A gutter with trapezoidal cross section is to be made from a long sheet of tin 8 in wide by turning up one third of its width on each side , what width a cross the top that will give a maximum capacity

Answer 1

Let the sides be turned up through an angle θ. Then the depth of the gutter will be 8/3 sinθ, and the cross-section of the gutter will have area

a = ((8/3) + 2*(8/3)cosθ)/2 * (8/3)sinθ
= 32/9 (sinθ + sin2θ)

for maximum area, we want da/dθ = 0, so

cosθ + 2cos2θ = 0
4cos^2θ + cosθ - 2 = 0
cosθ = 0.593

So, the width across the top for maximum area = 8/3 (1+2*0.593) = 5.83

Hmm. I was expecting the rectangular portion of the trapezoid to be a square, but that turns out not to be so. Oh, well. Still, check my math...

Answer 2

Oops. I messed up the length of the top base. It is 8/3 + 2*(8/3)cosθ

So, the area is

((8/3) + 8/3 + 2*(8/3)cosθ)/2 * (8/3)sinθ
= 32/9 (2sinθ + sin2θ)

For max area we want
cosθ + cos2θ = 0
2cos^2θ + cosθ - 1 = 0
(2cosθ-1)(cosθ+1) = 0
cosθ = 1/2

So, the top base is 8/3 + 2(8/3)(1/2) = 16/3.

Aha! The rectangular portion of the trapezoid is indeed a square!