一位德明政府中学（Dunman High）的校友计划捐款在正心园（Zheng Xin Yuan）中央建造两个锦鲤池。这两个池塘的形状分别为“D”和“H”，如下图所示（所有尺寸均以米为单位，且图表未按比例绘制）。 阴影区域表示池塘的水面。“D”形池塘由一个 $3h \times h$ 的矩形以及由两个半径分别为 $2r$ 和 $r$ 的同心半圆构成的弯曲区域组成。“H”形池塘由七个边长为 $h$ 的相同正方形组成。 证明这两个池塘的总表面积 $A$ m$^{2}$ 由下式给出： $A = 10h^{2} + \frac{3}{2}\pi r^{2}.$ 景观设计师建议 $rh^{2} = k$，其中 $k$ 为常数。随着 $r$ 和 $h$ 的变化，求 $A$ 取得驻值时 $\displaystyle \frac{r}{h}$ 的准确值。确定该驻点的性质，并在 $k = 1$ 时利用计算器求出 $A$ 的驻值。 曲线 $C$ 的方程为 $y = 2x + 5 + \frac{4}{x}.$ 通过求导，求出使 $C$ 上不存在对应点时 $y$ 的准确取值集合。

1. Find the surface area of the 'D' pond.

The 'D' pond consists of a rectangle and a semicircular annulus.
$\text{Area of rectangle} = 3h \times h = 3h^2$
$\text{Area of curved region} = \frac{1}{2}\pi(2r)^2 - \frac{1}{2}\pi r^2 = 2\pi r^2 - \frac{1}{2}\pi r^2 = \frac{3}{2}\pi r^2$
$\text{Area of 'D' pond} = 3h^2 + \frac{3}{2}\pi r^2$

2. Find the surface area of the 'H' pond and sum them up to show the final equation.

The 'H' pond consists of 7 identical squares of side $h$.
$\text{Area of 'H' pond} = 7 \times h^2 = 7h^2$

$\text{Total Area } A = \text{Area of 'D' pond} + \text{Area of 'H' pond}$
$A = \left(3h^2 + \frac{3}{2}\pi r^2\right) + 7h^2$
$A = 10h^2 + \frac{3}{2}\pi r^2 \text{ (shown)}$

1. Express $A$ in terms of a single variable $r$.

Given $rh^2 = k$, we have $h^2 = \frac{k}{r}$.
Substitute this into $A = 10h^2 + \frac{3}{2}\pi r^2$:
$A = 10\left(\frac{k}{r}\right) + \frac{3}{2}\pi r^2 = 10kr^{-1} + \frac{3}{2}\pi r^2$

2. Differentiate $A$ with respect to $r$ and find the condition for a stationary point.

$\frac{\mathrm{d}A}{\mathrm{d}r} = -10kr^{-2} + 3\pi r$
For a stationary value, $\frac{\mathrm{d}A}{\mathrm{d}r} = 0$:
$3\pi r - \frac{10k}{r^2} = 0$
$3\pi r^3 = 10k$

3. Find the exact value of $\frac{r}{h}$.

Substitute $k = rh^2$ back into the stationary condition:
$3\pi r^3 = 10(rh^2)$
Since lengths must be non-zero ($r > 0$), divide both sides by $r$:
$3\pi r^2 = 10h^2$
$\frac{r^2}{h^2} = \frac{10}{3\pi}$
Since lengths $r$ and $h$ must be positive:
$\frac{r}{h} = \sqrt{\frac{10}{3\pi}}$

4. Determine the nature of the stationary point using the second derivative.

$\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = 20kr^{-3} + 3\pi = \frac{20k}{r^3} + 3\pi$
Since $k > 0$ and $r > 0$, $\frac{\mathrm{d}^2A}{\mathrm{d}r^2} > 0$.
Therefore, the stationary point is a minimum.

5. Evaluate the stationary value of $A$ when $k = 1$.

When $k = 1$, the stationary condition $3\pi r^3 = 10$ gives:
$r = \sqrt[3]{\frac{10}{3\pi}} \approx 1.0198$
Substitute $k=1$ and $r \approx 1.0198$ into the expression for $A$:
$A = \frac{10}{r} + \frac{3}{2}\pi r^2$
$A \approx \frac{10}{1.0198} + \frac{3}{2}\pi (1.0198)^2$
$A \approx 9.8058 + 4.9029 = 14.7087$
$A \approx 14.7 \text{ (to 3 s.f.)}$

1. Find the first derivative of the given function.

Rewrite the equation as:
$y = 2x + 5 + 4x^{-1}$

Differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = 2 - 4x^{-2} = 2 - \frac{4}{x^2}$

2. Equate the derivative to zero to find the $x$-coordinates of the stationary points.

At the stationary points, $\frac{dy}{dx} = 0$:
$2 - \frac{4}{x^2} = 0$
$\frac{4}{x^2} = 2$
$x^2 = 2$
$x = \sqrt{2} \text{ or } x = -\sqrt{2}$

3. Calculate the corresponding $y$-coordinates of the stationary points.

Substitute the $x$-values into the original equation to find the corresponding $y$-values:

When $x = \sqrt{2}$:
$y = 2(\sqrt{2}) + 5 + \frac{4}{\sqrt{2}} = 2\sqrt{2} + 5 + 2\sqrt{2} = 5 + 4\sqrt{2}$

When $x = -\sqrt{2}$:
$y = 2(-\sqrt{2}) + 5 + \frac{4}{-\sqrt{2}} = -2\sqrt{2} + 5 - 2\sqrt{2} = 5 - 4\sqrt{2}$

4. Use the second derivative to determine the nature of the stationary points and define the empty range for $y$.

Use the second derivative test:
$\frac{d^2y}{dx^2} = 8x^{-3} = \frac{8}{x^3}$

When $x = \sqrt{2}$, $\frac{d^2y}{dx^2} = \frac{8}{(\sqrt{2})^3} > 0$, meaning the curve has a local minimum at $( \sqrt{2}, 5 + 4\sqrt{2} )$.
When $x = -\sqrt{2}$, $\frac{d^2y}{dx^2} = \frac{8}{(-\sqrt{2})^3} < 0$, meaning the curve has a local maximum at $( -\sqrt{2}, 5 - 4\sqrt{2} )$.

The function exhibits two separate continuous branches (for $x > 0$ and $x < 0$). The curve only exists for $y \ge 5 + 4\sqrt{2}$ and $y \le 5 - 4\sqrt{2}$.

Therefore, there are no points on $C$ for values of $y$ strictly between the local maximum and minimum:
$5 - 4\sqrt{2} < y < 5 + 4\sqrt{2}$