 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ib/paper-3/6d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

Let

c > 0

denote the stated unit cost of cylindrical wall, and assume

V > 0

. The circular cylinder and hemisphere give the volume constraint and total cost

V = π r^{2} h + \frac{2}{3} π r^{3}, C = 2 π cr h + 4 π c r^{2} .

Only the walls and cap have specified costs, so no separate base cost is included. Eliminating

h

gives

h = \frac{V}{π r ^{2}} - \frac{2}{3} r, C (r) = \frac{2 c V}{r} + \frac{8}{3} π c r^{2},

with

0 < r \leq (3 V / (2 π))^{1/3}

from

h \geq 0

. Now

C^{'} (r) = - 2 c V / r^{2} + (16/3) π cr

vanishes at

r^{3} = 3 V / (8 π)

. Moreover

C^{''} (r) = 4 c V / r^{3} + (16/3) π c > 0

and the stationary point is feasible, so it is the unique global minimum. The optimal dimensions are

r = (\frac{3 V}{8 π})^{1/3}, h = 2 r .

 New to topics? Read the docs here!