[已知半径为 $r$、圆心角为 $\theta$ 的圆扇形面积为 $\frac{1}{2}r^2\theta$。] 一位花商拥有一个形状为以 $O$ 为圆心的圆扇形 $OPQ$ 的温室，并打算围出一块矩形区域来种植玫瑰。 花园在下图中由一个固定的扇形 $OPQ$ 表示，其中 $OP = OQ = 10\text{ m}$ 且 $\angle POQ = \frac{\pi}{3}$ 弧度。矩形区域 $WXYZ$ 内接于扇形 $OPQ$，使得 $\angle ZOP = \theta$ 弧度。

1. Find the lengths of $ZY$ and $OY$ in terms of $\theta$.

In the right-angled $\triangle OZY$, $OZ = 10$ m (since $Z$ lies on the arc of the sector with radius 10) and $\angle ZOY = \theta$.
$\sin \theta = \frac{ZY}{10} \implies ZY = 10\sin\theta$
$\cos \theta = \frac{OY}{10} \implies OY = 10\cos\theta$

2. Find the length of $OX$ using $\triangle OWX$.

Since $WXYZ$ is a rectangle, $WX = ZY = 10\sin\theta$.
In the right-angled $\triangle OWX$, $\angle WOX = \frac{\pi}{3}$.
$\tan\left(\frac{\pi}{3}\right) = \frac{WX}{OX}$
$\sqrt{3} = \frac{10\sin\theta}{OX}$
$OX = \frac{10\sin\theta}{\sqrt{3}}$

3. Find the area of the rectangle and simplify to the required form using double angle identity.

The width of the rectangle is $XY = OY - OX = 10\cos\theta - \frac{10\sin\theta}{\sqrt{3}}$.
The area $A$ of the rectangle is given by $XY \times ZY$:
$A = \left(10\cos\theta - \frac{10\sin\theta}{\sqrt{3}}\right)(10\sin\theta)$
$A = 100\sin\theta\cos\theta - \frac{100}{\sqrt{3}}\sin^2\theta$
Using the identity $\sin 2\theta = 2\sin\theta\cos\theta$ and rationalising $\frac{100}{\sqrt{3}}$:
$A = 50(2\sin\theta\cos\theta) - 50\left(\frac{2\sqrt{3}}{3}\right)\sin^2\theta$
$A = 50\left(\sin 2\theta - \frac{2\sqrt{3}}{3}\sin^2\theta\right)$ (Shown)

1. Differentiate the area function $A$ with respect to $\theta$.

$A = 50\left(\sin 2\theta - \frac{2\sqrt{3}}{3}\sin^2\theta\right)$
$\frac{\mathrm{d}A}{\mathrm{d}\theta} = 50\left(2\cos 2\theta - \frac{2\sqrt{3}}{3}(2\sin\theta\cos\theta)\right)$
Using the double angle identity $2\sin\theta\cos\theta = \sin 2\theta$:
$\frac{\mathrm{d}A}{\mathrm{d}\theta} = 50\left(2\cos 2\theta - \frac{2\sqrt{3}}{3}\sin 2\theta\right)$

2. Set the derivative to zero and solve for $\theta$.

For a stationary point, $\frac{\mathrm{d}A}{\mathrm{d}\theta} = 0$:
$50\left(2\cos 2\theta - \frac{2\sqrt{3}}{3}\sin 2\theta\right) = 0$
$2\cos 2\theta = \frac{2\sqrt{3}}{3}\sin 2\theta$
$\tan 2\theta = \frac{2}{\frac{2\sqrt{3}}{3}} = \frac{6}{2\sqrt{3}} = \sqrt{3}$
Since the rectangle is inside a sector of angle $\frac{\pi}{3}$, $0 < \theta < \frac{\pi}{3}$, so $0 < 2\theta < \frac{2\pi}{3}$.
$2\theta = \frac{\pi}{3} \implies \theta = \frac{\pi}{6}$

3. Perform the second derivative test to verify it is a maximum point.

$\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} = 50\left(-4\sin 2\theta - \frac{4\sqrt{3}}{3}\cos 2\theta\right)$
When $\theta = \frac{\pi}{6}$ (so $2\theta = \frac{\pi}{3}$):
$\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} = 50\left(-4\sin\frac{\pi}{3} - \frac{4\sqrt{3}}{3}\cos\frac{\pi}{3}\right)$
$\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} = 50\left(-4\left(\frac{\sqrt{3}}{2}\right) - \frac{4\sqrt{3}}{3}\left(\frac{1}{2}\right)\right)$
Since both terms are negative, $\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} < 0$. Hence, the area is maximised when $\theta = \frac{\pi}{6}$.

