 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/11/c/iii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

The curved surface area excludes the two end disks, so

A = 2 π r x = 2 π tan α x (h - x)

A = 2 π tan α [\frac{h ^{2}}{4} - (x - \frac{h}{2})^{2}] .

Thus

x = \frac{h}{2}, A_{m a x} = \frac{π h ^{2} tan α}{2} .

It differs from the volume optimum because the objective contains one power of the radius rather than two.

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