Chapter 4.4, Problem 47E

To determine

To find the angle.

Answer to Problem 47E

The angle for the maximum volume is $\theta =\frac{\pi }{6}$

Explanation of Solution

Given information : The trapezoid of parallel side is $1$ ft and $1+2\sin \theta$ ft and height $\cos \theta$ .

Formula used: $V=\frac{1}{2}(s_1)(s_2)(h)$

Let the trapezoid of parallel side is $1$ ft and $1+2\sin \theta$ ft and height $\cos \theta$ .

  $\begin{array}{l}V=\frac{1}{2}(s_1)(s_2)(h)\\ V=\frac{1}{2}(\cos \theta )(1+1+2\sin \theta )(20)\\ V=\frac{1}{2}(\cos \theta )(2+2\sin \theta )(20)\\ V=(\cos \theta )(1+\sin \theta )(20)\end{array}$

For any function $f(x)$ the condition of minimum and maximum is obtained by finding $f\text{'}(x)=0$ , critical points and end points

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

Substitute $f\text{'}(x)=0$ in the above equation

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

  $\theta =\frac{\pi }{6},0<\theta <\frac{\pi }{2}$

  $\begin{array}{l}0=-\pi (y+3)(y-1)\\ y=1,y=-3\end{array}$

The critical point occurs at $y=1$ .

  $\frac{dV}{d\theta }>0,0<\theta <\frac{\pi }{6}$ and $\frac{dV}{dr}<0,\frac{\pi }{6}<\theta <\frac{\pi }{2}$ .

This corresponds to maximum volume.

Substitute of $\theta =\frac{\pi }{6}$

  $\begin{array}{l}V=(\cos \theta )(1+\sin \theta )(20)\\ V=(\cos \frac{\pi }{6})(1+\sin \frac{\pi }{6})(20)\\ V=15\sqrt{3}\end{array}$

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