(i) The perimeter of the sector is given by \(2r + r\theta = 50\). Solving for \(\theta\), we have:

\(\theta = \frac{1}{r}(50 - 2r)\).

The area of the sector is \(A = \frac{1}{2}r^2\theta\). Substituting \(\theta\), we get:

\(A = \frac{1}{2}r^2 \left( \frac{1}{r}(50 - 2r) \right) = 25r - r^2\).

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

\(\frac{dA}{dr} = 25 - 2r\).

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

\(25 - 2r = 0 \Rightarrow r = 12.5\).

Substitute \(r = 12.5\) into \(A = 25r - r^2\):

\(A = 25(12.5) - (12.5)^2 = 156.25\).

The second derivative \(\frac{d^2A}{dr^2} = -2\) is negative, indicating a maximum.