Fencing a School Garden Plot
A rectangular garden plot is to be fenced on three sides (one long side against a wall needs no fence). The fencing available is 48 metres. The gardener wants the enclosed area to be as large as possible but also insists the plot must be at least 10 m wide so two classes can work side by side. The plot's length is the side parallel to the wall. A second constraint: the plot must fit inside a courtyard whose maximum dimension in the length direction is 26 m.
1. Express the area in terms of the width, and find the width that maximises area with no other constraints.
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Let w = width (the two fenced sides perpendicular to the wall) and L = length parallel to the wall.
Fencing: 2w + L = 48, so L = 48 − 2w.
Area A = wL = w(48 − 2w) = 48w − 2w².
This is a downward parabola. Its vertex is at w = −b/(2a) = −48/(2·−2) = 12.
So w = 12 m gives L = 48 − 24 = 24 m and A = 288 m².
Check with completing the square: A = −2(w² − 24w) = −2[(w − 12)² − 144] = 288 − 2(w − 12)². Maximum 288 at w = 12. ✓
2. Apply both constraints (width at least 10 m, length at most 26 m) and state the best feasible dimensions and area.
Read the worked answer
Unconstrained optimum: w = 12, L = 24.
Constraint 1: w ≥ 10. Since 12 ≥ 10, this is satisfied at the optimum.
Constraint 2: L ≤ 26. At the optimum L = 24 ≤ 26, so this is also satisfied.
Therefore the unconstrained optimum is feasible: width 12 m, length 24 m, area 288 m².
For confirmation, if L were forced to 26, then 2w = 48 − 26 = 22, so w = 11 and A = 286 m², which is smaller. The constraints do not bind, so the answer is 12 m by 24 m with area 288 m².