(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.

(i) The area A of the track is composed of two rectangles and two semicircles. The area of the rectangles is given by 2xr and the area of the semicircles is \pi r^2. Therefore, the total area is:

\(A = 2xr + \pi r^2\)

Given the perimeter is 400 metres, we have:

\(2x + 2\pi r = 400\)

Solving for x gives:

\(x = 200 - \pi r\)

Substitute x into the area equation:

\(A = 2(200 - \pi r)r + \pi r^2\)

\(A = 400r - 2\pi r^2 + \pi r^2\)

\(A = 400r - \pi r^2\)

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

\(\frac{dA}{dr} = 400 - 2\pi r\)

Set \(\frac{dA}{dr} = 0\) to find the critical points:

\(400 - 2\pi r = 0\)

\(r = \frac{200}{\pi}\)

Substitute back to find x:

\(x = 200 - \pi \left(\frac{200}{\pi}\right) = 0\)

Thus, there are no straight sections.

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

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

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