Surface Area and Volume Problem Solving — Solutions

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1. Identifying formula needed

   1. Surface area. Cylinder SA = 2πr² + 2πrh (or just the curved + base if tank has no lid).
   2. Volume. Rectangular prism V = lwh.
   3. Surface area. Rectangular prism SA = 2(lw + lh + wh).
   4. Volume. Cylinder V = πr²h.
   5. Surface area. Rectangular prism SA = 2(lw + lh + wh).
   6. Volume. Triangular prism V = (½bh) × length.

2. Cylindrical tin can

   1. r = 5 cm, h = 14 cm. SA = 2π(5)² + 2π(5)(14) = 50π + 140π = 190π ≈ 596.90 cm²
   2. Cost = 596.90 × 0.0045 ≈ $2.686 ≈ $2.69
   3. Sheet area = 200 × 150 = 30 000 cm². Cans = 30 000 ÷ 596.90 ≈ 50.26 → 50 cans
   4. V = π(5)²(14) = 350π ≈ 1099.56 cm³ = 1099.56 mL ≈ 1.10 L
   5. Cost of soup = 1.0996 L × $0.80 ≈ $0.88

3. Painting a room

   1. Two long walls: 2 × (6 × 2.7) = 32.4 m². Two short walls: 2 × (4.5 × 2.7) = 24.3 m². Total = 56.7 m²
   2. 2 windows: 2 × (1.2 × 1.4) = 3.36 m². Door: 0.9 × 2.1 = 1.89 m². Total = 5.25 m²
   3. Net wall area = 56.7 − 5.25 = 51.45 m²
   4. Ceiling = 6 × 4.5 = 27 m²
   5. Total paint area (2 coats) = 2 × (51.45 + 27) = 2 × 78.45 = 156.9 m². Tins = 156.9 ÷ 12 = 13.075 → 14 tins. Cost = 14 × $34 = $476

4. Packaging comparison

   1. V = 10 × 6 × 10 = 600 cm³ ✓
   2. 600 = π(4.5)²h = 20.25πh. h = 600/(20.25π) ≈ 9.43 cm
   3. SA_A = 2(10×6 + 10×10 + 6×10) = 2(60+100+60) = 440 cm²
   4. SA_B = 2π(4.5)² + 2π(4.5)(9.43) = 40.5π + 84.87π = 125.37π ≈ 393.79 cm²
   5. Design B uses less material. Difference = 440 − 393.8 = 46.2 cm² less for Design B.

5. Concrete garden path

   1. Outer dimensions: (12 + 2) × (8 + 2) = 14 m × 10 m
   2. V_outer = 14 × 10 × 0.12 = 16.8 m³
   3. V_lawn = 12 × 8 × 0.12 = 11.52 m³
   4. V_path = 16.8 − 11.52 = 5.28 m³
   5. Cost = 5.28 × $320 = $1 689.60

6. Water tank design comparison

   1. Design X: V = π(2)²(4) = 16π ≈ 50.27 m³ = 50 265 litres
   2. Design Y: V = π(3)²(2) = 18π ≈ 56.55 m³ = 56 549 litres
   3. Design X: 50 265 L — just meets the requirement. Design Y: 56 549 L — meets it comfortably. Both designs meet the requirement, but Design Y provides more buffer.
   4. Design X SA = 2π(2)² + 2π(2)(4) = 8π + 16π = 24π ≈ 75.40 m². Design Y SA = 2π(3)² + 2π(3)(2) = 18π + 12π = 30π ≈ 94.25 m²
   5. Cost X = 75.40 × $85 = $6 409. Cost Y = 94.25 × $85 = $8 011. Design X is cheaper by $8 011 − $6 409 = $1 602

7. Gift wrapping

   1. SA = 2(25×18 + 25×10 + 18×10) = 2(450+250+180) = 2(880) = 1760 cm²
   2. 115% of 1760 = 1.15 × 1760 = 2024 cm²
   3. Sheet area = 70 × 55 = 3850 cm²
   4. 2024 ÷ 3850 = 0.526… → 1 sheet per box (one sheet is enough).
   5. 50 boxes × 1 sheet × $4.50 = $225.00

8. Grain silo

   1. V = π(3)²(8) = 72π ≈ 226.19 m³
   2. V_grain = π(3)²(6) = 54π ≈ 169.65 m³
   3. % filled = (54π / 72π) × 100 = 54/72 × 100 = 75.0%
   4. Mass = 169.65 × 800 = 135 720 kg = 135.7 tonnes
   5. Curved wall area = 2π(3)(8) = 48π ≈ 150.80 m². Cost = 150.80 × $12 ≈ $1 809.56

9. Pool liner

   1. r = 2.4 m. Base area = π(2.4)² = 5.76π ≈ 18.10 m²
   2. Curved wall = 2π(2.4)(1.2) = 5.76π ≈ 18.10 m²
   3. Total liner = 5.76π + 5.76π = 11.52π ≈ 36.19 m²
   4. Cost = 36.19 × $8.50 ≈ $307.62
   5. 90% depth = 0.9 × 1.2 = 1.08 m. V = π(2.4)²(1.08) = 6.2208π ≈ 19.54 m³ = 19 540 litres

10. Triangular prism chocolate bar

    1. Triangle area = ½×3×4 = 6 cm². V = 6 × 15 = 90 cm³
    2. Mass = 90 × 1.3 = 117 g
    3. SA = 2(6) + 3×15 + 4×15 + 5×15 = 12 + 45 + 60 + 75 = 192 cm²
    4. Cost = 192 × $0.002 = $0.384 ≈ $0.38 per bar
    5. Total volume = 500 × 90 = 45 000 cm³ = 45 litres