← Back to learning objective

Maximizing cross-sectional area of a trapezoidal gutter

A gutter is made by bending a 12-inch-wide sheet of metal into a trapezoidal shape. A strip of width 4 inches is left flat for the base, and the two remaining 4-inch-wide flaps on either side are folded upward at an angle θ\theta (measured from the vertical) to form the sides of the gutter. Let θ\theta be the angle each side makes with the vertical, so 0≤θ<π20 \le \theta < \tfrac{\pi}{2}.

The cross-section of the gutter is a trapezoid with a horizontal base of length 4 inches, and two slanted sides of length 4 inches each folded outward at angle θ\theta from the vertical. The cross-sectional area is given by

A(θ)=16sin⁡θ(1+cos⁡θ),0≤θ<π2.A(\theta) = 16\sin\theta(1+\cos\theta), \qquad 0 \le \theta < \frac{\pi}{2}.

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

Hint

Differentiate A(θ)=16sin⁡θ(1+cos⁡θ)A(\theta) = 16\sin\theta(1+\cos\theta) using the product rule: A′(θ)=16cos⁡θ(1+cos⁡θ)+16sin⁡θ(−sin⁡θ)A'(\theta) = 16\cos\theta(1+\cos\theta) + 16\sin\theta(-\sin\theta). Simplify to A′(θ)=16cos⁡θ+16cos⁡2θ−16sin⁡2θ=16cos⁡θ+16(2cos⁡2θ−1)A'(\theta) = 16\cos\theta + 16\cos^2\theta - 16\sin^2\theta = 16\cos\theta + 16(2\cos^2\theta - 1). Set A′(θ)=0A'(\theta) = 0 to get 32cos⁡2θ+16cos⁡θ−16=032\cos^2\theta + 16\cos\theta - 16 = 0, which simplifies to 2cos⁡2θ+cos⁡θ−1=02\cos^2\theta + \cos\theta - 1 = 0. Factor as (2cos⁡θ−1)(cos⁡θ+1)=0(2\cos\theta - 1)(\cos\theta + 1) = 0 to solve for cos⁡θ\cos\theta.

Step-by-step solution

Differentiate using the product rule: A′(θ)=16cos⁡θ(1+cos⁡θ)+16sin⁡θ(−sin⁡θ)=16cos⁡θ+16cos⁡2θ−16sin⁡2θA'(\theta) = 16\cos\theta(1+\cos\theta) + 16\sin\theta(-\sin\theta) = 16\cos\theta + 16\cos^2\theta - 16\sin^2\theta. Using the identity cos⁡2θ−sin⁡2θ=2cos⁡2θ−1\cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1, we have A′(θ)=16cos⁡θ+16(2cos⁡2θ−1)=16cos⁡θ+32cos⁡2θ−16A'(\theta) = 16\cos\theta + 16(2\cos^2\theta - 1) = 16\cos\theta + 32\cos^2\theta - 16. Setting A′(θ)=0A'(\theta) = 0 and dividing by 16 gives 32cos⁡2θ+16cos⁡θ−16=032\cos^2\theta + 16\cos\theta - 16 = 0, which simplifies to 2cos⁡2θ+cos⁡θ−1=02\cos^2\theta + \cos\theta - 1 = 0.

Factoring gives (2cos⁡θ−1)(cos⁡θ+1)=0(2\cos\theta - 1)(\cos\theta + 1) = 0, so cos⁡θ=12\cos\theta = \tfrac{1}{2} or cos⁡θ=−1\cos\theta = -1. Since 0≤θ<π/20 \le \theta < \pi/2, we have cos⁡θ>0\cos\theta > 0, so only cos⁡θ=1/2\cos\theta = 1/2 is valid, giving θ=π/3\theta = \pi/3. Checking the boundary (θ→0+\theta \to 0^+ gives A→0A \to 0) and the sign of A′(θ)A'(\theta) confirms this is a maximum. Substituting back: 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\tfrac{\sqrt{3}}{2}\cdot\tfrac{3}{2} = 12\sqrt{3} square inches.

Sign of the derivative just past the optimal angle

Hint
Step-by-step solution

Decreasing. Since θ=π/3\theta = \pi/3 is where the maximum area occurs, by the first derivative test, A′(θ)A'(\theta) is positive just before θ=π/3\theta = \pi/3 and negative just after. Thus the cross-sectional area is increasing before the optimal angle and decreasing after it. Therefore, just past the optimal angle, the area function is decreasing.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap