Plaza Fencing with a Mandatory Walkway and a Frontage Cap
A new-town authority has 120 m of surplus steel railing and wants to use all of it to fence a rectangular public plaza. One long side of the plaza will run along the blank wall of an existing community hall, so that side needs no railing; the other three sides must be fenced. Two design rules are already fixed: a 2 m wide paved walkway must run inside the plaza along the entire wall side, so that pedestrians can walk under the wall without stepping on the lawn, and the hall's frontage limits the plaza's length along the wall. The executive engineer's first sketch assumes the length can be whatever the mathematics allows; the horticulture wing has separately assumed a 50 m frontage limit, and both notes are now on the same file with the plaza still unfenced.
1. Ignoring the walkway and the frontage limit, find the dimensions that maximise the area of the fenced plaza and state the maximum area, showing why the optimum is where it is.
Read the worked answer
Let L = length along the hall wall, W = depth. Only three sides are fenced, so the constraint is L + 2W = 120, giving L = 120 − 2W. (Using all 120 m is correct - area increases until the constraint binds.)
Area A = L·W = (120 − 2W)W = 120W − 2W².
dA/dW = 120 − 4W = 0 → W = 30 m, hence L = 60 m and A = 1800 m². The second derivative is −4 < 0, so this is a maximum.
Cross-check by AM–GM: L·(2W) ≤ ((L + 2W)/2)² = 60² = 3600, so LW ≤ 1800, with equality exactly when L = 2W.
Answer: 60 m along the wall × 30 m deep, area 1800 m².
2. Now include the 2 m walkway running along the whole wall side and re-optimise. What are the new dimensions, and what is the usable lawn area?
Read the worked answer
The walkway removes 2 m of depth from the usable surface, so the quantity to maximise is now the lawn, not the whole fenced rectangle: lawn = L(W − 2), with L = 120 − 2W.
A(W) = (120 − 2W)(W − 2) = −2W² + 124W − 240 dA/dW = −4W + 124 = 0 → W = 31 m, so L = 120 − 62 = 58 m.
Lawn = 58 × 29 = 1682 m². The full fenced rectangle is 58 × 31 = 1798 m², which is 2 m² smaller than the earlier 1800 m². That is not an arithmetic error: changing the objective changes the optimum, and the same 120 m still fences only three sides.
Answer: 58 m along the wall × 31 m deep, giving a 29 m lawn depth and 1682 m² of lawn.
3. The hall's frontage permits at most 50 m along the wall. With the walkway retained, what should actually be built, and how much lawn is lost relative to the unconstrained optimum?
Read the worked answer
Write the lawn directly in terms of L. Since W = (120 − L)/2, lawn = L((120 − L)/2 − 2) = L(58 − L/2) = 58L − L²/2.
d/dL = 58 − L, which is positive for every L < 58. So on the admissible range L ≤ 50 the function is strictly increasing and the maximum lies on the boundary at L = 50.
L = 50 m → W = (120 − 50)/2 = 35 m → lawn = 50 × 33 = 1650 m².
Loss = 1682 − 1650 = 32 m², about 1.9 per cent of the lawn.
Answer: build 50 m × 35 m with a 33 m lawn depth and 1650 m² of lawn. The frontage cap costs 32 m². The optimum is a corner solution - setting the derivative to zero and reporting L = 58 m would be an infeasible answer, so the constraint must be checked against the stationary point, not assumed away.