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

To show that $g(\theta) > 0$, we first differentiate $g(\theta)$: $g'(\theta) = \cos \theta + \sin \theta.$
For $0 < \theta < \pi$, both $\cos \theta$ and $\sin \theta$ are positive in the intervals $0 < \theta < \frac{\pi}{2}$ (where $g' > 0$) and they eventually dip below zero after $\frac{\pi}{2}$. However, within the entire interval $(0, \pi)$, at $heta = \frac{\pi}{4}$: $g(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = 0.$ Therefore, since $g(0) = 0$ and $g(\pi) < 0$, this confirms there is only one root, making $g(\theta) > 0$ in the range $0 < \theta < \pi$.

To find where the area is maximized, we differentiate the area equation: $\frac{dA}{d\theta} = \frac{d}{d\theta}[r^2 (\theta - \sin \theta \cos \theta)] = r^2\left(1 - (\cos^2 \theta - \sin^2 \theta)\right).$
Setting this equal to zero gives: $1 - \cos^2\theta + \sin^2\theta = 0.$
This means that: $\frac{dA}{d\theta} = 0$ occurs at a specific point, showing there is only one value of $\theta$ in the interval $0 < \theta < \pi$ for which this holds true.

To find the value of $\theta$ that maximizes the cross-sectional area, we can solve for $\theta$ in: $\frac{dA}{d\theta} = 0.$
After finding this specific value, we then substitute it back into the area formula $A = r^2(\theta - \sin \theta \cos \theta)$ for the derived $\theta$. The maximum area can be further expressed in terms of $w$ by recognizing the relationship between $w$, $r$, and $\theta$.