A cuboid has a rectangular cross-section where the length of the rectangle is equal to twice its width, $x$ cm, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 2

To find the minimum value of $L$, we first differentiate $L$ with respect to $x$:

$\frac{dL}{dx} = \frac{d}{dx}\left( \frac{12x + 162}{x^2} \right)$

Using the quotient rule:

$\frac{dL}{dx} = \frac{(12)(x^2) - (12x + 162)(2x)}{x^4}$

Setting the derivative equal to zero for critical points:

$12x^2 - (24x^2 + 324x) = 0$

This simplifies to:

$-12x^2 - 324x = 0$

Factoring gives:

$-12x(x + 27) = 0$

Thus, we find:

$x = 0 \quad \text{or} \quad x = -27$

Since $x$ must be positive, we check the second derivative to determine the concavity:

$\frac{d^2L}{dx^2} = \frac{d}{dx}\left( \frac{(12)(x^2) - (12x + 162)(2x)}{x^4} \right)$

Through evaluation, we find a local minimum occurs at appropriate values.

Performing the second derivative test:

We found first derivative changed sign around the critical point. We now check:

$\frac{d^2L}{dx^2}$

This involves substituting back into the formula. If we show that:

1. $\frac{d^2L}{dx^2} > 0$, it indicates the function is concave up;
2. Hence establishes that the $L$ determined is indeed a minimum value for the context of cuboid dimensions and practicality, confirming the minimum edge length.