 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2/19/a/solution

Let the fixed positive volume be

V = ab c

. For an interior constrained extremum, Lagrange multipliers applied to the energy give

- \frac{2 A}{a ^{3}} + λb c = 0, - \frac{2 A}{b ^{3}} + λa c = 0, - \frac{2 A}{c ^{3}} + λab = 0.

Multiplying these equations by

a, b, c

shows that

2 A / a^{2} = 2 A / b^{2} = 2 A / c^{2} = λV

. Since the lengths are positive,

a = b = c = V^{1/3}

\frac{1}{a ^{2}} + \frac{1}{b ^{2}} + \frac{1}{c ^{2}} \geq 3 (\frac{1}{a ^{2} b ^{2} c ^{2}})^{1/3} = 3 V^{- 2/3},

with equality exactly for equal sides. Thus the optimal shape is a cube, and

E_{m i n} = 3 A V^{- 2/3}

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