 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/a/solution

S = 2 π r^{2} + 2 π r h

and the volume constraint is

π r^{2} h = V

L = 2 π r^{2} + 2 π r h - λ (π r^{2} h - V) .

The stationary equations are

4 π r + 2 πh - 2 λπ r h = 0, 2 π r - λπ r^{2} = 0.

For

r > 0

, the second gives

λ = 2/ r

; the first then gives

h = 2 r

. The constraint yields

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

and

S_{m i n} = 6 π (\frac{V}{2 π})^{2/3} .

To verify a global minimum, eliminate

h

to obtain

S (r) = 2 π r^{2} + 2 V / r

. It tends to infinity at both ends of

r > 0

and has

S^{''} (r) = 4 π + 4 V / r^{3} > 0

, so its unique stationary point is the global minimum.

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