Two vegetable beds against a wall
A school agriculture club has 72 m of fencing to make two equal rectangular vegetable beds side by side. The beds will share a common dividing fence, and the outer boundary uses three sides of fencing because the fourth side is formed by an existing straight wall. The club wants the total area of the two beds to be as large as possible. However, each bed must be at least 4 m wide, measured parallel to the wall, and at least 6 m long, measured away from the wall, because the irrigation lines and access path need that space. The club also has a rule that no single bed may have an area greater than 120 m², to keep crop trials separate. The president suggests making each bed 6 m wide and 18 m long, while the treasurer suggests 10 m wide and 12 m long. The teacher wants a calculus-based recommendation and a check that the practical constraints are satisfied.
1. Let x be the width of one bed measured along the wall and y be its length measured away from the wall. Write the fencing equation, the total-area function, and all constraints. Explain why the club cannot simply use the unconstrained maximum.
Read the worked answer
Fencing: two end fences of length y, one front fence of length 2x, and the shared divider of length y give 2x+3y=72, so y=(72-2x)/3. Total area A=2xy=2x(72-2x)/3=(144x-4x²)/3. Constraints: x≥4, y≥6, xy≤120, plus y=(72-2x)/3.
Unconstrained maximum: dA/dx=(144-8x)/3; setting zero gives x=18, y=12, A=432. Each bed then has area 18×12=216 m², which exceeds the 120 m² cap. Therefore the unconstrained maximum is infeasible.
2. Find the feasible dimensions that maximise total area, check both named proposals, and recommend a layout.
Read the worked answer
At a feasible optimum with the cap active, set xy=120 and combine with 2x+3y=72. Substitute y=120/x: 2x+360/x=72, so 2x²-72x+360=0, x²-36x+180=0. Then x=(36±24)/2, giving x=30 or x=6. For x=30, y=4, which violates y≥6. For x=6, y=20, which satisfies x≥4, y≥6 and xy=120. Total area=2×6×20=240 m², and fencing used=2(6)+3(20)=72 m.
President's 6 m by 18 m: each area=108 m², total=216 m²; fencing used=12+54=66 m, so 6 m unused and below cap. Treasurer's 10 m by 12 m: each area=120 m², total=240 m²; fencing used=20+36=56 m, so 16 m unused. Recommendation: use beds 6 m wide (along wall) by 20 m long. This reaches the maximum allowed area and uses all fencing; the treasurer's layout also reaches 240 m² but wastes fencing that cannot increase area because of the cap.