Chapter 7.6, Problem 94AYU

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, the builder bends this length up at an angle $\theta$. See the illustration. The area $A$ of the opening as a function of $\theta$ is given by

$A\left(\theta \right)\text{ = 16sin}\theta \left(\cos \theta \text{ + 1}\right)$, ${\text{0}}^{\circ }\text{ < }\theta {\text{ < 90}}^{\circ }$

(a) In calculus, you will be asked to find the angle that maximizes A by solving the equation

$\cos \left(2\theta \right)\text{ + cos}\theta {\text{ = 0, 0}}^{\circ }\text{ < }\theta {\text{ < 90}}^{\circ }$

Solve this equation for $\theta$.

(b) What is the maximum area $A$ of the opening?

(c) Graph $A\text{ = }A\left(\theta \right),{\text{ 0}}^{\circ }\le \theta \le {\text{ 90}}^{\circ }$ and find the angle $\theta$ that maximizes the area $A$. Also find the maximum area.

To determine

To find:

a. In calculus, you will be asked to find the angle $\theta$ that maximizes $A$ by solving the equation $\cos 2\theta +\cos \theta =0,0^{\circ }<\theta <{90}^{\circ }$. Solve this equation for $\theta$.

Answer to Problem 94AYU

a. $\theta =\frac{\pi }{3}$

Explanation of Solution

Given:

A rain gutter is to be constructed of aluminum sheets 12 inches wide. After marking off a length of 4 inches from each edge, the builder bends this length up at an angle $\theta$. See the illustration. The area $A$ of the opening as a function of $\theta$ is given by $A\left(\theta \right)=16\sin \theta \left(\cos \theta +1\right),0^{\circ }<\theta <{90}^{\circ }$.

Calculation:

a.

$\text{cos}2\theta =\text{ cos}2\theta -\text{ sin}2\theta$

$\text{cos}2\theta -\text{ sin}2\theta +\text{ cos}\theta =0$

$\text{sin}2\theta +\text{ cos}2\theta =1\Rightarrow \text{ sin}2\theta =1-\text{ cos}2\theta$

$\text{cos}2\theta -\left(1-\text{ cos}2\theta \right)+\text{ cos}\theta =0$

$\text{cos}2\theta -1+\text{ cos}2\theta +\text{ cos}\theta =0$

$2\text{cos}2\theta +\text{ cos}\theta -1=0$

$2\text{cos}2\theta -\text{ cos}\theta +2\text{cos}\theta -1=0$

$\text{cos}\theta \left(2\text{cos}\theta -1\right)+2\text{cos}\theta -1=0$

$\left(2\text{cos}\theta -1\right)\left(\text{cos}\theta +1\right)=0$

$2\text{cos}\theta -1=0,\text{ cos}\theta +1=0$

$\text{cos}\theta =\frac{1}{2},\text{ cos}\theta =-1$

$\theta =\frac{\pi }{3},\theta =\frac{5\pi }{3},\theta =\pi$

$\theta =\frac{\pi }{3}$

To determine

To find:

b. What is the maximum area $A$ of the opening?

Answer to Problem 94AYU

b. $12\sqrt{3}$

Explanation of Solution

Given:

A rain gutter is to be constructed of aluminum sheets 12 inches wide. After marking off a length of 4 inches from each edge, the builder bends this length up at an angle $\theta$. See the illustration. The area $A$ of the opening as a function of $\theta$ is given by $A\left(\theta \right)=16\sin \theta \left(\cos \theta +1\right),0^{\circ }<\theta <{90}^{\circ }$.

Calculation:

b. $A\left(\theta \right)=16\sin \theta \left(\cos \theta +1\right)$

$A\left(\theta \right)=16\sin \frac{\pi }{3}\left(\cos \frac{\pi }{3}+1\right)=\frac{16\sqrt{3}}{2}\left(\frac{1}{2}+1\right)=\frac{48}{4}\sqrt{3}=12\sqrt{3}$

To determine

To find:

c. Graph $A=A\left(\theta \right),0^{\circ }<\theta <{90}^{\circ }$, and find the angle $\theta$ that maximizes the area $A$. Also find the maximum area.

Answer to Problem 94AYU

c.

Explanation of Solution

Given:

A rain gutter is to be constructed of aluminum sheets 12 inches wide. After marking off a length of 4 inches from each edge, the builder bends this length up at an angle $\theta$. See the illustration. The area $A$ of the opening as a function of $\theta$ is given by $A\left(\theta \right)=16\sin \theta \left(\cos \theta +1\right),0^{\circ }<\theta <{90}^{\circ }$.

Calculation:

c.