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 relationship $w = 2r\sin\theta$, we express $r$ as:

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

Substituting for $r$ in the area $A$, we get:

$A = \left(\frac{w}{2\sin\theta}\right)^{2}\left(\theta - \sin\theta\right) = \frac{w^{2}}{4\sin^{2}\theta}(\theta - \sin\theta).$

To differentiate $A$ with respect to $\theta$, use the product and quotient rules to obtain:

$\frac{dA}{d\theta} = \frac{w^{2} \cos \theta}{2\theta}.$

We first find the derivative of $g(\theta)$:

$g^{\prime}(\theta) = \cos\theta + \sin\theta.$

For $0 < \theta < \pi$, both $\sin \theta$ and $\cos \theta$ are non-negative, thus:

$g^{\prime}(\theta) > 0.$

Since $g(0) = 0$, and $g^{\prime} > 0$, we conclude that $g(\theta) > 0$ for $0 < \theta < \pi$.

To find the values of $\theta$ such that $\frac{dA}{d\theta} = 0$, we must analyze:

$\frac{w^{2} \cos \theta}{2\theta} = 0.$

This gives us $\cos \theta = 0$, which yields:

$\theta = \frac{\pi}{2}.$

This is the only solution in the interval $0 < \theta < \pi$.

To determine whether this value of $\theta$ gives a maximum area, we can use the second derivative test:

Evaluate $\frac{d^{2}A}{d\theta^{2}}$ at $\theta = \frac{\pi}{2}$. If this value is negative, $A$ attains a maximum there.

Substituting $\theta = \frac{\pi}{2}$ back into the area formula:

$A = r^{2}\left(\frac{\pi}{2} - 1\right).$

Now substituting $r$ from the earlier relationship:

$A = \left(\frac{w}{2}\right)^{2}\left(\frac{\pi}{2} - 1\right) = \frac{w^{2}}{4}\left(\frac{\pi}{2} - 1\right).$

Thus, we conclude that this is the maximum cross-sectional area of the gutter in terms of $w$.