The Pool Deck Scale Model
A hotel is building a rectangular pool deck around a rectangular pool. The pool measures 12 m by 25 m. The deck has uniform width x meters on all four sides, and the total outer deck footprint (pool plus deck) must fit within a paved area of 600 m². The architect drafts a scale model at 1:50 and measures the outer rectangle on the model as 0.36 m by 0.62 m. A contractor says the deck width should be 2 m because that makes the outer dimensions 16 m by 29 m, which "looks about right." The project manager needs to know the maximum deck width allowed by the 600 m² constraint, whether the model's dimensions are consistent with that width, and how the surface area and volume of a railing scale between model and reality.
1. Set up and solve for the maximum uniform deck width x allowed by the 600 m² footprint limit. Does the contractor's 2 m width fit?
Read the worked answer
Outer dimensions: (12 + 2x) by (25 + 2x). Constraint: (12 + 2x)(25 + 2x) ≤ 600. Expand: 300 + 24x + 50x + 4x² ≤ 600 → 4x² + 74x + 300 ≤ 600 → 4x² + 74x − 300 ≤ 0 → 2x² + 37x − 150 ≤ 0.
Solve 2x² + 37x − 150 = 0: x = [−37 ± √(37² − 4·2·(−150))] / (2·2) = [−37 ± √(1369 + 1200)] / 4 = [−37 ± √2569] / 4. √2569 ≈ 50.69. Positive root: x ≈ (−37 + 50.69)/4 ≈ 13.69/4 ≈ 3.42 m.
So the maximum uniform width is about 3.42 m (the quadratic opens upward, so the feasible interval is 0 ≤ x ≤ 3.42).
The contractor's 2 m width gives outer area (12+4)(25+4) = 16 × 29 = 464 m² ≤ 600, so it fits - but it is not the maximum. The manager could widen the deck up to ~3.42 m and still stay within 600 m². Note 2 m is well within the limit; the contractor's "looks about right" happens to be feasible but is not justified by the constraint.
2. The scale model shows an outer rectangle of 0.36 m by 0.62 m at 1:50. Are these model dimensions consistent with a real outer rectangle, and what real deck width does the model imply? Then explain how surface area and railing volume scale.
Read the worked answer
At 1:50, multiply model lengths by 50: 0.36 m × 50 = 18 m; 0.62 m × 50 = 31 m.
Real outer rectangle implied: 18 m by 31 m. Compare to pool 12 by 25: the deck width would be (18 − 12)/2 = 3 m and (31 − 25)/2 = 3 m. Consistent! The model implies a uniform 3 m deck.
Check against the 600 m² limit: 18 × 31 = 558 m² ≤ 600. ✓ So the model is feasible and corresponds to x = 3 m (just under the 3.42 m maximum).
Scaling rules:
- Lengths scale by the linear factor k = 50 (real = model × 50).
- Areas scale by k² = 2500. A model deck area of A_model corresponds to 2500 · A_model in reality.
- Volumes (e.g., railing material) scale by k³ = 125,000. A railing volume V_model corresponds to 125,000 · V_model.
So the model's 18×31 outer rectangle is consistent with a 3 m deck, and any area or volume computed on the model must be multiplied by 2500 or 125,000 respectively to get real quantities.