Tank Geometry: Cylinder Versus Square-Base Cuboid
A fabrication workshop has won an order for a closed cylindrical steel tank to hold 20 m³ of edible oil. The site clearance fixes the cylindrical height at 4.0 m, and the client's available floor area is 22 m². The workshop can instead build a closed rectangular tank with a square base and the same 4.0 m height. Steel plate costs ₹620 per m², and the purchase rule is to add 6% wastage to the net calculated surface area before costing. A competing supplier has quoted a standard cylinder of radius 1.5 m and height 4.0 m, describing it as 'about 28 m³, more value for the same money'. The client is willing to relax the height downwards if that reduces plate area for the same 20 m³ capacity, provided the footprint stays within 22 m². The workshop manager wants a defensible recommendation on shape, dimensions and cost before signing the contract.
1. For the cylindrical design at height 4.0 m, find the required radius, the closed surface area, and the plate cost including wastage. Check the floor-space limit.
Read the worked answer
Volume: πr²h = 20, with h = 4. r² = 20 ÷ (4π) = 1.59155, so r = 1.2616 m (say 1.26 m).
Footprint = πr² = 5.00 m², well within the 22 m² limit.
Closed surface area:
- Two ends: 2πr² = 2 × 5.00 = 10.00 m²
- Curved surface: 2πrh = 2π × 1.2616 × 4 = 31.71 m²
- Net area = 41.71 m²
With 6% wastage: 41.71 × 1.06 = 44.21 m². Cost = 44.21 × 620 = ₹27,410 (approximately).
Note the check: a cylinder with r = 1.26 m and h = 4 m actually holds π × 1.59155 × 4 = 20.0 m³, confirming the radius.
2. Compare this with the square-base cuboid of the same height and volume. Which design should the workshop recommend on plate cost alone?
Read the worked answer
Cuboid: base side s, height 4, volume s² × 4 = 20, so s² = 5 and s = 2.236 m. Footprint = 5.00 m², identical to the cylinder.
Surface area:
- Top and bottom: 2s² = 10.00 m²
- Four sides: 4sh = 4 × 2.236 × 4 = 35.78 m²
- Net area = 45.78 m²
With 6% wastage: 48.53 m². Cost = 48.53 × 620 = ₹30,089.
Comparison: the cylinder saves 45.78 − 41.71 = 4.07 m² of net plate, about ₹2,679, or 8.9% of the cuboid cost, with the same footprint.
Recommendation: the cylinder is cheaper on plate alone. However, plate cost is not total cost: the cuboid has 12 welded edges against two circular seams for the cylinder, so welding, cutting and inspection costs would widen the cylinder's advantage. The cylinder should be recommended, subject to the workshop's rolling capacity and the client accepting a circular base ring.
3. Evaluate the competitor's 1.5 m × 4 m cylinder and state the dimensions you would specify if the client insists on minimising plate area for the same 20 m³.
Read the worked answer
Competitor's tank: volume = π × 1.5² × 4 = 28.27 m³, which is 41% more than the 20 m³ required. Surface area = 2π(2.25) + 2π(1.5)(4) = 14.14 + 37.70 = 51.84 m²; with 6% wastage, 54.95 m²; cost = ₹34,067. The client would pay about ₹6,657 (24%) more for capacity that cannot be used, so 'more cubic metres for the same money' is not value.
For a closed cylinder of fixed volume V, area A = 2πr² + 2V/r. Setting dA/dr = 4πr − 2V/r² = 0 gives V = 2πr³, i.e. h = 2r. Here r = (20/2π)^(1/3) = 1.471 m and h = 2.942 m; area = 40.79 m²; cost with wastage = 40.79 × 1.06 × 620 = ₹26,807, saving only about ₹600 (2.2%) against the 4 m tank.
Judgement: specify r = 1.26 m, h = 4 m for simplicity, since the saving from the theoretical optimum is small and a wider, squatter tank raises handling and base-ring issues. Reject the 1.5 m quote.