Playground Design Constraints
A rectangular playground is to be built against a straight wall, using 60 m of fencing for the other three sides. The two side fences are perpendicular to the wall, and the fence parallel to the wall is opposite it. Safety rules require the enclosed area to be at least 350 m² so children have room to run, and the side length must be at least 8 m so equipment fits. The designer also wants the diagonal distance across the playground to be at most 25 m to keep sight lines short for supervisors.
1. Express the area and the diagonal in terms of the side length x, and find the range of x satisfying the area and side constraints.
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Variables. Let x be each side length perpendicular to the wall. The fence parallel to the wall has length 60 − 2x. Area A(x) = x(60 − 2x) = 60x − 2x².
Area constraint. 60x − 2x² ≥ 350 → 2x² − 60x + 350 ≤ 0 → x² − 30x + 175 ≤ 0. Roots: x = (30 ± √(900 − 700)) ÷ 2 = (30 ± √200) ÷ 2 ≈ (30 ± 14.14) ÷ 2, giving x ≈ 7.93 and x ≈ 22.07. So 7.93 ≤ x ≤ 22.07.
Side constraint. x ≥ 8, and 60 − 2x > 0 → x < 30.
Combined range. 8 ≤ x ≤ 22.07 (approximately).
2. Apply the diagonal constraint and determine the final feasible interval for x.
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Diagonal. d² = x² + (60 − 2x)². Require d² ≤ 625.
Expand. x² + 3,600 − 240x + 4x² ≤ 625 → 5x² − 240x + 2,975 ≤ 0.
Roots. Discriminant = 240² − 4 × 5 × 2,975 = 57,600 − 59,500 = −1,900 < 0.
Interpretation. The quadratic 5x² − 240x + 2,975 opens upward with no real roots, so it is always positive. The minimum occurs at x = 240 ÷ 10 = 24, where d² = 5(576) − 240(24) + 2,975 = 2,880 − 5,760 + 2,975 = 95, so d ≈ 9.75 m. Every feasible x gives d² ≥ 95, and the constraint d² ≤ 625 fails only for large x. Check endpoints: at x = 8, d² = 64 + 44² = 64 + 1,936 = 2,000 > 625. At x = 22.07, d² ≈ 487 + 15.86² ≈ 487 + 251.5 ≈ 738.5 > 625. The diagonal constraint fails across the entire area-feasible range.
Conclusion. No x satisfies all constraints; the designer must relax the diagonal limit (to about 44.7 m at x = 8) or the area requirement.
3. The designer decides to keep the diagonal limit at 25 m and drop the area requirement. Find the x-interval that satisfies the side and diagonal constraints, and state the maximum area achievable.
Read the worked answer
Solve d² ≤ 625. 5x² − 240x + 2,975 ≤ 0. Discriminant = −1,900 < 0, so the quadratic is always positive. This means d² > 625 for every x - the diagonal constraint cannot be met at all with 60 m of fencing.
Check the minimum diagonal. At x = 24, the fence parallel to the wall is 60 − 48 = 12 m, and d = √(24² + 12²) = √720 ≈ 26.8 m, which exceeds 25 m. The smallest possible diagonal with 60 m of fencing is about 26.8 m.
Conclusion. The diagonal constraint is infeasible for any rectangle with 60 m of three-sided fencing. The designer must lengthen the fencing to at least about 56 m of diagonal-limited configuration or raise the diagonal limit to at least 26.8 m. The maximum area, unconstrained by the diagonal, is at x = 15: A = 15 × 30 = 450 m².