$ABCDEF$ is the cross-sectional plan of a temporary spectator stand, where $AB$ is parallel to $FC$ and $ED$ and $AE$ is parallel to $CD$.
Given that $AB = x$, $ED = 4x$, $FE = y$, $AF = 2y$ and $BC = 10\text{ m}$.

1. Find the horizontal and vertical distances between points $B$ and $C$.

Horizontal distance between $B$ and $C$ is $FC - AB = 4x - x = 3x$.
Vertical distance between $B$ and $C$ is $AF = 2y$.

2. Apply Pythagoras' theorem to the right-angled triangle formed by $B$, $C$, and their horizontal/vertical projections.

$(3x)^2 + (2y)^2 = BC^2 9x^2 + 4y^2 = 10^2 9x^2 + 4y^2 = 100$

3. Rearrange the equation to express $y$ in terms of $x$.

$4y^2 = 100 - 9x^2 y^2 = \frac{100 - 9x^2}{4} y = \frac{\sqrt{100 - 9x^2}}{2}$ (since $y > 0$)

1. Find the expression for $y$ in terms of $x$ (from part a).

By drawing a vertical line from $B$ down to $FC$, a right-angled triangle is formed with horizontal side $4x - x = 3x$ and vertical side $2y$.
Using Pythagoras' theorem:
$(3x)^2 + (2y)^2 = 10^2$
$9x^2 + 4y^2 = 100$
$y = \frac{1}{2}\sqrt{100 - 9x^2}$

2. Express the total area $S$ in terms of $x$ and $y$ by splitting the figure.

The cross-section can be split into the trapezium $ABCF$ and the rectangle $FCDE$.
Area of trapezium $ABCF = \frac{1}{2}(AB + FC) \times AF$
$= \frac{1}{2}(x + 4x)(2y) = 5xy$
Area of rectangle $FCDE = ED \times FE$
$= 4x \times y = 4xy$
Total area $S = 5xy + 4xy = 9xy$

3. Substitute $y$ into the area expression to obtain the final formula.

Substitute $y = \frac{1}{2}\sqrt{100 - 9x^2}$ into $S = 9xy$:
$S = 9x \left( \frac{1}{2}\sqrt{100 - 9x^2} \right)$
$S = \frac{9}{2}x\sqrt{100 - 9x^2}$ (shown)

1. Differentiate $S$ with respect to $x$ using the product and chain rules.

$S = \frac{9}{2}x(100 - 9x^2)^{\frac{1}{2}}$

$\frac{dS}{dx} = \frac{9}{2}(100 - 9x^2)^{\frac{1}{2}} + \frac{9}{2}x \left[ \frac{1}{2}(100 - 9x^2)^{-\frac{1}{2}}(-18x) \right]$
$\frac{dS}{dx} = \frac{9}{2}\sqrt{100 - 9x^2} - \frac{81x^2}{2\sqrt{100 - 9x^2}}$

2. Set the derivative to zero to find the stationary point.

For a stationary value, $\frac{dS}{dx} = 0$:
$\frac{9}{2}\sqrt{100 - 9x^2} - \frac{81x^2}{2\sqrt{100 - 9x^2}} = 0$
$\frac{9}{2}\sqrt{100 - 9x^2} = \frac{81x^2}{2\sqrt{100 - 9x^2}}$

3. Solve for $x^2$.

Multiply both sides by $2\sqrt{100 - 9x^2}$:
$9(100 - 9x^2) = 81x^2$
$900 - 81x^2 = 81x^2$
$162x^2 = 900$
$x^2 = \frac{900}{162} = \frac{50}{9}$

4. Substitute $x^2$ back into the expression for $S$ to find the stationary value.

Since $x > 0$ (as it represents a length), $x = \frac{\sqrt{50}}{3} = \frac{5\sqrt{2}}{3}$.
Substitute $x$ and $x^2$ into $S$:
$S = \frac{9}{2} \left( \frac{5\sqrt{2}}{3} \right) \sqrt{100 - 9\left(\frac{50}{9}\right)}$
$S = \frac{15\sqrt{2}}{2} \sqrt{100 - 50}$
$S = \frac{15\sqrt{2}}{2} \sqrt{50}$
$S = \frac{15\sqrt{2}}{2} (5\sqrt{2})$
$S = \frac{75 \times 2}{2} = 75$

1. Use the first derivative test to determine the nature of the stationary point

From part (c), $\frac{dS}{dx} = \frac{450 - 81x^2}{\sqrt{100 - 9x^2}}$ and the stationary point is at $x = \frac{\sqrt{50}}{3} \approx 2.357$.

Testing values around $x = 2.357$:
When $x = 2$, $\frac{dS}{dx} = \frac{450 - 81(2)^2}{\sqrt{100 - 9(2)^2}} = \frac{126}{8} > 0$
When $x = 3$, $\frac{dS}{dx} = \frac{450 - 81(3)^2}{\sqrt{100 - 9(3)^2}} = \frac{-279}{\sqrt{19}} < 0$

2. Conclude the nature of the stationary value

Since the gradient $\frac{dS}{dx}$ changes from positive to negative as $x$ increases through the stationary point, the stationary value is a maximum.