The Garden Geometry
A rectangular garden is 20 meters long and 12 meters wide. The owner wants to install a gravel path of uniform width around the garden, outside its borders. The path will increase the total area (garden plus path) to 1.5 times the garden's original area. The gravel costs $8 per square meter, and the owner has a budget of $1,200 for the path.
1. Determine the width of the path that will make the total area 1.5 times the original garden area. Then calculate the cost of the gravel for the path and check if it is within the budget.
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Original area = 20 * 12 = 240 m². Desired total area = 1.5 * 240 = 360 m². Let x be the path width. The outer dimensions are (20 + 2x) and (12 + 2x). So (20+2x)(12+2x) = 360. Expand: 240 + 40x + 24x + 4x² = 360 → 4x² + 64x + 240 = 360 → 4x² + 64x - 120 = 0 → divide by 4: x² + 16x - 30 = 0. Solve: x = [-16 ± √(256 + 120)]/2 = [-16 ± √376]/2 = [-16 ± 2√94]/2 = -8 ± √94. Since x > 0, x = √94 - 8 ≈ 9.695 - 8 = 1.695 m. So path width ≈ 1.70 m.
Area of path = total area - garden area = 360 - 240 = 120 m². Cost = 120 * $8 = $960. Budget $1,200, so within budget.
2. If the owner wants to maximize the path width while staying within the $1,200 budget, what is the maximum possible width? (Assume the path is uniform and the cost is proportional to area.)
Read the worked answer
Budget allows area of path = $1,200 / $8 = 150 m². Total area = 240 + 150 = 390 m². So (20+2x)(12+2x) = 390 → 240 + 64x + 4x² = 390 → 4x² + 64x - 150 = 0 → divide by 2: 2x² + 32x - 75 = 0. Solve: x = [-32 ± √(1024 + 600)]/4 = [-32 ± √1624]/4 = [-32 ± 2√406]/4 = -8 ± (√406)/2. √406 ≈ 20.149, so x ≈ -8 + 10.0745 = 2.0745 m. So maximum width ≈ 2.07 m.