let the width of the drainage channel at the top be 2w

then Area of cross-section A = h(2w + 6)/0.5

Also from the trapezoid 6^2 = h^2 + (3-w)^2
h= ?(6^2 – (3-w)^2)

Substituting for h in A
A = (?(6^2 – (3-w)^2))*(2w+6)/2

To obtain max A find the value of x when A’ = 0
A’ = (-w)(2w + 6)/(?(6^2 – (3-w)^2)) + (?(6^2 – (3-w)^2)) = 0
A’ = -2w^2 -6w + 36 – (3 -w)^2
A’ = -2w^2 – 6w +36 – 9 – w^2 +6w = 0
A’ = -3w^2 + 27 = 0
w^2 = 9
w = 3

So the max area of cross section occurs x = 6

So the trapezoid is a square

Bet there are not many drainage channel engineers out there that could have replied to this.