(i) To find the area of rectangle OPQR, we need to express y in terms of x. Using similar triangles, we have:

\(\frac{y}{16-x} = \frac{12}{16}\)

Solving for y, we get:

\(y = 12 - \frac{3}{4}x\)

The area A of rectangle OPQR is given by:

\(A = xy = x(12 - \frac{3}{4}x) = 12x - \frac{3}{4}x^2\)

(ii) To find the stationary value, we differentiate A with respect to x:

\(\frac{dA}{dx} = 12 - \frac{6}{4}x\)

Setting \(\frac{dA}{dx} = 0\), we solve:

\(12 - \frac{6}{4}x = 0\)

\(x = 8\)

Substituting \(x = 8\) back into the area formula:

\(A = 12(8) - \frac{3}{4}(8)^2 = 96 - 48 = 48\)

Since the second derivative \(\frac{d^2A}{dx^2} = -\frac{6}{4}\) is negative, the stationary point is a maximum.