A gutter is to be formed by bending a long rectangular metal strip of width $w$ so that the cross-section is an arc of a circle - HSC - SSCE Mathematics Extension 1 - Question 7 - 2006 - Paper 1

From the geometry of the problem, the relationship between $r$, $w$, and $heta$ can be established as follows:

$w = 2r \sin \theta \, \Rightarrow \, r = \frac{w}{2 \sin \theta}.$

Now substituting this expression of $r$ into the area equation derived in part (a), we get:

$A = \bigg(\frac{w}{2 \sin \theta}\bigg)^2 \bigg(\theta - \frac{\text{sin} \theta}{2}\bigg).$

To find $\frac{dA}{d\theta}$, we can use product and quotient rules on the differentiated form. After differentiation, we shall simplify to obtain:

$\frac{dA}{d\theta} = \frac{w^2 \text{cos} \theta (\text{sin} \theta - \theta \text{cos} \theta)}{2\theta^2}.$

First, we express $g(\theta)$ as defined in the question:

$g(\theta) = \text{sin} \theta - \theta \text{cos} \theta.$

To find $g'(\theta)$, we differentiate:

$g'(\theta) = \text{cos} \theta - (\text{cos} \theta - \theta \text{sin} \theta) = \text{sin} \theta.$

Within the interval $0 < \theta < \pi$, we know that $g'(\theta) > 0$, which indicates that $g(\theta)$ is an increasing function. Evaluating at $\theta = 0$ gives $g(0) = 0$. Therefore,

$g(\theta) > 0 \, \text{for} \, 0 < \theta < \pi.$

To find critical points for $\frac{dA}{d\theta}$, we set it to zero:

$\frac{w^2 \text{cos} \theta (\text{sin} \theta - \theta \text{cos} \theta)}{2\theta^2} = 0.$

Since $w^2$ and $2\theta^2$ are always positive in the given interval, we can deduce:

$\text{cos} \theta = 0 \, \text{or} \, \text{sin} \theta - \theta \text{cos} \theta = 0.$

The first condition, $\text{cos} \theta = 0$, gives $\theta = \frac{\pi}{2}$. The second condition needs more inspection:

Solving $\text{sin} \theta = \theta \text{cos} \theta$ shows there is another root in the interval $0 < \theta < \pi$. Therefore, we conclude there is exactly one value of $\theta$ in this interval.

To determine the maximum area, we can analyze the second derivative sign test. If the value of $\theta = \frac{\pi}{2}$ gives us a maximum for $\frac{dA}{d\theta}$, we can substitute this value back into the area formula from part (a):

A_{max} = r^2 \bigg(\frac{\pi}{2} - \frac{1}{2}igg).

Substituting for $r$ in terms of $w$ gives us the final maximum cross-sectional area in terms of $w$.

Thus, the maximum area is:

$A_{max} = \frac{w^2}{4} \bigg(\frac{\pi}{2} - \frac{1}{2}\bigg).$