如图所示为 $3y = 7x + 6$ 和 $x + 5y = 12$ 图像的一部分。$T$ 是 $3y = 7x + 6$ 上的一点，且 $U$ 是 $x + 5y = 12$ 上的一点。$V$ 和 $W$ 是 $x$ 轴上的两个点，使得 $TUVW$ 构成一个矩形。 已知 $TW = h$ 个单位。

1. Identify the $y$-coordinates of $T$ and $U$

Since $TUVW$ is a rectangle with $W$ and $V$ on the $x$-axis, $TW$ is vertical and $TU$ is horizontal. Given $TW = h$, the $y$-coordinate of both $T$ and $U$ is $h$.

2. Find the $x$-coordinate of $T$

Substitute $y = h$ into the equation for $T$, $3y = 7x + 6$:
$3h = 7x + 6$
$7x = 3h - 6$
$x = \frac{3h - 6}{7}$

3. Find the $x$-coordinate of $U$

Substitute $y = h$ into the equation for $U$, $x + 5y = 12$:
$x + 5h = 12$
$x = 12 - 5h$

1. Find the $x$-coordinates of $T$ and $U$ from part (i)

Since $T$ and $U$ have $y$-coordinate $h$ (as $TW = h$ and $TUVW$ is a rectangle on the $x$-axis):
For $T$: $3(h) = 7x_T + 6 \implies x_T = \frac{3h - 6}{7}$
For $U$: $x_U + 5(h) = 12 \implies x_U = 12 - 5h$

2. Express the width of the rectangle $WV$ in terms of $h$

Width $WV = x_U - x_T$
$WV = (12 - 5h) - \left(\frac{3h - 6}{7}\right)$

3. Calculate the area of the rectangle and simplify to show the given expression

Area $A = \text{width} \times \text{height} = WV \times h$
$A = \left[ 12 - 5h - \frac{3h - 6}{7} \right] h$
$A = \left[ \frac{84 - 35h - 3h + 6}{7} \right] h$
$A = \left[ \frac{90 - 38h}{7} \right] h$
$A = \frac{90}{7}h - \frac{38}{7}h^2$ (Shown)

1. Differentiate $A$ with respect to $h$

$\frac{\mathrm{d}A}{\mathrm{d}h} = \frac{90}{7} - \frac{76}{7}h$

2. Find the value of $h$ at the stationary point

For a stationary value, $\frac{\mathrm{d}A}{\mathrm{d}h} = 0$.
$\frac{90}{7} - \frac{76}{7}h = 0$
$76h = 90$
$h = \frac{45}{38}$

3. Calculate the stationary value of $A$

$A = \frac{90}{7}\left(\frac{45}{38}\right) - \frac{38}{7}\left(\frac{45}{38}\right)^2$
$A = \frac{4050}{266} - \frac{2025}{266} = \frac{2025}{266} \text{ (or } 7\frac{163}{266} \text{)}$

4. Determine the nature of the stationary value using the second derivative test

$\frac{\mathrm{d}^2A}{\mathrm{d}h^2} = -\frac{76}{7}$
Since $\frac{\mathrm{d}^2A}{\mathrm{d}h^2} < 0$, the stationary value is a maximum.