Do circular wheels maximize mechanical efficiency?

The width is probably meant into the third dimension, like thickness? This means that we can just look at a 2D problem with a surface mass density $\rho$.

The question does not state whether there is friction involved. If there was no friction, every body would slide down exactly the same. This way the question is not interesting in any way.

Since all the “wheels” are supposed to have the surface area and width, they all have the same mass. So one can reword the question perhaps like this: If we redistribute the mass from a circular shape to something else, what changes? The inertia tensor will change. We are only interested in the one moment of inertia which is along the rotation axis. This is given by the following: $$ I = \iint \mathrm d^2 x \, \rho (\vec x) \, |\vec x^2| = \int \mathrm d x_1 \int \mathrm dx_2 \,\rho (\vec x) \, (x_1^2 + x_2^2) =\int \mathrm d r \, r\int \mathrm d \phi \,\rho (r, \phi) \, r^2 \,.$$ We assume $\rho$ to be constant within the surface and zero outside of it. And there you can see why the circle is the best option: If you have a compact circle, the factor $r^3$ will only become as large as the radius of the circle. If you distort the same surface area to an ellipse, for some angles $\phi$ you will not have that large of $r^3$, that “saves” you a bit of inertia. However, for other angles you need to integrate over $r$ further, and that $r^3$ bit becomes “expensive”.

So no matter how you deviate from the circle with the constraint that the total surface area is the same, you will have make $I$ larger than before. Therefore the circle has the smallest moment of inertia and can rotate the fastest given the same driving moment.

An ellipse with the same surface area has a larger circumference. Therefore it does not need to rotate as fast as the circle to cover the same distance. However, I think that the effect of $I$ becoming larger will outweigh the reduction in revolutions needed to reach the bottom of the slope.

A complication arises as the shape gets further from circular symmetry. When it gains sufficient angular velocity, any non-circular object can 'jump' off the incline. See A jumping cylinder on an inclined plane. On leaving the plane the object becomes a projectile. Whether this speeds the descent or not might be difficult to determine.

Another complication is that the circle is the only shape which has neutral stability. It will roll continuously on any incline, no matter how small the angle. For all other shapes, the stability is positive. On a horizontal plane or small incline it will rock about an equilibrium position. It will only start rolling continuously if the angle is large enough.

The statement is false as stated. The following shape will be faster:

I don't prove this fact, but it's hopefully clear that by choosing the distribution of holes I can make the average density as a function of $r$ the same as that of a sphere, which you know is faster.

You might have more luck if you restrict the shape to be convex. But to make a well-formed problem you still have to specify what you mean by "reaches the bottom of the incline faster". For example, if you mean the time taken for the contact point to move from the top to the bottom of the slope then a long stick with the same length of the slope which has nearly fallen over will clearly be the fastest. (This might sound like a joke, but it shows that if you want a mathematically rigorous solution you have to make a mathematically precise problem!)

There are other issues too... if there are no dissipative forces and the slope is long enough then any non-circular object will eventually start bouncing down the slope rather than rolling. How do you want to handle these cases?