一位照明顾问打算为圣诞节建造一个巨型的八面体灯光结构。该结构的棱由 12 根粗细可忽略不计的直钢棒制成，这些钢棒在端点处焊接在一起，形成两个全等的直四棱锥。如上图所示，每个棱锥的高度为 $3y$ 米，底面是边长为 $2x$ 米的正方形。已知所用钢棒的总长度为 480 米，

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 $x = 20$.