Fencing a School Kitchen Garden within Budget
A school in Belagavi wants a rectangular kitchen garden against a straight compound wall 28 m long. The wall forms one side; 44 m of fencing is available for the other three sides. Safety rules require the width perpendicular to the wall to be at least 7 m for wheelchair access, and a drain limits the length parallel to the wall to at most 26 m. Fencing costs ₹350 per metre, and a gate costs ₹2,200. The eco-club has a budget of ₹18,000. One proposal is 7 m by 30 m; another is 11 m by 22 m. The club wants the feasible layout with the greatest area and asks whether the area-maximising layout fits the budget. Assume the gate is a single fixed cost and that all 44 m of fencing must be used unless the budget forces a reduction.
1. Express the garden area A in terms of the width x, state the feasible interval for x from the access and drain rules, and find the area-maximising dimensions.
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Let x be the width perpendicular to the wall and L the length along the wall. The fencing covers two widths and one length, so 2x + L = 44, giving L = 44 − 2x.
Constraints: x ≥ 7 for access, and L ≤ 26 gives 44 − 2x ≤ 26, so x ≥ 9. Also L ≥ 0 gives x ≤ 22. Hence x is in [9, 22].
Area A = x(44 − 2x) = 44x − 2x^2. A' = 44 − 4x, so the critical point is x = 11. A'' = −4 < 0, so this is a maximum. Then L = 44 − 22 = 22 m and A = 11 × 22 = 242 m².
Endpoint checks: x = 9 gives L = 26 and A = 234 m²; x = 22 gives L = 0 and A = 0. The maximum is therefore 11 m by 22 m, area 242 m².
2. Check the two proposals and the budget for the 11 m by 22 m layout. Is the area-maximising layout feasible?
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The 7 m by 30 m proposal uses 2(7) + 30 = 44 m of fencing, but its length 30 m exceeds the 26 m drain limit, so it is infeasible. The 11 m by 22 m proposal satisfies 11 ≥ 7 and 22 ≤ 26 and uses 44 m of fencing, so it is feasible and, from the optimisation, maximises area.
Cost of 44 m of fencing = 44 × 350 = ₹15,400. Adding the gate gives ₹15,400 + ₹2,200 = ₹17,600. This is ₹400 below the ₹18,000 budget.
The area-maximising 11 m by 22 m layout is therefore feasible, with a small contingency margin.
3. If fencing cost rises to ₹370 per metre, is 11 m by 22 m still within budget? Find the minimum reduction in total fence length needed, and state one revised feasible layout.
Read the worked answer
At ₹370 per metre, the 44 m layout costs 44 × 370 = ₹16,280 for fencing. With the gate, the total is ₹16,280 + ₹2,200 = ₹18,480, which is ₹480 over budget.
The fencing budget is ₹18,000 − ₹2,200 = ₹15,800, so the maximum fencing length is 15,800/370 = 42.7027 m. The reduction needed is 44 − 42.7027 = 1.2973 m, about 1.30 m.
One revised layout keeps width x = 11 m and sets length L = 42.7027 − 22 = 20.7027 m. This satisfies x ≥ 7 and L ≤ 26, uses about 42.70 m of fencing, and gives area 11 × 20.7027 = 227.73 m². The 11 m by 22 m layout is no longer within budget.