Chapter 5, Problem 80RE

A.

To determine

Volume of trough as a function of $\frac{\theta }{2}$

Answer to Problem 80RE

Volume of trough is $\sin \frac{\theta }{2}\cos \frac{\theta }{2}$

Explanation of Solution

Given:

The given diagram is:

  

From the figure

  $\begin{array}{l}BD=AD\\ AB=2AD\mathrm{.....}\left(1\right)\end{array}$

In triangle ADC,

  $\begin{array}{l}\sin \frac{\theta }{2}=\frac{AD}{\frac{1}{2}}\\ AD=\frac{1}{2}\sin \frac{\theta }{2}\end{array}$

From (1)

  $\begin{array}{l}AB=2AD\\ AB=\sin \frac{\theta }{2}\mathrm{......}\left(2\right)\end{array}$

In triangle ADC,

  $\begin{array}{l}\cos \frac{\theta }{2}=\frac{CD}{\frac{1}{2}}\\ CD=\frac{1}{2}\cos \frac{\theta }{2}\mathrm{......}\left(3\right)\end{array}$

Now, area of triangle ABC,

  $\begin{array}{l}=\frac{1}{2}AB\times CD\\ =\frac{1}{2}\sin \frac{\theta }{2}\times \frac{1}{2}\cos \frac{\theta }{2}\\ =\frac{1}{4}\sin \frac{\theta }{2}\cos \frac{\theta }{2}\end{array}$

And volume = area $\times$ length

  $\begin{array}{l}=\frac{1}{4}\sin \frac{\theta }{2}\cos \frac{\theta }{2}\times 4\\ =\sin \frac{\theta }{2}\cos \frac{\theta }{2}\end{array}$

So, volume of trough is $\sin \frac{\theta }{2}\cos \frac{\theta }{2}$ .

B.

To determine

Volume of trough as a function of $\theta$ and value of $\theta$ for maximum volume.

Answer to Problem 80RE

Volume of trough is $\frac{1}{2}\sin \theta$ cubic meters

And it is maximum when $\theta =\frac{\pi }{2}$

Explanation of Solution

Given:

The given diagram is:

  

Consider the following identity

  $\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$

So, volume simplifies to

  $\begin{array}{l}\text{Volume}=\frac{1}{2}\left(2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right)\\ =\frac{1}{2}\sin \theta \end{array}$

Now, volume is maximum when $\sin \theta$ is maximum i.e. 1

  $\begin{array}{l}\sin \theta =1\\ \Rightarrow \theta =\frac{\pi }{2}\end{array}$

So, volume of trough is $\frac{1}{2}\sin \theta$ cubic meters

And it is maximum when $\theta =\frac{\pi }{2}$