(i) The height of the triangle is 3x. The total area of the cardboard is given by:

\(10y + \frac{1}{2} \times 8x \times 3x = 200\)

Solving for \(y\):

\(10y = 200 - 12x^2\)

\(y = \frac{200 - 24x^2}{10x}\)

(ii) The volume \(V\) is given by the area of the base times the height:

\(V = \frac{1}{2} \times 8x \times 3x \times y = 240x - 28.8x^3\)

(iii) Differentiate \(V\) with respect to \(x\):

\(\frac{dV}{dx} = 240 - 86.4x^2\)

Set \(\frac{dV}{dx} = 0\) to find the stationary point:

\(240 - 86.4x^2 = 0\)

\(x = \frac{1}{3}\)

(iv) Differentiate again to determine the nature of the stationary point:

\(\frac{d^2V}{dx^2} = -172.8x\)

Since \(\frac{d^2V}{dx^2} < 0\), the stationary value is a maximum.