Constrained Cylindrical Can Design
A beverage company wants a cylindrical can holding 2000 cm³. The metal sheet costs ₹0.05 per cm², and the can is made from a fixed sheet; the production line can form cans with radius r between 3 cm and 6 cm only. The design engineer writes the surface area as S(r)=2πr²+4000/r and sets S'(r)=0. The solution comes out r≈6.83 cm, which is outside the machine's range. A junior engineer suggests 'rounding down' to r=6 cm and keeping the height fixed at the value that gives 2000 cm³. The plant manager wants to know whether r=6 cm is actually the cheapest allowed can, or whether the true minimum is at the other endpoint r=3 cm, and how much the choice matters in rupee terms. The can must be closed at both ends, and no overlap or waste is to be assumed.
1. Show that the stationary point of S(r)=2πr²+4000/r is r=(1000/π)^(1/3)≈6.83 cm, and determine whether S is increasing or decreasing on [3,6].
Read the worked answer
Differentiate: S'(r)=4πr−4000/r². Set S'(r)=0: 4πr=4000/r², so 4πr³=4000 and r³=1000/π. Hence r=(1000/π)^(1/3). Numerically, 1000/π≈318.31, and 318.31^(1/3)≈6.83 cm. To decide monotonicity on [3,6], evaluate at r=6: S'(6)=4π(6)−4000/36≈75.40−111.11=−35.71<0. Since S' is continuous and has no zero in [3,6] (the only positive zero is 6.83), S'<0 throughout [3,6]. Therefore S is strictly decreasing on the allowed interval. A decreasing function on [3,6] takes its minimum at the right endpoint r=6, not at r=3 and not at the stationary point outside the interval.
2. Find the dimensions and surface area at r=6, and compare the material cost with the r=3 endpoint.
Read the worked answer
Volume: πr²h=2000. At r=6, h=2000/(π·36)=2000/(36π)≈17.68 cm. Surface area S=2πr²+2πrh=2π(36)+2π(6)(17.68). More directly, 2πrh=2πr·2000/(πr²)=4000/r, so S(6)=72π+4000/6≈226.19+666.67=892.86 cm². At r=3, h=2000/(9π)≈70.74 cm and S(3)=18π+4000/3≈56.55+1333.33=1389.88 cm². The r=6 can uses about 497.02 cm² less metal. At ₹0.05 per cm², the saving is 497.02×0.05≈₹24.85 per can. Since S is decreasing on [3,6], r=6 is indeed the cheapest allowed can; the other endpoint r=3 is far more expensive because the tall narrow can has a large side area.
3. The junior suggests keeping h fixed at the value that goes with r=6.83 while using r=6. Explain why this cannot produce a 2000 cm³ can and what constraint it violates.
Read the worked answer
At r=6.83, h=2000/(π(6.83)²)≈2000/146.6≈13.64 cm. If the line uses r=6 with the same h=13.64, the volume is π(6²)(13.64)=π·36·13.64≈1542 cm³, not 2000 cm³. The volume constraint πr²h=2000 is violated. The surface-area formula S(r) already eliminated h using h=2000/(πr²); you cannot change r while holding the old h and still stay on the constraint curve. To compare allowed cans correctly, h must be recomputed for each r. At r=6 the correct height is about 17.68 cm, which is taller than the junior's 13.64 cm. This is why the constrained minimum must be found by evaluating S(r) at r=6, not by plugging r=6 into a formula that still assumes the unconstrained height.