Problem 83 - Precalculus : Building Concepts and Connections

Question

A rectangular sandbox is to be enclosed with a fence on three of its sides and a brick wall on the fourth side. If 24 feet of fencing material is available, what dimensions will yield an enclosed region with an area of 70 square feet?

Step-by-Step Solution

Answer

The dimensions that will yield an enclosed area of 70 sq ft are approximately \(4.39 \, ft\) and \(15.22 \, ft\).

Step 1: Identifying the Variables

Two sides of the fence can be seen as having a length of x feet each and the third side having a length of y feet. Hence, the rectangular area can be formulated as \(A = x*y\).

Step 2: Building the Equation

According to the problem, 24 feet of fencing material are available. This means \(2*x + y = 24\) or \(y = 24 - 2x\). The area is given as 70 sq ft so \(x*y = 70\) or \(x*(24-2x) = 70\).

Step 3: Solve Polynomial Equation

Transform area equation into a form of a quadratic equation: \(2x^2 - 24x + 70 = 0\). This quadratic equation can be solved by either factoring, completing the square or using the quadratic formula.

Step 4: Solve for x

In our case, the best way to solve it might be using the quadratic formula \(x = [24 ± sqrt((24)^2 - 4*2*70)] / (2*2)\). After calculating the values, we only consider the positive root because lengths cannot be negative. So, \(x \approx 4.39 \, ft\).

Step 5: Solve for y

Plug in \(x\) into the perimeter equation to find \(y\). Hence, \(y = 24 - 2*4.39 \approx 15.22 \, ft\).

Key Concepts

Rectangular Geometry

A rectangular shape is fundamental in geometry because it has opposing sides that are equal in length and includes four right angles. Rectangular geometry primarily deals with the properties and measures of rectangles, such as:

• Width (shorter sides)
• Length (longer sides)
• Area
• Perimeter

For any given rectangle, the **area** is the amount of space it covers and is calculated using the formula \( A = x \, \times \, y \), where \(x\) and \(y\) are the length and width.
In this specific fencing problem, understanding rectangular geometry helps to figure out various dimensions that satisfy both perimeter and area constraints. Each variable in the problem is thus crucial for determining the configuration or layout of the rectangle.

Fencing Problem

The fencing problem is a classic task in applied mathematics, requiring optimal usage of material for enclosing spaces. Here, it involves enclosing a rectangular area with a limited amount of fencing material on three sides, while one side is supported by an existing wall.
The problem can be broken down into simpler steps:

• Identify the known quantities: fencing length and the area to be enclosed.
• Translate the problem into mathematical expressions, such as \(2x + y = 24\), which corresponds to the perimeter of the three fenced sides.
• Use these expressions to solve for any unknown variables, thereby determining the values that will meet the given conditions.

In the given problem setup, one goal is to efficiently use all available fencing material, ensuring that the predefined area requirement (in this case, 70 square feet) is met.

Area and Perimeter

Understanding the concepts of area and perimeter is crucial for solving real-world problems involving dimensions and boundaries.
**Area** measures the space within a boundary: for a rectangle, this is found using \(A = x \, \times \, y\). It indicates how much surface the rectangle covers.
**Perimeter** is the total distance around a boundary. In problems with partial enclosures like the one given, the perimeter formula may change. The use of given expressions like \(2x + y = 24\) ensures the proper allocation of material to cover three sides.
The goal in this exercise is balancing the need for a specific area while adhering to the perimeter restrictions involving limited resources. Solving the quadratic equation derived from these conditions allows for finding the dimensions needed to satisfy both area and perimeter constraints.