Chapter 5, Problem 118RE

To determine

To find:The volume of the trough as a function of $\theta$ and to find the value of $\theta$ such that the volume is maximum.

Answer to Problem 118RE

The required volume is $\frac{\sin \theta }{2}$ and the value of the angle is $90°$ .

Explanation of Solution

Given information:

Length of the trough for feeding cattle = 4 meters

Length of the two equal sides of the isosceles triangle $=\frac{1}{2}$ meter

The angle between the equal sides $=\theta$

Formula used:

The double angle formula:

  $\sin 2u=2\sin u\cos u$

Consider that the volume of the trough be V and length of the trough be h.

The figure of the cross-section isosceles triangle can be shown below:

  

From the above figure it is seen that $\cos \frac{\theta }{2}=\frac{\text{Adjacent}}{\text{Hypotenuse}}$ .

Substitute $h$ for Adjacent and $\frac{1}{2}$ for Hypotenuse in the above expression.

  $\begin{array}{c}\cos \frac{\theta }{2}=\frac{h}{\frac{1}{2}}\\ \cos \frac{\theta }{2}\times \frac{1}{2}=\frac{h}{\cancel{\frac{1}{2}}}\times \cancel{\frac{1}{2}}\\ \frac{1}{2}\cos \frac{\theta }{2}=h\end{array}$

Also, $\sin \frac{\theta }{2}=\frac{\text{Opposite}\text{side}}{\text{Hypotenuse}}$

Substitute $\frac{b}{2}$ for Opposite side and $\frac{1}{2}$ for Hypotenuse in the above expression.

  $\begin{array}{c}\sin \frac{\theta }{2}=\frac{\frac{b}{2}}{\frac{1}{2}}\\ \sin \frac{\theta }{2}=\frac{b}{\cancel{2}}\times \frac{\cancel{2}}{1}\\ \sin \frac{\theta }{2}=b\end{array}$

Know that the surface area of an isosceles triangle is, $A=\frac{b\times h}{2}$ .

Substitute $\sin \frac{\theta }{2}$ for b and $\frac{1}{2}\cos \frac{\theta }{2}$ for h in the above expression and simplify.

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

It is known that volume of the trough can be determined by the formula $V=A\times h$ .

Here, $h=4$

Substitute 4 for h and $\frac{1}{4}\sin \frac{\theta }{2}\cos \frac{\theta }{2}$ for A in the formula $V=A\times h$ and simplify.

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

Multiply and divide the right hand side of the above expression by 2.

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

Use the double angle formula on the above expression that is, $\sin 2u=2\sin u\cos u$ .

  $V=\frac{\sin \theta }{2}$

For the volume to be maximum the value of $\sin \theta$ has to be maximum that is, $\sin \theta =1$ .

From the expression $\sin \theta =1$ can be written as,

  $\begin{array}{c}\sin \theta =1\\ \cancel{{\sin }^{-1}}\left(\cancel{\sin }\theta \right)={\sin }^{-1}\left(1\right)\\ \theta =90°\end{array}$

Hence, the required volume is $\frac{\sin \theta }{2}$ and the value of the angle is $90°$ .