometric shapes. This manipulation is often the first step of many optimization problems, setting the stage for using techniques like quadratic functions to find solutions.