What is the most efficient shape of magnet? If you had a fixed volume (x millimetres cubed) of neodymium magnet 'material' of the equivalent strength of magnetism, what shape would produce the strongest poles? Cube? Disc? Long cylinder? 
Or is the strength more to do with how the magnet is magnetised during production?
Thanks
 A: An object where the size of the local magnetization $\mathbf{M}$ is fixed ($|\mathbf{M}| =M =$ Const.) can be thought of alternatively as an object where on the boundary a surface current $\mathbf{K}=\mathbf{M}\times \mathbf{n}$ flows. In the last formula, $\mathbf{n}$ is the local normal vector of the object's boundary. Additionally, an internal current $\mathbf{J}=\nabla \times \mathbf{M}$ will flow if the magnetization $\mathbf{M}$ varies in direction. $\mathbf{B}$ is a function of these two types of current via the Biot-Savart law.
Given a volume $V$ of a material with magnetization $M$, we may bound from below the strongest obtainable (local) magnetic field strength $B_{max}(M,V)$ by considering rod configurations. The previous discussion has revealed that a rod of certain length and radius which is magnetized along its axis produces the same field as a solenoid carrying a surface current $M$ going around the cylindrical boundary. A solenoid which with a smaller length $L$ (and larger cross-sectional area $A=V/L$) will not "concentrate" the magnetic field as good as a longer and slimmer solenoid. If we let $L^2/A=L^3/V$ grow large (towards the limit of an infinitely long solenoid) the field inside becomes $B=\mu_0 M$. So we have derived that for all volumes $V$
$$B_{max}(M,V)\geq \mu_0 M.$$
I think it can be argued that this inequality is in fact an equality (for all $V$) and the maximal field strength -or should we say: "supremal" field strength?- is obtained by taking these ever longer rod-shapes. I'll edit this answer when I have that question settled.
Remark: this answer of course presupposes a system size for which a continuum mechanics or mean-field approach to this question is appropriate.
A: If we take "equivalent strength of magnetism" to be local magnetization (and same direction) and "strongest pole" to be  total magnetic dipole moment, all volumes are equally as strong, because magnetic dipole moment is simply the sum over the entire volume of local magnetization.
A: The strength of a permanent magnet depends firstly from its constituents which are the magnetic dipole moments of the involved subatomic particles. In the same way two bar magnets have a stronger magnetic field all the aligned in the same direction magnetic dipole moments of the involved particles produce the common field.
But it is not possible to create a common field of any strength. Any material could be magnetized and any material gas his magnetic saturation behind which it is not possible to increase the magnetic field.

If you had a fixed volume (x millimetres cubed) of neodymium magnet 'material' of the equivalent strength of magnetism, what shape would produce the strongest poles...

From the above explained it seems to be clear that having a given amount of magnetic material in a first step one has to try to shape it in a long bar magnet and in the case one reach the saturation of its magnetic field then only to make the magnet broader.

Or is the strength more to do with how the magnet is magnetised during production?

Of course it has. The finer one could melt to powder the material of your choose the better one could align the magnetic dipole moments of the involved subatomic particles in an external magnetic field.
