A Cylinder That Must Fit Under a Height Limit
A fabricator must build a closed cylindrical storage tank of volume 2.00 m³. The side-wall material costs ₹300 per square metre, while the top and bottom discs cost ₹500 per square metre because they are stamped. The tank must fit under a fixed ceiling, so its height cannot exceed 1.50 m. The design variable is the base radius r; the height h is then determined by the volume. A junior drafter ignores the ceiling and minimizes cost using calculus, obtaining a height of about 1.92 m. The senior asks whether that design is acceptable and, if not, what the cheapest feasible dimensions are. Welds, material thickness, and cutting costs are ignored. The cylinder is closed, and both discs use the same material. The fabricator wants an exact cost comparison before approving a revised drawing.
1. Derive the unconstrained minimum-cost dimensions and check them against the height limit.
Read the worked answer
The volume constraint is [V=\pi r^2h=2.00,\qquad h=\frac{2}{\pi r^2}.] The cost is the disc cost plus the side cost: [C=500(2\pi r^2)+300(2\pi rh).] Substitute (h): [C=1000\pi r^2+600\pi r\cdot\frac{2}{\pi r^2}=1000\pi r^2+\frac{1200}{r}.] Differentiate: [\frac{dC}{dr}=2000\pi r-\frac{1200}{r^2}.] Set to zero: [2000\pi r^3=1200,\qquad r^3=\frac{1200}{2000\pi}=0.19099,\qquad r=0.576\ \text{m}.] Then [h=\frac{2}{\pi(0.576)^2}=1.92\ \text{m}.] The second derivative is positive, so this is a minimum. But (h=1.92\ \text{m}) exceeds the 1.50 m ceiling, so the unconstrained design is not feasible.
2. Find the cheapest feasible dimensions and the corresponding cost.
Read the worked answer
The height limit requires [h=\frac{2}{\pi r^2}\le1.50,] so [r^2\ge\frac{2}{1.50\pi}=0.4244,\qquad r\ge0.6515\ \text{m}.] The unconstrained cost curve decreases up to (r=0.576\ \text{m}) and increases afterward. On the feasible interval (r\ge0.6515\ \text{m}), the cost is therefore increasing, so the minimum occurs at the boundary: [r=0.6515\ \text{m},\qquad h=1.50\ \text{m}.] The cost is [C=1000\pi(0.4244)+\frac{1200}{0.6515}.] [C=1333+1842\approx\text{₹}3175.] The ceiling binds: the cheapest feasible tank is wider and shorter than the unconstrained optimum.
3. If the ceiling were raised to 2.00 m, how much money would the unconstrained design save, and why is the answer not found by simply setting h = 2.00 m?
Read the worked answer
At a 2.00 m ceiling, the unconstrained optimum (h=1.92\ \text{m}) becomes feasible because it is below the ceiling. Its cost is [C=1000\pi(0.576)^2+\frac{1200}{0.576}] [C=1000\pi(0.331)+2083=1040+2083\approx\text{₹}3123.] The saving relative to the feasible ceiling-limited design is about [3175-3123=\text{₹}52,] roughly 1.6%.
Setting (h=2.00\ \text{m}) would force a particular radius and would not minimize cost. Once the ceiling is above 1.92 m, the constraint is inactive, and the optimum remains the interior stationary point at (r=0.576\ \text{m}), not a boundary point.