Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 19

Let $h=x^2+y^2+z-x$. Using Lagrange multipliers with $\nabla f=\lambda \nabla h$ givesThe third equation fixes $\lambda =1$. The first two then reduce to $2x-ay=1$ and $2y=ax$. If $a^2\ne 4$, their solution and the constraint giveAt $a=2$, the second equation requires $y=x$ and makes the first inconsistent; at $a=-2$, it requires $y=-x$ and gives the same inconsistency. Thus there is no constrained stationary point when $a=\pm 2$. The constraint gradient never vanishes because its vertical component is one, so the multiplier condition has not excluded any regular constrained stationary points.

Eliminate $z$ using the constraint: $z=x-x^2-y^2$. The restricted objective is the quadratic polynomial $F(x,y)=axy+x-x^2-y^2$, with constant Hessian matrixIts eigenvalues are $-2+a$ and $-2-a$. HenceFor $|a|<2$, both eigenvalues are negative and the quadratic polynomial is strictly concave, so the maximum is global on the entire constraint surface. For $|a|>2$, the opposite signs give increasing and decreasing directions. At $a=\pm 2$ there is no stationary point, as already found, rather than a degenerate extremum to classify. No parameter value produces a constrained local minimum.

Write ${ϵ}_k=E_0(k+\frac{1}{2})$ and use Lagrange multipliers $\alpha ,\beta$ for the particle-number and energy constraints. The stationary equations areThus $n_k=e^{-\alpha }{e}^{-\beta {ϵ}_k}$. Finite normalization over infinitely many levels requires $\beta >0$. Put $q=e^{-\beta E_0}$, so $0<q<1$. The geometric series givesThereforeFor the energy constraint, use $n_k=N(1-q)q^k$ and $\sum k{q}^k=q/(1-q{)}^2$, obtained by differentiating the geometric series. This yieldsRecognizing the hyperbolic cotangent givesThe strict positivity of all occupations requires $E>N{E}_0/2$; the paper's positivity assumptions alone do not guarantee this feasibility condition. For admissible energy the multiplier is uniquely $\beta =E_0^{-1}\ln [(2E+N{E}_0)/(2E-N{E}_0)]>0$. Equality $E=N{E}_0/2$ is only the limiting state with all occupation in level zero, outside the stipulated strictly positive family; lower energy is impossible.

For rigor beyond formal multiplier equations, $p_k=n_k/N$ has fixed mean $E/(N{E}_0)-1/2$. The objective differs from $N$ times its information entropy only by the fixed constant $N-N\ln N$. The maximum entropy distribution on the nonnegative integers is uniquely geometric at that mean, confirming the stationary family and its maximum character.