Open-Top Can: Optimising Sheet Area Under a Cutting-Table Constraint
A small fabrication unit in Nacharam must supply 200 open-top cylindrical stainless-steel cans, each holding exactly 500 cm³ (0.5 litre). The sheet is 0.8 mm SS-304 at Rs 620 per square metre, and the cutting table can only shear circles whose diameters are whole centimetres, so the radius comes in 0.5 cm steps. Height is not stepped and can be set freely. The supervisor's thumb rule has always been 'radius equals height', but the order is quoted at a fixed price, so metal cost decides the margin. The laser welder needs a 1 cm overlap strip along the vertical seam, and the base disc is cut with 0.5 cm edge allowance on the diameter. The workshop currently buys 12 m² of sheet per batch and claims 8 per cent scrap. A trainee proposes simply buying 15 per cent more sheet than the calculated area, rather than re-computing the geometry.
1. Derive the radius-to-height relation that minimises sheet area for an open-top can of fixed volume, compute the ideal dimensions for 500 cm³, and say whether the supervisor's rule is right, lucky or generally wrong.
Read the worked answer
Let r be radius and h height, with V = 500 cm³ fixed.
Open top: S = πr² + 2πrh, and h = V/(πr²), so S(r) = πr² + 2V/r.
Setting dS/dr = 2πr − 2V/r² = 0 gives r³ = V/π. The second derivative, 2π + 4V/r³, is positive, so this is a minimum.
Substituting back: h = V/(πr²) = (V/π)^(1/3) = r. The rule is therefore correct for an open-top can - but correct by accident of the open top, not as a general truth. For a closed can, S = 2πr² + 2V/r, giving r³ = V/(2π) and h = 2r; applying 'r = h' there wastes about 6 per cent of the sheet.
Numbers: r = (500/3.1416)^(1/3) = (159.2)^(1/3) = 5.42 cm, h = 5.42 cm, and S = 3πr² = 276.8 cm².
Answer: r = h = 5.42 cm for the open-top can, and the rule must not be carried over to lidded work.
2. Using the cutting-table step, the seam overlap and the base allowance, compute the sheet needed for 200 cans and evaluate the trainee's 15 per cent uplift.
Read the worked answer
The nearest allowed radius is 5.5 cm. With r fixed, h = 500/(π × 5.5²) = 500/95.03 = 5.26 cm.
Area per can: lateral surface 2πrh = 2π × 5.5 × 5.26 = 181.8 cm²; seam overlap 1 × 5.26 = 5.3 cm²; base disc π(5.5 + 0.25)² = π × 33.06 = 103.9 cm². Total ≈ 291.0 cm², i.e. 0.0291 m² per can.
Check the rounding penalty: at r = 5.5 the ideal area is π(5.5²) + 2 × 500/5.5 = 95.0 + 181.8 = 276.8 cm², identical to the continuous optimum, because S is flat near its minimum. There is no reason to fight the table.
200 cans: 200 × 0.0291 = 5.82 m². With 8 per cent scrap, 6.29 m², about Rs 3,900 at Rs 620/m². The present 12 m² order is nearly double that.
The trainee's 15 per cent uplift gives 5.82 × 1.15 = 6.69 m² - 0.40 m², roughly Rs 250 per batch, worse than a proper nesting calculation, and it hides the seam and edge allowances instead of costing them explicitly. Reject the uplift.
3. The customer now asks for the same 500 cm³ can with a lid. Work out the new ideal dimensions, compare with reusing the existing 5.5 cm radius, and recommend.
Read the worked answer
Closed can: S = 2πr² + 2πrh with h = V/(πr²), so S = 2πr² + 2V/r. Then dS/dr = 4πr − 2V/r² = 0, giving r³ = V/(2π) = 79.6, so r = 4.30 cm and h = 2r = 8.60 cm.
Minimum area: 2π(4.30²) + 2 × 500/4.30 = 116.2 + 232.6 = 348.8 cm².
Reusing the existing 5.5 cm body and simply adding a lid, keeping h = 5.26 cm: 2π(5.5²) + 181.8 = 190.1 + 181.8 = 371.9 cm², which is 6.6 per cent more metal, about 23 cm² per can.
Judgement: 6.6 per cent of 348.8 cm² over 200 cans is about 0.46 m² per batch, roughly Rs 290 extra, against the cost and lead time of a new die, gauge and welding fixture. For a 200-off order, reuse the 5.5 cm tooling. If the lidded can becomes a repeat product, the 4.30 cm radius pays for itself in a few batches. Note that the old 'r = h' habit is now wrong - the geometry must decide, not custom.