 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/19/b/solution

The base lies in the plane

z = R - h

. Its circular rim lies on the sphere, so

r^{2} + (R - h)^{2} = R^{2}, 0 \leq h \leq 2 R .

Maximize the cone volume

V_{c} = π r^{2} h /3

using the Lagrange multiplier expression

L = \frac{π}{3} r^{2} h - λ [r^{2} + (R - h)^{2} - R^{2}] .

For a nondegenerate cone, the

r

λ = πh /3

. The

h

π r^{2} /3 + 2 λ (R - h) = 0

, hence

r^{2} = 2 h (h - R)

. Combining this with

r^{2} = 2 R h - h^{2}

gives

h = \frac{4 R}{3}, r = \frac{2 2}{3} R .

Indeed, eliminating

r

gives

V_{c} (h) = π h^{2} (2 R - h) /3

, which vanishes at the endpoints and has a unique positive interior maximum at

4 R /3

. Thus

V_{c, m a x} = \frac{32 π R ^{3}}{81}, \frac{V _{c, m a x}}{( 4/3 ) π R ^{3}} = \frac{8}{27} .

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