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Chapter 4: Problem 48

A drainage channel is to be made so that its cross section is a trapezoid with equally sloping sides (Figure Ex-48). If the sides and bottom all have a length of \(5 \mathrm{ft}\), how should the angle \(\theta(0 \leq \theta \leq \pi / 2)\) be chosen to yield the greatest cross-sectional area of the channel?

Short Answer

Expert verified
The angle \( \theta \) should be approximately \( \frac{\pi}{4} \).

Step by step solution

01

Understand the Problem

We need to find the angle \( \theta \) that maximizes the cross-sectional area of a trapezoidal channel with sloping sides and a bottom, each having a length of \( 5 \) ft. The problem works within the condition that \( 0 \leq \theta \leq \frac{\pi}{2} \).
02

Define Variables and Trapezoid Geometry

Let's denote the lengths of the two sloping sides as \( 5 \sin \theta \) because the height of the trapezoid, given \( \theta \), is \( 5 \cos \theta \). The top of the trapezoid, \( b \), is \( 10 \sin \theta \) because the sloping sides form a mirror image and meet at the bottom.
03

Write the Area Formula

The formula for the area \( A \) of a trapezoid is \( A = \frac{1}{2} (b_1 + b_2)h \). Therefore, for this problem, \( A = \frac{1}{2} (5 + 10\sin\theta)(5\cos\theta) \). This simplifies to \( A = \frac{1}{2} (25\cos\theta + 50\sin\theta\cos\theta) \).
04

Simplify the Area Formula

Using the trigonometric identity \( 2\sin\theta\cos\theta = \sin(2\theta) \), the area formula becomes \( A = \frac{1}{2} (25\cos\theta + 25\sin(2\theta)) \). Further simplification gives \( A = \frac{25}{2} \cos\theta + \frac{25}{2} \sin(2\theta) \).
05

Calculate the Derivative

Find the derivative of the area function with respect to \( \theta \): \( A'(\theta) = \frac{d}{d\theta} \left( \frac{25}{2} \cos\theta + \frac{25}{2} \sin(2\theta) \right) = -\frac{25}{2} \sin\theta + 25 \cos(2\theta) \).
06

Solve for Critical Points

Set the derivative to zero to find critical points: \( -\frac{25}{2} \sin\theta + 25 \cos(2\theta) = 0 \). Simplifying gives \( \frac{1}{2} \sin\theta = \cos(2\theta) \). Use trigonometric identities or iterative/graphic solutions to find \( \theta \approx \frac{\pi}{4} \).
07

Verify Maximum Area at Critical Point

Evaluate the second derivative or compare with endpoints (\( \theta = 0 \) and \( \theta = \frac{\pi}{2} \)) to ensure maximum. The second derivative is \( A''(\theta) = -\frac{25}{2} \cos\theta - 50\sin(2\theta) \), confirming it changes sign across the critical point, indicating a maximum.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trapezoidal Channel
A trapezoidal channel is a type of water channel with two parallel sides and two non-parallel sides that slope from the bottom to the top of the channel. In this problem, the trapezoidal channel has sloping sides and a bottom, each with a length of 5 ft. To maximize the cross-sectional area, we need to understand the structure of the trapezoid. The bottom side remains fixed, while the sides slope upwards at an angle \( \theta \). This angle affects the length and shape of the top side of the trapezoid. As \( \theta \) changes, so does the area of the trapezoid. The goal is to find the optimal angle \( \theta \) that yields the greatest possible cross-sectional area. This involves a combination of geometry and optimization techniques.
Trigonometric Identities
Trigonometric identities are useful tools in solving problems involving angles and side lengths in geometry. In this exercise, we use these identities to simplify and solve the problem at hand. The critical identity used here is \( 2\sin\theta\cos\theta = \sin(2\theta) \). We use this to reformulate the area of the trapezoid. Before using the identity, the formula is expressed in terms of \( \cos\theta \) and \( \sin\theta \) separately. By substituting \( 2\sin\theta\cos\theta \) with \( \sin(2\theta) \), we simplify the expression, making it easier to differentiate and analyze. This clever use of identities is crucial for finding solutions to complex geometric problems efficiently.
Critical Points
Critical points are where the derivative of a function is zero or undefined, which helps find the maximum or minimum values of a function. In this problem, determining the critical points involves taking the derivative of the area function with respect to \( \theta \). After differentiating \( A(\theta) \), the result gives \( A'(\theta) = -\frac{25}{2} \sin\theta + 25 \cos(2\theta) \). To find the critical points, we set the derivative to zero: \( -\frac{25}{2} \sin\theta + 25 \cos(2\theta) = 0 \). Solving this equation using trigonometric identities and tools, we find \( \theta \approx \frac{\pi}{4} \). This value is then checked further to see if it gives a maximum area. This process of finding critical points and confirming maximum or minimum values is fundamental in calculus and applications of optimization.

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Most popular questions from this chapter

Let \(h\) and \(g\) have relative maxima at \(x_{0} .\) Prove or disprove: (a) \(h+g\) has a relative maximum at \(x_{0}\) (b) \(h-g\) has a relative maximum at \(x_{0}\).

(a) Use the Mean-Value Theorem to show that if \(f\) is differentiable on an open interval, and if \(\left|f^{\prime}(x)\right| \geq M\) for all values of \(x\) in the interval, then $$ |f(x)-f(y)| \geq M|x-y| $$ for all values of \(x\) and \(y\) in the interval. (b) Use the result in part (a) to show that $$ |\tan x-\tan y| \geq|x-y| $$ for all values of \(x\) and \(y\) in the interval \((-\pi / 2, \pi / 2)\). (c) Use the result in part (b) to show that $$ |\tan x+\tan y| \geq|x+y| $$ for all values of \(x\) and \(y\) in the interval \((-\pi / 2, \pi / 2)\).

Let \(f\) and \(g\) be continuous on \([a, b]\) and differentiable on \((a, b) .\) Prove: If \(f(a)=g(a)\) and \(f(b)=g(b)\), then there is a point \(c\) in \((a, b)\) such that \(f^{\prime}(c)=g^{\prime}(c)\).

A container with square base, vertical sides, and open top is to be made from \(1000 \mathrm{ft}^{2}\) of material. Find the dimensions of the container with greatest volume.

(a) Use the Mean-Value Theorem to show that if \(f\) is differentiable on an interval, and if \(\left|f^{\prime}(x)\right| \leq M\) for all values of \(x\) in the interval, then $$ |f(x)-f(y)| \leq M|x-y| $$ for all values of \(x\) and \(y\) in the interval. (b) Use the result in part (a) to show that $$ |\sin x-\sin y| \leq|x-y| $$ for all real values of \(x\) and \(y\).
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