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Maximizing trapezoidal cross-sectional area

A metal sheet 12 inches wide is bent to form a trough with an isosceles trapezoidal cross-section. The sheet is bent into three equal 4-inch segments: the middle segment forms the flat bottom of the trough, and the two outer segments are bent upward at an angle θ\theta from the horizontal, forming the slanted sides. Each side segment has length 4 inches.

The cross-sectional area of the trough (a trapezoid with bottom base 4, and two slanted sides of length 4 making angle θ\theta with the horizontal, where 0<θ<π20 < \theta < \dfrac{\pi}{2}) is given by

A(θ)=16sin⁡θ(1+cos⁡θ).A(\theta) = 16\sin\theta(1+\cos\theta).

Find the angle θ\theta (in radians) that maximizes the cross-sectional area A(θ)A(\theta), and find the maximum area.

Hint

Differentiate A(θ)A(\theta) using the product rule, set A′(θ)=0A'(\theta) = 0, and simplify using the identity cos⁡2θ−sin⁡2θ=2cos⁡2θ−1\cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1. Then solve for θ\theta in (0,π/2)(0, \pi/2) and confirm it gives a maximum.

Step-by-step solution

Differentiate using the product rule: A′(θ)=16cos⁡θ(1+cos⁡θ)+16sin⁡θ(−sin⁡θ)=16(cos⁡θ+cos⁡2θ−sin⁡2θ)A'(\theta) = 16\cos\theta(1+\cos\theta) + 16\sin\theta(-\sin\theta) = 16(\cos\theta + \cos^2\theta - \sin^2\theta). Using sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1-\cos^2\theta, this simplifies to A′(θ)=16(2cos⁡2θ+cos⁡θ−1)A'(\theta) = 16(2\cos^2\theta + \cos\theta - 1).

Setting A′(θ)=0A'(\theta)=0 gives 2cos⁡2θ+cos⁡θ−1=02\cos^2\theta + \cos\theta - 1 = 0, which factors as (2cos⁡θ−1)(cos⁡θ+1)=0(2\cos\theta - 1)(\cos\theta+1) = 0, so cos⁡θ=1/2\cos\theta = 1/2 or cos⁡θ=−1\cos\theta = -1. Since 0<θ<π/20<\theta<\pi/2, only cos⁡θ=1/2\cos\theta = 1/2 applies, giving θ=π/3\theta = \pi/3. Checking sign changes of A′(θ)A'(\theta) (positive before, negative after) confirms a maximum, and substituting back gives A(π/3)=16sin⁡(π/3)(1+cos⁡(π/3))=16⋅32⋅32=123A(\pi/3) = 16\sin(\pi/3)(1+\cos(\pi/3)) = 16\cdot\frac{\sqrt3}{2}\cdot\frac{3}{2} = 12\sqrt3 square inches.

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