Fencing a Rectangular Garden with a Fixed Budget
A school gardening club has 48 metres of fencing and a budget of $300. They want to enclose a rectangular vegetable plot against an existing straight wall, so only three sides need fencing: two equal widths and one length parallel to the wall. Fencing costs $6 per metre, and they also need to buy soil at $4 per square metre for the enclosed area. A club member claims that the plot can be made as large as 300 square metres while staying within budget. The club advisor is not sure this is possible and asks for a careful analysis before any materials are ordered.
1. Express the enclosed area A as a function of the width w, and find the width that maximises the area given the 48 m of fencing.
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Let the width be w (two sides) and the length parallel to the wall be L. Fencing: 2w + L = 48, so L = 48 - 2w.
Area: A(w) = w(48 - 2w) = 48w - 2w².
This is a downward parabola. The maximum occurs at w = -b/(2a) = -48/(2·(-2)) = 12.
So w = 12 m gives L = 48 - 24 = 24 m, and A = 12 × 24 = 288 m².
Check: w must satisfy 0 < w < 24 for positive length. The vertex lies inside this interval, so 288 m² is the true maximum.
2. Determine whether the club can afford the maximum-area plot. If not, find the largest area they can actually afford and the corresponding dimensions.
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Cost for the maximum plot: fencing = 48 × $6 = $288, soil = 288 × $4 = $1,152. Total = $1,440, far above $300. So the 300 m² claim is impossible on budget.
Budget constraint: fencing cost 6(2w + L) plus soil cost 4wL ≤ 300.
Using L = 48 - 2w for the fencing limit, cost = 6·48 + 4w(48 - 2w) = 288 + 192w - 8w².
Set 288 + 192w - 8w² ≤ 300 → -8w² + 192w - 12 ≤ 0 → 8w² - 192w + 12 ≥ 0 → w² - 24w + 1.5 ≥ 0.
Roots: w = [24 ± √(576 - 6)]/2 = [24 ± √570]/2 ≈ [24 ± 23.87]/2, giving w ≈ 0.065 or w ≈ 23.93.
So affordable widths are w ≤ 0.065 or w ≥ 23.93. The second case gives L = 48 - 2w ≈ 0.13 m - a very long, thin strip. Area ≈ 23.93 × 0.13 ≈ 3.1 m². The first case gives a tiny plot too.
Conclusion: with $300 the club cannot build any large plot; the maximum affordable area is roughly 3 m², so they must raise more funds or reduce fencing length.