= Solution
The first <Brillouin zone> is the square
$$
-\frac\pi a\leq k_x<\frac\pi a,\qquad -\frac\pi a\leq k_y<\frac\pi a.
$$
At half filling, translation of the <wavevector> by $\mathbf Q=(\pi/a,\pi/a)$ exchanges the two band <energies>:
$$
E(\mathbf k+\mathbf Q)-\epsilon_0=-[E(\mathbf k)-\epsilon_0].
$$
Thus the same energy-pair argument as in part c gives
$$
\boxed{\mu(T)=\epsilon_0.}
$$
At zero <temperature>, the <Fermi surface> satisfies $\cos(k_xa)+\cos(k_ya)=0$. Within the square <Brillouin zone> this is the diamond
$$
\boxed{|k_x|+|k_y|=\frac\pi a,}
$$
whose vertices are $(\pm\pi/a,0)$ and $(0,\pm\pi/a)$. The occupied region is the diamond's interior, containing the band minimum at the zone centre. It has half the <Brillouin zone> area, consistent with one <electron> per atom and two <spin> states per <wavevector>.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2018-square-and-rectangular-fermi-surfaces.png]
{title=<Brillouin zones> and half-filled <Fermi surfaces> for square and rectangular models}
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