Chapter 9.6, Problem 6MRPS

Solution

a.

To calculate:The value of $\theta$ maximize the volume of trough.

  $\theta ={90}^{\circ }$

Given information: Predict that $\theta ={90}^{\circ }$ because volume depends on area, area depends on sine, and sine has a maximum at ${90}^{\circ }$ will boost the volume.

Conclusion:

  $\theta ={90}^{\circ }$

b.

To calculate: The volume of the trough as a function of $\theta$ .

The body has a triangular prism form.

  $V=216.\sin \theta$

Given information: The area of one of the triangular faces.

Calculation:

Take note of the body's triangular prism-like shape.

The area of the base bb and the height $h$ of that prism can be multiplied to determine its volume $V.$

  $V=h.b$

Height is given equal $12feet$

  $V=12.b$

Area A of triangle found by using formula

  $\begin{array}{c}A=\frac{1}{2}ab\sin \theta \\ A=\frac{1}{2}\mathrm{.6.6.}\sin \theta \\ A=18.\sin \theta \end{array}$

Substitute $b=18.\sin \theta$ the formula volume.

  $\begin{array}{l}V=\mathrm{12.18.}\sin \theta \\ V=216.\sin \theta \end{array}$

Conclusion:

The body has a triangular prism form.

  $V=216.\sin \theta$

c.

To calculate:The volume changes as $\theta$ increases from $0^{\circ }$ to ${180}^{\circ }$

Volume increases from $0^{\circ }$ to ${90}^{\circ }$ . In ${90}^{\circ }$ reaches it’s maximum.

Given information: Function that describes the volume is

  $V=216\sin \theta$

Calculation:

$\begin{array}{l}\theta \\ 0^{\circ }\\ {30}^{\circ }\\ {45}^{\circ }\\ {60}^{\circ }\\ {90}^{\circ }\end{array}$

$\begin{array}{l}V\\ 0\\ 108\\ 152.74\\ 187.06\\ 216\end{array}$

$\begin{array}{l}\theta \\ {100}^{\circ }\\ {120}^{\circ }\\ {150}^{\circ }\\ {160}^{\circ }\\ {180}^{\circ }\end{array}$

$\begin{array}{l}V\\ 212.72\\ 187.06\\ 108\\ 73.88\\ 0\end{array}$

Take note of the table's indication that $V$ rises from $0$ to $90.$

Circumference $90.$ reaches its highest point.

The drop continues from ${90}^{\circ }$ to ${180}^{\circ }$ .

Conclusion:

Volume increases from $0^{\circ }$ to ${90}^{\circ }$ . In ${90}^{\circ }$ reaches it’s maximum.

d.

To calculate:The value of $\theta$ maximize the volume.

The outcome supports our prediction.

  $\theta ={90}^{\circ }$ ; as predicted

Given information: Using the table and earlier computations

Calculation:

The volume is highest when, it can be concluded $\theta ={90}^{\circ }$ .

Conclusion:

The outcome supports our prediction.

  $\theta ={90}^{\circ }$ ; as predicted