(i) The perimeter of the window is given by the sum of the rectangle's width \(2r\), twice the height \(2h\), and the semicircle's circumference \(\pi r\). Thus, the equation is:

\(2h + 2r + \pi r = 8\)

Solving for \(h\):

\(2h = 8 - 2r - \pi r\)

\(h = 4 - r - \frac{1}{2} \pi r\)

(ii) The area \(A\) consists of the rectangle and the semicircle. The rectangle's area is \(2rh\) and the semicircle's area is \(\frac{1}{2} \pi r^2\). Therefore:

\(A = 2rh + \frac{1}{2} \pi r^2\)

Substitute \(h = 4 - r - \frac{1}{2} \pi r\) into the area equation:

\(A = 2r(4 - r - \frac{1}{2} \pi r) + \frac{1}{2} \pi r^2\)

\(A = 8r - 2r^2 - \pi r^2 + \frac{1}{2} \pi r^2\)

\(A = 8r - 2r^2 - \frac{1}{2} \pi r^2\)

(iii) To find the stationary value, differentiate \(A\) with respect to \(r\):

\(\frac{dA}{dr} = 8 - 4r - \pi r\)

Set \(\frac{dA}{dr} = 0\):

\(8 - 4r - \pi r = 0\)

\(8 = r(4 + \pi)\)

\(r = \frac{8}{4 + \pi} \approx 1.12\)

(iv) To determine if this is a maximum or minimum, find the second derivative:

\(\frac{d^2A}{dr^2} = -4 - \pi\)

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