深度为 $d$ 的输水通道的横截面 $PQRS$ 如下图所示。已知 $PQ+QR+RS=2a$，且 $PQ$ 和 $RS$ 各自与直线 $QR$ 的夹角均为 $\theta$。

1. Express the slant lengths $PQ$ and $RS$, and the base $QR$ in terms of $d$, $a$, and $\theta$.

From right-angled trigonometry, the slant height $PQ = RS = \frac{d}{\sin\theta} = d\csc\theta$.
Given $PQ + QR + RS = 2a$:
$d\csc\theta + QR + d\csc\theta = 2a$
$QR = 2a - 2d\csc\theta$

2. Express the cross-sectional area as the sum of a central rectangle and two side right-angled triangles.

Area of the central rectangle $= QR \times d = (2a - 2d\csc\theta)d$

The horizontal projection (base) of the slant side $PQ$ is $\frac{d}{\tan\theta} = d\cot\theta$.
Area of the two identical side right-angled triangles $= 2 \times \left( \frac{1}{2} \times d\cot\theta \times d \right) = d^2\cot\theta$

3. Sum the areas and factorise to show the final result.

Total Area $A = \text{Area of rectangle} + \text{Area of triangles}$
$A = (2a - 2d\csc\theta)d + d^2\cot\theta$
$A = 2ad - 2d^2\csc\theta + d^2\cot\theta$
$A = 2ad + d^2(\cot\theta - 2\csc\theta)$

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

Given $A = 2ad + d^2(\cot\theta - 2\csc\theta)$, we differentiate $A$ with respect to $\theta$:
$\frac{dA}{d\theta} = d^2(-\csc^2\theta - 2(-\csc\theta\cot\theta))$
$\frac{dA}{d\theta} = d^2(2\csc\theta\cot\theta - \csc^2\theta)$

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

To find the maximum area, set $\frac{dA}{d\theta} = 0$. We can factor the derivative:
$d^2(2\csc\theta\cot\theta - \csc^2\theta) = 0$
$d^2\csc^2\theta \left( \frac{2\cos\theta}{\sin\theta} \cdot \sin\theta - 1 \right) = 0$
$d^2\csc^2\theta (2\cos\theta - 1) = 0$
Since $0 < \theta < \pi$ for an interior angle of the cross-section, $\csc\theta \neq 0$.
$2\cos\theta - 1 = 0 \implies \cos\theta = \frac{1}{2}$
$\theta = \frac{\pi}{3}$

3. Verify the nature of the stationary point using the second derivative test.

Using the factored form $\frac{dA}{d\theta} = d^2\csc^2\theta(2\cos\theta - 1)$, we apply the product rule to find the second derivative:
$\frac{d^2A}{d\theta^2} = d^2 \left[ -2\csc^2\theta\cot\theta(2\cos\theta - 1) + \csc^2\theta(-2\sin\theta) \right]$
At $\theta = \frac{\pi}{3}$, $2\cos\theta - 1 = 0$, so the first term vanishes:
$\frac{d^2A}{d\theta^2} = d^2 \left[ 0 + \left(\csc^2\frac{\pi}{3}\right)\left(-2\sin\frac{\pi}{3}\right) \right]$
$\frac{d^2A}{d\theta^2} = d^2 \left( \frac{4}{3} \right) (-\sqrt{3}) = -\frac{4\sqrt{3}}{3}d^2$
Since $d^2 > 0$, $\frac{d^2A}{d\theta^2} < 0$, confirming that $A$ is maximized when $\theta = \frac{\pi}{3}$.

4. Substitute the optimal angle back into the area function to find the maximum value.

Substitute $\theta = \frac{\pi}{3}$ into the original equation for $A$:
$A_{\max} = 2ad + d^2\left(\cot\frac{\pi}{3} - 2\csc\frac{\pi}{3}\right)$
$A_{\max} = 2ad + d^2\left( \frac{1}{\sqrt{3}} - 2\left(\frac{2}{\sqrt{3}}\right) \right)$
$A_{\max} = 2ad + d^2\left( \frac{1 - 4}{\sqrt{3}} \right)$
$A_{\max} = 2ad - \frac{3}{\sqrt{3}}d^2$
$A_{\max} = 2ad - \sqrt{3}d^2$