The diagram shows a glass window $ABCDEF$, consisting of a rectangle $ABEF$ of height $3x\text{ cm}$ and width $y\text{ cm}$ and a trapezium $BCDE$ in which $CD = x\text{ cm}$ and $BC = 2y\text{ cm}$. $ABC$ is a straight line and $DE = 10\sqrt{2}\text{ cm}$. Given that $x$ can vary,

1. Express $y$ in terms of $x$ using the trapezium $BCDE$.

Draw a horizontal line from $D$ to $BE$, meeting $BE$ at $G$. Then $DG = BC = 2y$ and $EG = BE - BG = 3x - x = 2x$.
In right-angled triangle $EGD$:
$EG^2 + DG^2 = DE^2$
$(2x)^2 + (2y)^2 = (10\sqrt{2})^2$
$4x^2 + 4y^2 = 200$
$x^2 + y^2 = 50$
$y^2 = 50 - x^2$
$y = \sqrt{50 - x^2}$ (since $y > 0$)

2. Find the total area $S$ in terms of $x$ and $y$.

Area of rectangle $ABEF = 3x \times y = 3xy$
Area of trapezium $BCDE = \frac{1}{2}(BE + CD) \times BC = \frac{1}{2}(3x + x)(2y) = 4xy$
Total area $S = 3xy + 4xy = 7xy$

3. Substitute $y$ to show the final expression for $S$.

Substitute $y = \sqrt{50 - x^2}$ into $S = 7xy$:
$S = 7x(\sqrt{50 - x^2})$ (Shown)

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

$S = 7x(50 - x^2)^{\frac{1}{2}}$
$\frac{dS}{dx} = 7(50 - x^2)^{\frac{1}{2}} + 7x \cdot \frac{1}{2}(50 - x^2)^{-\frac{1}{2}}(-2x)$
$\frac{dS}{dx} = 7(50 - x^2)^{\frac{1}{2}} - \frac{7x^2}{(50 - x^2)^{\frac{1}{2}}}$

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

For a stationary value, $\frac{dS}{dx} = 0$.
$7(50 - x^2)^{\frac{1}{2}} - \frac{7x^2}{(50 - x^2)^{\frac{1}{2}}} = 0$
$7(50 - x^2)^{\frac{1}{2}} = \frac{7x^2}{(50 - x^2)^{\frac{1}{2}}}$
$7(50 - x^2) = 7x^2$
$50 - x^2 = x^2$
$2x^2 = 50$
$x^2 = 25$

3. Solve for $x$, rejecting the negative value since $x$ represents a length.

$x = 5$ or $x = -5$
Since $x > 0$ (length), $x = 5$.

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

Let's test values of $x$ around $x = 5$:
When $x = 4.9$, $\frac{dS}{dx} = 7(50 - 4.9^2)^{\frac{1}{2}} - \frac{7(4.9)^2}{(50 - 4.9^2)^{\frac{1}{2}}} \approx 35.69 - 32.94 > 0$
When $x = 5.1$, $\frac{dS}{dx} = 7(50 - 5.1^2)^{\frac{1}{2}} - \frac{7(5.1)^2}{(50 - 5.1^2)^{\frac{1}{2}}} \approx 34.28 - 37.35 < 0$
Since the gradient changes from positive to negative, $S$ is a maximum when $x = 5$.