(i) Given \(z = 3x + 2y\) and \(xy = 600\), we can express \(y\) in terms of \(x\):

\(y = \frac{600}{x}\).

Substitute \(y\) in the expression for \(z\):

\(z = 3x + 2\left(\frac{600}{x}\right) = 3x + \frac{1200}{x}\).

(ii) To find the stationary value, differentiate \(z\) with respect to \(x\):

\(\frac{dz}{dx} = 3 - \frac{1200}{x^2}\).

Set \(\frac{dz}{dx} = 0\) to find critical points:

\(3 - \frac{1200}{x^2} = 0\).

\(\frac{1200}{x^2} = 3\).

\(x^2 = 400\).

\(x = 20\) (since \(x\) is positive).

Substitute \(x = 20\) back into the expression for \(z\):

\(z = 3(20) + \frac{1200}{20} = 60 + 60 = 120\).

To determine the nature, find the second derivative:

\(\frac{d^2z}{dx^2} = \frac{2400}{x^3}\).

At \(x = 20\), \(\frac{d^2z}{dx^2} = \frac{2400}{20^3} > 0\), indicating a minimum.