Pregunta
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9. John has 120 feet of fencing and wishes to enclose a rectangular area. What is the maximum area? Define variable and show the equation generated by the question.

Ask by Schofield Hardy. in the United States
Mar 14,2025

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Let \( x \) be the width and \( y \) be the length of the rectangle. The perimeter equation is \( 2x + 2y = 120 \). Solving for \( y \) gives \( y = 60 - x \). The area is \( A = x(60 - x) = 60x - x^2 \). To maximize the area, set \( x = 30 \), making the rectangle a square with sides of 30 feet. The maximum area is 900 square feet.

Solución

**Step 1: Define the Variables** Let \( x \) be the length of one side (width) of the rectangle and \( y \) be the length of the adjacent side (length). **Step 2: Write the Perimeter Equation** Since the total amount of fencing is 120 feet, the perimeter \( P \) of the rectangle is given by \[ 2x + 2y = 120. \] **Step 3: Express \( y \) in Terms of \( x \)** Divide the equation by 2 to simplify: \[ x + y = 60 \quad \Longrightarrow \quad y = 60 - x. \] **Step 4: Write the Area Equation** The area \( A \) of the rectangle is given by \[ A = x \times y. \] Substitute the expression for \( y \): \[ A = x(60 - x) = 60x - x^2. \] **Step 5: Determine the Value of \( x \) that Maximizes the Area** The area function is a quadratic equation: \[ A(x) = -x^2 + 60x. \] This is a downward-opening parabola (since the coefficient of \( x^2 \) is negative), so the maximum area is at the vertex. The \( x \)-coordinate of the vertex for a quadratic \( ax^2 + bx + c \) is given by: \[ x = -\frac{b}{2a}. \] Here, \( a = -1 \) and \( b = 60 \): \[ x = -\frac{60}{2(-1)} = \frac{60}{2} = 30. \] **Step 6: Find \( y \) and Calculate the Maximum Area** Substitute \( x = 30 \) back into the equation for \( y \): \[ y = 60 - 30 = 30. \] Thus, the rectangle is a square with side lengths of 30 feet. The maximum area is: \[ A = 30 \times 30 = 900 \text{ square feet}. \] **Summary of the Equations:** - Perimeter: \(2x + 2y = 120\). - Area: \(A = x(60 - x) = 60x - x^2\). The maximum area that can be enclosed is \(900\) square feet.

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The Deep Dive

To solve this problem, let’s define the length of the rectangle as \( l \) and the width as \( w \). The perimeter of the rectangle is given by the formula \( P = 2l + 2w \). Since John has 120 feet of fencing, we have the equation \( 2l + 2w = 120 \) or simplified to \( l + w = 60 \). The area \( A \) of the rectangle can be represented as \( A = l \times w \). By substituting \( w = 60 - l \) into the area equation, we get \( A = l(60 - l) = 60l - l^2 \). To find the maximum area, we can complete the square or use calculus, leading us to determine that the maximum area of 1,800 square feet occurs when both \( l \) and \( w \) are 30 feet. Now, let’s talk about a pivotal moment in history: the problem of maximizing area with fixed perimeter isn’t just a math exercise—it’s a principle often used in agriculture and city planning! Ancient civilizations utilized similar geometry to optimize land use, ensuring that they made the best use of their resources. Think of how land layout today continues to draw on those foundational principles, refining them with modern technology for everything from farming techniques to urban design. For real-world application, imagine you’re a landscape designer! To create the most efficient garden or lawn space, knowing how to maximize area within a specific boundary can turn a basic layout into a flourishing haven. Whether you’re working with a small backyard or planning a community park, applying these principles helps you optimize nature’s canvas, ensuring your greenspace gives the greatest possible return for maintenance and aesthetics. Happy gardening!

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