A rectangular plot of land has a perimeter of 100 meters. Find the maximum possible area of this plot.
["Maximizing the Area of a Rectangular Plot: A 100-Meter Perimeter Case", "When planning land development—whether for a garden, building, or farm—it’s essential to know how to maximize usable area within strict fencing constraints. One common real-world problem is determining the largest possible area a rectangular plot can yield with a fixed perimeter, such as 100 meters. In this article, we explore the mathematical principles behind this optimization and show how a rectangle with a perimeter of 100 meters achieves its maximum possible area when it’s a perfect square.", "---", "### Understanding the Perimeter Constraint", "The perimeter ( P ) of a rectangle is given by the formula:\n[\nP = 2 \ imes (length + width)\n]", "Given:\n[\n2 \ imes (L + W) = 100 \quad \ ext{meters}\n]", "Simplify:\n[\nL + W = 50 \quad \ ext{meters}\n]", "Our goal is to maximize the area ( A ) of the rectangle, defined as:\n[\nA = L \ imes W\n]", "---", "### Expressing Area in Terms of One Variable", "From the perimeter equation, solve for ( W ):\n[\nW = 50 - L\n]", "Substitute into the area formula:\n[\nA = L \ imes (50 - L) = 50L - L^2\n]", "Now, the area is a quadratic function:\n[\nA(L) = -L^2 + 50L\n]", "This is a downward-opening parabola (since the coefficient of ( L^2 ) is negative), and its maximum value occurs at the vertex.", "---", "### Finding the Maximum Area Using the Vertex Formula", "For a quadratic equation ( y = ax^2 + bx + c ), the vertex (maximum) occurs at:\n[\nL = -\frac{b}{2a}\n]", "Here, ( a = -1 ), ( b = 50 ), so:\n[\nL = -\frac{50}{2 \ imes (-1)} = \frac{50}{2} = 25 \ ext{ meters}\n]", "Then, substitute back to find width:\n[\nW = 50 - L = 50 - 25 = 25 \ ext{ meters}\n]", "Thus, the rectangle with maximum area is a square with sides of 25 meters.", "---", "### Calculating the Maximum Area", "[\nA_{\ ext{max}} = 25 \ imes 25 = 625 \ ext{ square meters}\n]", "---", "### Why a Square Maximizes the Area", "Mathematically and geometrically, among all rectangles with a fixed perimeter, the square provides the largest area. This is because spreading the perimeter equally (25 m by 25 m) spreads the length and width as evenly as possible, minimizing “wasted” space.", "Any deviation from equal sides increases the difference and reduces the total area. For example:", "- If width = 30 m, then length = 20 m → Area = 600 m² (less than 625 m²)\n- If width = 40 m, length = 10 m → Area = 400 m² (even smaller)", "---", "### Practical Implications", "- Land Use Efficiency: Covering 100 meters of fencing, a 25 m × 25 m square plot provides the largest usable land.\n- Design Flexibility: This principle applies to farming, landscaping, shed construction, and urban development.\n- Mathematical Insight: A fundamental truth in optimization: symmetry and balance yield maximum efficiency under symmetric constraints.", "---", "### Conclusion", "For a rectangular plot with a perimeter of 100 meters, the maximum possible area is achieved when the plot is a square with each side measuring 25 meters. This yields a total area of 625 square meters—the most efficient use of fencing available. Whether planning a garden, construction site, or agricultural field, building a square maximizes space and sustainability.", "---", "Key Takeaways:\n- Fixed perimeter → maximize area via symmetry\n- For Perimeter = 100 m, max Area = 625 m² when Dimensions = 25 m × 25 m\n- Use the vertex formula ( L = \frac{P}{4} ) for quick optimization\n- Square is always optimal among rectangles of equal perimeter", "Optimize your land—build a square for the best return on fencing investment!"]
