Fencing a Rectangular Community Garden Against a Wall
A village committee has 60 m of fencing to enclose a rectangular community garden. One side of the plot lies along an existing straight wall, so only the other three sides need fencing. Let x metres be the length of each of the two sides perpendicular to the wall. The committee also plans a 1 m wide path outside the plot along the wall; this uses no fencing. Fencing costs ₹250 per metre, and the committee has a budget of ₹16,000. One member proposes a plot of area 500 m², while another wants the largest possible area. The wall is long enough for any reasonable layout. Before approving the layout, the committee must know the dimensions that maximise the area, whether the proposed 500 m² plot is feasible, and whether the fencing cost fits the budget.
1. Express the garden area A in terms of x, state the valid domain, and find the dimensions that maximise A. Also check whether the fencing cost for those dimensions is within the ₹16,000 budget.
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The side parallel to the wall has length 60 − 2x, so A(x)=x(60−2x)=60x−2x². The practical domain is 0<x<30, because both x and 60−2x must be positive. This is a downward parabola; the maximum occurs at x=−b/(2a)=−60/(2·(−2))=15. Then A(15)=15×30=450 m², and the plot is 15 m by 30 m. Fencing used is 60 m, costing 60×250=₹15,000, which is ₹1,000 below the budget. So the maximum-area layout is feasible.
2. A member claims a 500 m² plot is possible. Set up the corresponding quadratic equation and use its discriminant to decide. Explain what the result says about the claim.
Read the worked answer
Set x(60−2x)=500, giving 2x²−60x+500=0, or x²−30x+250=0. The discriminant is D=(−30)²−4·1·250=900−1000=−100, which is negative. No real x satisfies the equation. Geometrically, the horizontal line A=500 lies above the parabola's maximum value 450, so it never meets the area graph. The claim is impossible with 60 m of fencing; the largest possible area is 450 m².