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 $A = 10h^2 + \tfrac{3}{2}\pi r^2.$ The landscape designer proposes that $h$ and $r$ be related by the equation $rh^2 = k$, where $k$ is a constant. As $r$ and $h$ vary, find the exact value of $\dfrac{r}{h}$ 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 $y = 2x + 5 + \dfrac{4}{x}$. By differentiation, find the exact set of values of $y$ for which there are no points on C.

1. Find the total surface area of the rectangular and square components of the ponds.

$\text{Area of H pond} = 7 \times h^2 = 7h^2$
$\text{Area of rectangular part of D pond} = 3h \times h = 3h^2$
$\text{Total straight-edged area} = 7h^2 + 3h^2 = 10h^2$

2. Find the area of the curved component of the D pond (the semi-circular annulus).

$\text{Area of outer semi-circle} = \frac{1}{2}\pi(2r)^2 = 2\pi r^2$
$\text{Area of inner semi-circle} = \frac{1}{2}\pi(r)^2$
$\text{Area of annulus} = 2\pi r^2 - \frac{1}{2}\pi r^2 = \frac{3}{2}\pi r^2$

3. Sum the areas to obtain the final expression for the total surface area $A$.

$A = 10h^2 + \frac{3}{2}\pi r^2$ (shown)

1. Express the total area $A$ in terms of a single variable, $r$.

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

2. Differentiate $A$ with respect to $r$ and set it to zero for a stationary point.

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

3. Substitute $k = rh^2$ back to find the exact value of the ratio $\frac{r}{h}$.

$3\pi r^3 = 10(rh^2)$
Since $r \neq 0$, divide both sides by $r$:
$3\pi r^2 = 10h^2$
$\frac{r^2}{h^2} = \frac{10}{3\pi}$
Since $r$ and $h$ represent physical lengths, they must be positive. Thus:
$\frac{r}{h} = \sqrt{\frac{10}{3\pi}}$

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

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

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

When $k = 1$, the stationary condition is:
$3\pi r^3 = 10(1) \implies r = \left(\frac{10}{3\pi}\right)^{\frac{1}{3}} \approx 1.0199$
Substitute $r$ and $k=1$ into the expression for $A$:
$A = \frac{10(1)}{r} + \frac{3}{2}\pi r^2$
$A = \frac{10}{1.0199} + \frac{3}{2}\pi (1.0199)^2$
$A \approx 9.8048 + 4.9018 = 14.7066$
$A = 14.7 \text{ m}^2 \text{ (to 3 s.f.)}$

1. Differentiate the equation of the curve with respect to $x$.

$y = 2x + 5 + 4x^{-1}$
$\frac{dy}{dx} = 2 - 4x^{-2} = 2 - \frac{4}{x^2}$

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

$2 - \frac{4}{x^2} = 0$
$x^2 = 2 \implies x = \sqrt{2} \text{ or } x = -\sqrt{2}$

3. Calculate the corresponding $y$-coordinates and determine the nature of each stationary point using the second derivative test.

$\frac{d^2y}{dx^2} = \frac{8}{x^3}$
For $x = \sqrt{2}$, $y = 2\sqrt{2} + 5 + \frac{4}{\sqrt{2}} = 5 + 4\sqrt{2}$.
Since $\frac{d^2y}{dx^2} = \frac{8}{(\sqrt{2})^3} > 0$, $( \sqrt{2}, 5 + 4\sqrt{2} )$ is a local minimum.

For $x = -\sqrt{2}$, $y = -2\sqrt{2} + 5 + \frac{4}{-\sqrt{2}} = 5 - 4\sqrt{2}$.
Since $\frac{d^2y}{dx^2} = \frac{8}{(-\sqrt{2})^3} < 0$, $( -\sqrt{2}, 5 - 4\sqrt{2} )$ is a local maximum.

4. Determine the set of values of $y$ for which there are no points on the curve.

Given that the local maximum is at $y = 5 - 4\sqrt{2}$ and the local minimum is at $y = 5 + 4\sqrt{2}$, the graph takes all values except those strictly between these two turning points.

Thus, there are no points on $C$ for:
$5 - 4\sqrt{2} < y < 5 + 4\sqrt{2}$