Answer:
θ = π/3 radians
Step-by-step explanation:
The average width* of the gutter will be ...
(10 cm)(1 + cos(θ))
and its depth will be ...
(10 cm)sin(θ)
so the cross-sectional area will be the product ...
area = (10 cm)²(1 +cos(θ))sin(θ)
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This will be a maximum when its derivative is zero:
d(area)/dθ = 0 = (100 cm²)(-sin(θ))sin(θ) +(1 +cos(θ))cos(θ)
0 = cos(θ) -sin(θ)² +cos(θ)² . . . . . . simplify, divide by 100 cm²
2cos(θ)² +cos(θ) -1 = 0 . . . . . . . use 1-cos² for sin²
(2cos(θ)-1)(cos(θ) +1) = 0 . . . . . . factor
The solution of interest is ...
cos(θ) = 1/2 ⇒ θ = π/3
The gutter will carry the maximum amount of water when the sides are bent up through an angles of π/3 radians.
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* The gutter is in the shape of a trapezoid. Its top (opening) dimension is (10 cm)(1+2cos(θ)), and its bottom width is 10 cm. The area of a trapezoid is half the sum of these values, multiplied by the height. Half the sum of the widths is ...
((10 +20cos(θ)) +10)/2 = 10 +10cos(θ) = 10(1 +cos(θ)) . . . . cm
