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

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- Solution iii

Solution

The curved surface area excludes the two end disks, so

A = 2 π r x = 2 π tan α x (h - x)

. Completing the square gives

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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