Posted by soffy on .
The cross-section of a drain is a trapezoid. The sides and the bottom of the trapezoid each have a length of 10 feet. Determine the angle θ such that the drain will have maximal cross-sectional area.
For lack of specificity, I will let θ be the angle between the upper base and a side. I expect it will end up 90°, making a square cross-section, but let's wee what happens.
The height of the trapezoid will be 10sinθ, so the area is
a = 10sinθ (10+10+2*10cosθ)
= 10sinθ (20+20cosθ)
= 200sinθ (1+cosθ)
da/dθ = 200cosθ (1+cosθ) + 200sinθ(-sinθ)
= 200(cos^2θ + cosθ - sin^2θ)
da/dθ = 0 when cosθ = 1/2 or -1.
Looks like the max area occurs at θ=π/3
Looks like my gut feeling was wrong. Good thing, as it would have been a boring trapezoid.
calculus - correction -
Oops. I forgot to divide by 2 when getting the average of the bases, but that does not affect the answer in this case.