4. Calculate the exact maximum area.

Substitute $\theta = \frac{\pi}{6}$ into the expression for $A$:
$A = 50\left(\sin\left(2\left(\frac{\pi}{6}\right)\right) - \frac{2\sqrt{3}}{3}\sin^2\left(\frac{\pi}{6}\right)\right)$
$A = 50\left(\sin\frac{\pi}{3} - \frac{2\sqrt{3}}{3}\left(\frac{1}{2}\right)^2\right)$
$A = 50\left(\frac{\sqrt{3}}{2} - \frac{2\sqrt{3}}{3}\left(\frac{1}{4}\right)\right)$
$A = 50\left(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{6}\right)$
$A = 50\left(\frac{3\sqrt{3}}{6} - \frac{\sqrt{3}}{6}\right) = 50\left(\frac{2\sqrt{3}}{6}\right) = \frac{50\sqrt{3}}{3}$

1. Formulate the expression for the total cost $C$ in terms of the area $A$.

The area of the sector $OPQ$ is $\frac{1}{2}r^2\alpha = \frac{1}{2}(10)^2\left(\frac{\pi}{3}\right) = \frac{50\pi}{3}$.

The area left for marigolds is the remaining area: $\frac{50\pi}{3} - A$.

The total cost $C$ is given by:
$C = 5A + 4\left(\frac{50\pi}{3} - A\right) = A + \frac{200\pi}{3}$

2. Find the coordinates of the turning point using the maximum value of $A$.

From part (ii), the maximum value of $A$ occurs at $\theta = \frac{\pi}{6}$ and is $A_{\text{max}} = \frac{50\sqrt{3}}{3}$.

Substituting this into the expression for $C$ gives the maximum cost:
$C_{\text{max}} = \frac{50\sqrt{3}}{3} + \frac{200\pi}{3} = \frac{50\sqrt{3} + 200\pi}{3}$

Hence, the turning point is a maximum at $\left(\frac{\pi}{6}, \frac{50\sqrt{3} + 200\pi}{3}\right)$.

3. Determine the domain of $\theta$ and the coordinates of the end points.

Based on the geometry of the sector, the inscribed rectangle is constrained by $0 \le \theta \le \frac{\pi}{3}$.

At $\theta = 0$, $A = 0$, so $C = \frac{200\pi}{3}$.
At $\theta = \frac{\pi}{3}$, evaluating $A$ gives $50\left(\sin\frac{2\pi}{3} - \frac{2\sqrt{3}}{3}\sin^2\frac{\pi}{3}\right) = 50\left(\frac{\sqrt{3}}{2} - \frac{2\sqrt{3}}{3}\left(\frac{3}{4}\right)\right) = 0$, so $C = \frac{200\pi}{3}$.

The end points are $\left(0, \frac{200\pi}{3}\right)$ and $\left(\frac{\pi}{3}, \frac{200\pi}{3}\right)$.

4. Describe the required sketch.

Draw a set of axes for $C$ against $\theta$. The graph is an inverted U-shape starting from the point $\left(0, \frac{200\pi}{3}\right)$, smoothly rising to the maximum turning point at $\left(\frac{\pi}{6}, \frac{50\sqrt{3} + 200\pi}{3}\right)$, and symmetrically falling back down to the end point at $\left(\frac{\pi}{3}, \frac{200\pi}{3}\right)$.

1. Formulate the total maintenance cost $C$ in terms of $\theta$ and set up the inequality.

The total area of the sector $OPQ$ is:
$\text{Area} = \frac{1}{2}r^2\alpha = \frac{1}{2}(10)^2\left(\frac{\pi}{3}\right) = \frac{50\pi}{3} \text{ m}^2$
The maintenance cost, $C$, involves $5/m$^2$ for the rectangular area $A$ and $4/m$^2$ for the remaining sector area:
$C = 5A + 4\left(\frac{50\pi}{3} - A\right) = A + \frac{200\pi}{3}$
To find when the cost is less than $220, we set up the inequality:
$A + \frac{200\pi}{3} < 220$
$50\left(\sin 2\theta - \frac{2\sqrt{3}}{3}\sin^2\theta\right) + \frac{200\pi}{3} < 220$

2. Solve for the boundary values of $\theta$ using a graphing calculator.

Using a graphing calculator, we find the values of $\theta$ that satisfy the equation $50\left(\sin 2\theta - \frac{2\sqrt{3}}{3}\sin^2\theta\right) + \frac{200\pi}{3} = 220$.
The points of intersection are:
$\theta \approx 0.11403 \quad \text{and} \quad \theta \approx 0.93316$

3. Determine the valid range of $\theta$ by interpreting the roots alongside the geometric domain.

The physical domain of the inscribed rectangle requires $\theta$ to be within $0 < \theta < \frac{\pi}{3}$.
From the shape of the graph of $C$ sketched in part (iii) (an inverted U-shape starting from a minimum value of $\approx 209.4$ at the boundaries), the cost is strictly less than $220 in the intervals outside the two roots.
Hence, restricting to 3 significant figures, the range of values for $\theta$ is:
$0 < \theta < 0.114 \quad \text{or} \quad 0.933 < \theta < \frac{\pi}{3}$