Chapter 7.3, Problem 105E

Sawing a Wooden Beam A rectangular beam is to be cut from a cylindrical log of diameter 20 in.

1. (a) Show that the cross-sectional area of the beam is modeled by the function

$A(\theta )=200\sin 2\theta$where θ is as shown in the figure.

1. (b) Show that the maximum cross-sectional area of such a beam is 200 in². [Hint: Use the fact that sin u achieves its maximum value at u = π/2.]

To determine

To show: The cross sectional area of the beam of diameter 20in is $A\left(\theta \right)=200\sin 2\theta$.

Explanation of Solution

Formula used:

Double-Angle Formulae:

Double-Angle Formulae:

Formula for sine: $\sin 2x=2\sin x\cos x$.

Proof:

Suppose the rectangle be named as ABCD.

Denote the diagonal of the rectangle as AC.

Consider the right angled triangle ABC, so by Pythagoras theorem, find the length l.

$\begin{array}{l}\cos \theta =\frac{BC}{AC}\\ \cos \theta =\frac{BC}{20}\\ 20\cos \theta =BC=l\end{array}$

Use Pythagoras theorem to find the breadth b.

$\begin{array}{l}\sin \theta =\frac{AB}{AC}\\ \sin \theta =\frac{AB}{20}\\ 20\sin \theta =AB=b\end{array}$

Compute the cross-sectional area of the beam of diameter 20in.

$\begin{array}{c}\text{Area}=lb\\ =\left(20\cos \theta \right)\left(20\sin \theta \right)\\ =400\cos \theta \sin \theta \left(\because \sin 2x=2\sin x\cos x\right)\\ =200\sin 2\theta \end{array}$

Hence, it is verified that the cross sectional area of the beam of diameter 20in is $A\left(\theta \right)=200\sin 2\theta$.

To determine

To show: The maximum cross-sectional area of such a beam is 200 ${\text{in}}^2$.

Explanation of Solution

Proof:

The cross sectional area of the beam of diameter 20in is $A\left(\theta \right)=200\sin 2\theta$.

Notice that $\sin 2\theta$ has the maximum value 1, when $\theta =\frac{\pi }{4}$.

Therefore, the maximum cross-sectional area of such a beam is ${\text{200in}}^2$, when $\theta =\frac{\pi }{4}$.