Answer:
θ = 60°
Step-by-step explanation:
The cross sectional area of the trapezoid shape will be that of a trapezoid with bases of 10 cm and (10 cm + 2·(10 cm)·cos(θ)) and height (10 cm)·sin(θ).
That area in cm² is ...
A = (1/2)(b1 +b2)h = (1/2)(10 + (10 +20cos(θ))(10sin(θ)
A = 100sin(θ)(1 +cos(θ))
A graphing calculator shows this area to be maximized when ...
θ = π/3 radians = 60°
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<em>A</em> will be maximized when its derivative with respect to θ is zero. That derivative can be found to be 2cos(θ)² +cos(θ) -1, so the solution reduces to ...
cos(θ) = 1/2
θ = arccos(1/2) = π/3
