A Dunman High alumnus plans to donate money to build two koi ponds in the middle of Zheng Xin Yuan. The ponds are in the shape of "D" and "H" as shown below (all dimensions are in metres and the diagrams are not drawn to scale). The shaded areas represent the pond surface. The "D" pond is made up of a $3h$ by $h$ rectangle and a curved area formed from two concentric semi-circles of radii $2r$ and $r$ respectively. The "H" pond is made up of seven identical squares of side $h$. Show that the total surface area of the two ponds, $A$ m$^{2}$, is given by $A = 10h^{2} + \frac{3}{2}\pi r^{2}.$ The landscape designer proposes that $rh^{2} = k$, where $k$ is a constant. As $r$ and $h$ vary, find the exact value of $\displaystyle \frac{r}{h}$ at the stationary value of $A$. Determine the nature of this stationary point and use a calculator to evaluate the stationary value of $A$ if $k = 1$. The equation of a graph $C$ is given by $y = 2x + 5 + \frac{4}{x}.$ By differentiation, find the exact set of values of $y$ for which there are no points on $C$.

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:
$5 - 4\sqrt{2} < y < 5 + 4\sqrt{2}$