Chapter 5.4, Problem 47E

To determine

To find the maximum volume of the angle.

Answer to Problem 47E

The maximum volume, the angle should be $\theta =\frac{\pi }{6}.$

Explanation of Solution

Given information: Given curves $1\text{ft and }1+2\sin \theta \text{ft}\text{.}$

Calculation:

Trapezoid of parallel side is $1\text{ft and }1+2\sin \theta \text{ft and}$ height $\text{cos}\theta .$

Volume of the trapezoid is:

  $\begin{array}{l}V=\frac{1}{2}\left(\cos \theta \right)\left(1+1+2\sin \theta \right)\left(20\right)\\ =\left(\cos \theta \right)\left(1+\sin \theta \right)\left(20\right)\end{array}$

For any function $f\left(x\right)$ the condition of minimum and maximum is obtained by find $f^{\prime }\left(x\right)=0$ ,

Critical points and end points,

  $\begin{array}{l}V=\left(\cos \theta \right)\left(1+\sin \theta \right)\left(20\right),0<\theta <\frac{\pi }{2}\\ \frac{dV}{d\theta }=20\left(\cos \theta \right)\left(\sin \theta \right)+20\left(1+\sin \theta \right)\left(-\sin \theta \right)\\ =20\left(1-{\sin }^2\theta \right)-20{\sin }^2\theta +20\sin \theta \\ =40-40{\sin }^2\theta +20\sin \theta \end{array}$

Critical points occur at

  $\begin{array}{l}{V}^{\prime }=0\\ 40-40{\sin }^2\theta +20\sin \theta =0\\ 2{\sin }^2\theta -\sin \theta -2=0\\ -20\left(2\sin \theta -1\right)\left(\sin \theta +1\right)=0\\ \sin \theta =-1,\sin \theta =\frac{1}{2}\\ \theta =\frac{\pi }{6},\theta =\frac{3\pi }{6}\end{array}$

  $\text{Since }0<\theta <\frac{\pi }{2}$ Critical point occurs at $\theta =\frac{\pi }{6}.$

Here $\frac{dV}{d\theta }>0\text{for}0<\theta <\frac{\pi }{6}\text{and }\frac{dV}{d\theta }<0\text{ for}\frac{\pi }{6}<\theta <\frac{\pi }{2}$ this corresponds to minimum distance.

For $\theta =\frac{\pi }{6}$

  $\begin{array}{l}V=\left(\cos \theta \right)\left(1+\sin \theta \right)\left(20\right)\\ =15\sqrt{3}\end{array}$

Thus, the maximum volume, the angle should be $\theta =\frac{\pi }{6}.$