[已知棱锥的体积为 $\frac{1}{3} \times (\text{base area}) \times (\text{height})$。] 一个垂直高度为 $h$ m 的正棱锥，其底面是边长为 $2x$ m 的正方形，体积为 $\frac{8}{3}\text{ m}^3$。

1. Set up the equation for the volume of the pyramid based on its given dimensions.

$\text{Volume} = \frac{1}{3} \times \text{base area} \times \text{height}$
$\frac{8}{3} = \frac{1}{3} \times (2x)^2 \times h$
$\frac{8}{3} = \frac{1}{3}(4x^2)h$

2. Solve the equation to make $h$ the subject.

$\frac{4x^2 h}{3} = \frac{8}{3}$
$4x^2 h = 8$
$h = \frac{8}{4x^2}$
$h = \frac{2}{x^2}$

1. Find the height $h$ in terms of $x$ (from part i)

$V = \frac{1}{3} \times (2x)^2 \times h = \frac{8}{3}$
$4x^2 h = 8 \implies h = \frac{2}{x^2}$

2. Find the slant height of the triangular faces

Let the slant height be $l$. Applying Pythagoras' theorem to the right-angled triangle formed by the vertical height and the apothem (half the base length):
$l^2 = h^2 + x^2$
$l = \sqrt{\left(\frac{2}{x^2}\right)^2 + x^2} = \sqrt{\frac{4}{x^4} + x^2}$

3. Form the equation for the total surface area $S$

The total surface area consists of the square base and 4 identical triangular faces:
$S = \text{Area of base} + 4 \times (\text{Area of one triangular face})$
$S = (2x)^2 + 4\left(\frac{1}{2} \times 2x \times l\right)$
$S = 4x^2 + 4xl$
$S = 4x^2 + 4x\sqrt{\frac{4}{x^4} + x^2}$

4. Factor out the expression to match the required form

Factor out $x^2$ from the terms inside the square root. Since $x > 0$ for a length, $\sqrt{x^2} = x$:
$S = 4x^2 + 4x\sqrt{x^2\left(\frac{4}{x^6} + 1\right)}$
$S = 4x^2 + 4x(x)\sqrt{1 + \frac{4}{x^6}}$
$S = 4x^2 + 4x^2\sqrt{1 + \frac{4}{x^6}}$
$S = 4x^2 \left[ 1 + \sqrt{\left(1 + \frac{4}{x^6}\right)} \right] \text{ (shown)}$

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

Expand the expression for $S$ slightly to facilitate differentiation:
$S = 4x^2 + 4x^2\left(1 + 4x^{-6}\right)^{\frac{1}{2}}$

Using the product rule on the second term:
$\frac{\mathrm{d}S}{\mathrm{d}x} = 8x + 8x\left(1 + 4x^{-6}\right)^{\frac{1}{2}} + 4x^2 \left[ \frac{1}{2}\left(1 + 4x^{-6}\right)^{-\frac{1}{2}} \left(-24x^{-7}\right) \right]$

$\frac{\mathrm{d}S}{\mathrm{d}x} = 8x + 8x\left(1 + 4x^{-6}\right)^{\frac{1}{2}} - \frac{48x^{-5}}{\sqrt{1 + 4x^{-6}}}$

2. Equate the derivative to zero and rearrange to isolate the radical term.

For a stationary value, $\frac{\mathrm{d}S}{\mathrm{d}x} = 0$:
$8x + 8x\left(1 + 4x^{-6}\right)^{\frac{1}{2}} - \frac{48x^{-5}}{\sqrt{1 + 4x^{-6}}} = 0$

Divide the entire equation by 8:
$x + x\left(1 + 4x^{-6}\right)^{\frac{1}{2}} - \frac{6x^{-5}}{\sqrt{1 + 4x^{-6}}} = 0$

Multiply throughout by $x^5\sqrt{1 + 4x^{-6}}$ to eliminate the fraction and negative indices:
$x^6\sqrt{1 + 4x^{-6}} + x^6\left(1 + 4x^{-6}\right) - 6 = 0$

$x^6\sqrt{1 + \frac{4}{x^6}} + x^6 + 4 - 6 = 0$

$x^6\sqrt{1 + \frac{4}{x^6}} = 2 - x^6$

3. Square both sides to solve for $x$.

Square both sides of the equation:
$x^{12}\left(1 + \frac{4}{x^6}\right) = (2 - x^6)^2$

$x^{12} + 4x^6 = 4 - 4x^6 + x^{12}$

Subtract $x^{12}$ from both sides:
$4x^6 = 4 - 4x^6$
$8x^6 = 4$
$x^6 = \frac{1}{2}$

Since $x$ represents half the side length of the pyramid's base, $x > 0$:
$x = (0.5)^{\frac{1}{6}} \approx 0.890898$

$x \approx 0.89$ (correct to 2 d.p.)