(i) The perimeter of the plate is given by the sum of the sides of the rectangle and the arc of the quarter-circle: \(2x + 2y + \frac{\pi x}{2} = 60\). Solving for \(y\), we have:

\(2y = 60 - 2x - \frac{\pi x}{2}\)

\(y = 30 - x - \frac{\pi x}{4}\)

(ii) The area of the plate consists of the area of the rectangle and the area of the quarter-circle. The area of the rectangle is \(xy\) and the area of the quarter-circle is \(\frac{1}{4} \pi x^2\). Thus,

\(A = xy + \frac{\pi x^2}{4}\)

Substituting \(y = 30 - x - \frac{\pi x}{4}\) into the equation, we get:

\(A = x(30 - x - \frac{\pi x}{4}) + \frac{\pi x^2}{4}\)

\(A = 30x - x^2 - \frac{\pi x^2}{4} + \frac{\pi x^2}{4}\)

\(A = 30x - x^2\)

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

\(\frac{dA}{dx} = 30 - 2x\)

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

\(30 - 2x = 0\)

\(x = 15 \text{ cm}\)

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

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

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