A lighting consultant intends to construct a giant light structure for Christmas that is octahedral in shape. The edges of this structure are made of 12 straight pieces of steel rods of negligible thickness that are welded at their ends to form 2 congruent right pyramids. Each pyramid has a height of $3y$ metres and a square base of side $2x$ metres as shown in the diagram above. Given that the total length of steel rods used is 480 metres,

1. Express the length of one slant edge in terms of $x$.

Let $L$ be the length of one slant edge.
The structure consists of 4 base edges and 8 slant edges (4 for the top pyramid, 4 for the bottom).
The total length of the steel rods is 480 metres:
$4(2x) + 8L = 480$
$8x + 8L = 480$
$L = 60 - x$

2. Relate the vertical height, slant edge, and base dimensions using Pythagoras' theorem.

The base is a square of side $2x$. The distance from the center of the base to a corner is half the length of the diagonal:
$\frac{1}{2}\sqrt{(2x)^2 + (2x)^2} = \frac{1}{2}\sqrt{8x^2} = \sqrt{2}x$
Applying Pythagoras' theorem to the right-angled triangle formed by the vertical height $3y$, the center-to-corner distance $\sqrt{2}x$, and the slant edge $L$:
$(3y)^2 + (\sqrt{2}x)^2 = L^2$

3. Substitute the expression for $L$ and simplify to show the required result.

Substitute $L = 60 - x$ into the equation:
$9y^2 + 2x^2 = (60 - x)^2$
$9y^2 + 2x^2 = 3600 - 120x + x^2$
$9y^2 = 3600 - 120x - x^2$

Taking the square root (and noting that the height $3y$ must be positive):
$3y = \sqrt{3600 - 120x - x^2}$ (shown)

1. Deduce the slant height of the triangular faces and show the total surface area function.

Let the slant height of each triangular face be $h_s$.
By Pythagoras' theorem on the right-angled triangle formed by the vertical height of a pyramid ($3y$) and half the base side ($x$):
$h_s^2 = (3y)^2 + x^2$

From part (i), $(3y)^2 = 3600 - 120x - x^2$, so:
$h_s^2 = (3600 - 120x - x^2) + x^2 = 3600 - 120x$
$h_s = \sqrt{3600 - 120x}$

The structure consists of $2 \times 4 = 8$ congruent triangular faces.
Total Surface Area, $A = 8 \times \left(\frac{1}{2} \times \text{base} \times h_s\right)$
$A = 8 \times \left(\frac{1}{2} \times 2x \times \sqrt{3600 - 120x}\right)$
$A = 8x\sqrt{3600 - 120x} \quad \text{(Shown)}$

2. Differentiate the total surface area with respect to x.

To find the maximum total surface area, we differentiate $A$ with respect to $x$.
Using the Product Rule and Chain Rule on $A = 8x(3600 - 120x)^{\frac{1}{2}}$:
$\frac{dA}{dx} = 8(3600 - 120x)^{\frac{1}{2}} + 8x \left[ \frac{1}{2}(3600 - 120x)^{-\frac{1}{2}}(-120) \right]$
$\frac{dA}{dx} = 8\sqrt{3600 - 120x} - \frac{480x}{\sqrt{3600 - 120x}}$
$\frac{dA}{dx} = \frac{8(3600 - 120x) - 480x}{\sqrt{3600 - 120x}}$
$\frac{dA}{dx} = \frac{28800 - 1440x}{\sqrt{3600 - 120x}}$

3. Equate the first derivative to zero to find the stationary value of x.

For maximum area, set $\frac{dA}{dx} = 0$:
$\frac{28800 - 1440x}{\sqrt{3600 - 120x}} = 0$
$28800 - 1440x = 0$
$1440x = 28800$
$x = 20$

4. Verify that the stationary value gives a maximum area using the First Derivative Test.

We use the First Derivative Test to verify the nature of the stationary point at $x = 20$:
When $x = 19$, $\frac{dA}{dx} = \frac{28800 - 1440(19)}{\sqrt{3600 - 120(19)}} = \frac{1440}{\sqrt{1320}} > 0$
When $x = 21$, $\frac{dA}{dx} = \frac{28800 - 1440(21)}{\sqrt{3600 - 120(21)}} = \frac{-1440}{\sqrt{1080}} < 0$

Since the gradient $\frac{dA}{dx}$ changes from positive to negative as $x$ increases through $20$, the total surface area is maximised when $x = 20$.