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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 94 (page 496)

Constructing a Rain Gutter A rain gutter is to be constructed of aluminum sheets $12$ inches wide. After marking off a length of $4$ inches from each edge, this length is bent up at an angle $胃$ . See the illustration. The area $A$ of the opening as a function of $胃$ is given by
$A (胃) = 16 \sin 胃 (\cos 胃 + 1) 0 掳 < 胃 < 90 掳$
(a) In calculus, you will be asked to find the angle $胃$ that maximizes $A$ by solving the equation
$\cos (2 胃) + \cos 胃 = 0, 0 掳 < 胃 < 90 掳$
Solve this equation for $胃$
(b)What is the maximum area $A$ of the opening?
(c)Graph $A = A (胃), 0 掳鈮� 胃鈮� 90 掳$ , and find the angle $胃$ that maximizes the area $A$ . Also find the maximum area. Compare the results to the answer found earlier.

Short Answer

Expert verified

The value is $胃 = 60 掳$ , the graph is

Step by step solution

01

Part (a) Step 1: Given information

Given the area of opening as $A (胃) = 16 \sin 胃 (\cos 胃 + 1) 0 掳 < 胃 < 90 掳$

02

Part (a) Step 2: Use the trigonometric identity and calculate

Using identity, we get

$\cos (2 胃) + \cos 胃 = 0 - 2 \cos (\frac{2 胃 + 胃}{2}) \cos (\frac{2 胃 - 胃}{2}) = 0 - 2 \cos (\frac{3 胃}{2}) \cos (\frac{胃}{2}) = 0$

03

Part (a) Step 3: Solve separately

Solving separately, we get

$\cos (\frac{3 胃}{2}) = 0 \frac{3 胃}{2} = \arccos 0 \frac{3 胃}{2} = 90 掳, 270 掳胃 = 60 掳, 180 掳$

role="math" localid="1646474942226" $\cos (\frac{胃}{2}) = 0 \frac{胃}{2} = \arccos 0 \frac{胃}{2} = 90 掳, 270 掳胃 = 180 掳胃 = 60^{鈭�}$

04

Part (b) Step 1: Given information

Given the equation $A (胃) = 16 \sin 胃 (\cos 胃 + 1)$

05

Part (b) Step 2: Substituting value from eqn(1) and calculating

Calculating, we get

$A (胃) = 16 \sin 胃 (\cos 胃 + 1) A (60 掳) = 16 \sin 60 掳 (\cos 60 掳 + 1) A (60 掳) = 16 (\frac{\sqrt{3}}{2}) (\frac{1}{2} + 1) A (60 掳) = 16 (\frac{\sqrt{3}}{2}) (\frac{3}{2}) A (60 掳) = 12 \sqrt{3}$

06

Part (c) Step 1: Given information

Given $A = A (胃), 0 掳鈮� 胃鈮� 90 掳$

07

Part (c) Step 2: Sketching the graph

The graph is

As shown in the graph, the maximum area $A = 12 \sqrt{3}$ is achieved at $胃 = 60 掳$ , identical to the answer in (b).

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Most popular questions from this chapter

Rewrite in terms of sine and cosine functions: $c o t (胃) s e c (胃)$

Projectile Motion The range R of a projectile propelled downward from the top of an inclined plane at an angle $胃$ to the inclined plane is given by

$R (胃) = \frac{2 v_{0}^{2} \sin 鈦� 胃 \cos 鈦� (胃鈭� 蠒)}{g \cos^{2} 鈦� 蠒}$

where $v_{o}$ is the initial velocity of the projectile, $鈭�$ is the angle the plane makes with respect to the horizontal, and g is the acceleration due to gravity.

(a) Show that for fixed $v_{o}$ and $鈭�$ the maximum range down the incline is given by

$R_{max} = \frac{v_{0}^{2}}{g (1 鈭� \sin 鈦� 蠒)}$

(b) Determine the maximum range if the projectile has an initial velocity of $50$ meters/second, the angle of the plane is $鈭� = 35^{o}$ , and $g = 9.8$ meters/second².

Suppose that f and g are two functions with the same domain. If $f (x) = g (x)$ for every x in the domain, the equation is called a(n) ____. Otherwise, it is called a(n) _____ equation.

A rectangle is inscribed in a semicircle of radius $1$ . See the illustration.

Part (a): Express the area A of the rectangle as a function of the angle $胃$ shown in the illustration.

Part (b): Show that $A (胃) = \sin (2 胃)$ .

Part (c): Find the angle $胃$ that results in the largest area A.

Part (d): Find the dimensions of this largest rectangle.

In Problems 57鈥� 80, solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$

$\sec^{2} 胃 + \tan 胃 = 0$

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