The figure below shows a right-angled triangular prism.
The height and base of the triangular faces of the prism are $3x$ meters and $4x$ meters respectively. The length of the prism is $L$ meters.

1. Express the volume of the prism in terms of $x$ and $L$ and equate to 240.

Area of triangular base $= \frac{1}{2} \times 4x \times 3x = 6x^2$
Volume $= \text{Base Area} \times \text{Length} = 6x^2 L$
$6x^2 L = 240$

2. Express $L$ in terms of $x$.

$L = \frac{240}{6x^2} = \frac{40}{x^2}$

3. Find the length of the hypotenuse of the triangular face.

Hypotenuse $= \sqrt{(3x)^2 + (4x)^2} = \sqrt{9x^2 + 16x^2} = \sqrt{25x^2} = 5x$

4. Formulate the expression for the total surface area $A$ and substitute $L$.

Total Surface Area $A = 2 \times (\text{Area of triangular base}) + \text{Area of 3 rectangular faces}$
$A = 2(6x^2) + (3x)L + (4x)L + (5x)L$
$A = 12x^2 + 12xL$
Substitute $L = \frac{40}{x^2}$:
$A = 12x^2 + 12x\left(\frac{40}{x^2}\right)$
$A = 12x^2 + \frac{480}{x}$ (Shown)

1. Differentiate $A$ with respect to $x$

$A = 12x^2 + 480x^{-1}$
$\frac{dA}{dx} = 24x - 480x^{-2}$

2. Set the first derivative to zero to find the stationary value of $x$

For a stationary value, $\frac{dA}{dx} = 0$:
$24x - \frac{480}{x^2} = 0$
$24x^3 = 480$
$x^3 = 20$
$x = \sqrt[3]{20} \approx 2.7144$

3. Find the second derivative to determine the nature of the stationary value

$\frac{d^2A}{dx^2} = 24 - 480(-2)x^{-3}$
$\frac{d^2A}{dx^2} = 24 + \frac{960}{x^3}$

4. Evaluate the second derivative at the stationary point

Substitute $x^3 = 20$ into the second derivative:
$\frac{d^2A}{dx^2} = 24 + \frac{960}{20} = 24 + 48 = 72$
Since $\frac{d^2A}{dx^2} > 0$, the stationary value of $A$ is a minimum.