[Omega]/[chi] = (b + nd)d/{b + 2d [root](n² + 1)}.
Inserting the value of b,
m = [Omega]/[chi] = {2d[root](n² + 1) - nd}/
{4d [root](n² + 1) - 2nd} = ½d.
That is, with given side slopes, the section is least for a given
discharge when the hydraulic mean depth is half the actual depth.
A simple construction gives the form of the channel which fulfils this
condition, for it can be shown that when m = ½d the sides of the
channel are tangential to a semicircle drawn on the water line.