χ = b + 2d √ (n2 + 1).
Both Ω and χ are to be minima. Differentiating, and equating to zero.
(db/dd + n) d + b + nd = 0,
db/dd + 2 √ (n2 + 1) = 0;
eliminating db/dd,
{n − 2√ (n2 + 1)} d + b + nd = 0;
b = 2 {√ (n2 + 1) − n} d.
But
Ω / χ = (b + nd) d / {b + 2d √ (n2 + 1)}.
Inserting the value of b,
m = Ω/χ = {2d √ (n2 + 1) − nd} / {4d √ (n2 + 1) − 2nd} = 1⁄2 d.
That is, with given side slopes, the section is least for a given discharge when the hydraulic mean depth is half the actual depth.