Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 19

Let the fixed positive volume be $V=abc$. For an interior constrained extremum, Lagrange multipliers applied to the energy giveMultiplying these equations by $a,b,c$ shows that $2A/a^2=2A/b^2=2A/c^2=\lambda V$. Since the lengths are positive, $a=b=c=V^{1/3}$.

The arithmetic-geometric mean inequality proves this is the global minimum:with equality exactly for equal sides. Thus the optimal shape is a cube, and $\boxed{E_{\min }=3A{V}^{-2/3}}$.

Put $q=2.4048$ and let the fixed constraint value be $C>0$. For $V=\pi R^{2}H$, the Lagrange multiplier equations giveDividing removes $\lambda$ and yieldsTo establish a minimum, rather than merely a stationary ratio, set $u=q^2/R^2$ and $v={\pi }^2/H^2$. Then $u+v=C$ and $V={\pi }^2{q}^2/(u\sqrt{v})$. The denominator squared is $u^2(C-u)$; its derivative is $u(2C-3u)$, giving a unique maximum at $u=2C/3$, $v=C/3$. The volume tends to infinity at either endpoint. This proves the stated ratio is the unique global minimum, as in the optimal cylinder for a reciprocal-square constraint.

A trigonometric simplification avoids repeated differentiation:for sufficiently small nonzero $x$. Continuity of cosine then gives

This is the defining difference quotient for the derivative of sine at $a$. ThereforeThe result holds for every real constant $a$.

For every fixed real $a$, the base is positive once $x>|a|$. Taking its logarithm givesThe Taylor series of the logarithm makes the bracket $2a/x+O(x^{-3})$, so this logarithm tends to $2a$. Continuity of the exponential givesFor $a=0$ the base is identically one, consistently giving the same answer.