Yogurt Tub Geometry Under Shelf and Die Constraints
A small dairy in Kagawa wants a cylindrical yogurt tub with a resealable metal lid. The supplier charges by metal area, and the factory allots exactly 600 cm² of sheet metal per tub for the curved side, the base and the lid together; seams and waste are ignored, and the lid is a full disc of the same radius as the base. The tub must fit a supermarket shelf with total height no more than 12.0 cm, while the label printer needs at least 8.0 cm of height. The design team proposes choosing the radius that maximises volume and then rounding it to the nearest 0.5 cm, because the lid-stamping die comes in 0.5 cm steps. A trainee warns that this plan may not give the best saleable tub and that the height limits must be checked first. Let r be the base radius and h the height, both in centimetres.
1. Derive V(r) and find the radius that maximises volume when only the metal-area constraint applies. State the corresponding height and explain why it is a maximum.
Read the worked answer
Metal area: S = 2πr² + 2πrh = 600, so h = 300/(πr) − r.
Volume: V = πr²h = πr²(300/(πr) − r) = 300r − πr³.
Differentiate: dV/dr = 300 − 3πr². Setting this to zero gives r² = 100/π, so r = 10/√π ≈ 5.64 cm.
Then h = 300/(π × 5.64) − 5.64 ≈ 17.0 − 5.64 = 11.3 cm, i.e. h ≈ 2r.
This is a maximum because d²V/dr² = −6πr < 0 for r > 0. The volume is V ≈ 300(5.64) − π(5.64)³ ≈ 1692 − 564 ≈ 1128 cm³.
2. Impose 8.0 ≤ h ≤ 12.0 cm and the 0.5 cm die steps. Which radius should the factory use, and what volume does it give? Show the feasible interval and test the admissible die sizes.
Read the worked answer
Height limits: 8 ≤ 300/(πr) − r ≤ 12.
From h ≤ 12: πr² + 12πr − 300 ≥ 0, which gives r ≥ 5.47 cm. From h ≥ 8: πr² + 8πr − 300 ≤ 0, which gives r ≤ 6.56 cm.
So the feasible radii are 5.47 ≤ r ≤ 6.56 cm, and the die admits r = 5.5, 6.0 and 6.5 cm.
V(5.5) = 1650 − π(166.4) ≈ 1127 cm³; h ≈ 11.9 cm. V(6.0) = 1800 − π(216) ≈ 1121 cm³; h ≈ 9.9 cm. V(6.5) = 1950 − π(274.6) ≈ 1087 cm³; h ≈ 8.2 cm.
Best: r = 5.5 cm, V ≈ 1127 cm³. The continuous optimum 5.64 cm is excluded by the die, but rounding it to 5.5 happens to land on the best admissible value here.
3. Management reduces the shelf limit to h ≤ 10.0 cm. Re-evaluate the trainee's warning that 'rounding the continuous optimum is always safe'. What radius is now best?
Read the worked answer
With h ≤ 10: πr² + 10πr − 300 ≥ 0 gives r ≥ 5.98 cm, while h ≥ 8 still gives r ≤ 6.56 cm. The admissible die sizes are now only 6.0 and 6.5 cm.
V(6.0) ≈ 1121 cm³ and V(6.5) ≈ 1087 cm³, so r = 6.0 cm is best.
Rounding the unconstrained optimum 5.64 cm to 5.5 cm would give a tub of height about 11.9 cm, which violates the 10 cm shelf limit. So the trainee is right: feasibility must be applied before rounding, and the best step choice can move when a constraint changes.