(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.

(i) To find \(\frac{dy}{dx}\), differentiate \(y = \frac{1}{x-3} + x\).

The derivative of \(\frac{1}{x-3}\) is \(\frac{-1}{(x-3)^2}\) and the derivative of \(x\) is 1.

Thus, \(\frac{dy}{dx} = \frac{-1}{(x-3)^2} + 1\).

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = \frac{-1}{(x-3)^2} + 1\).

The derivative of \(\frac{-1}{(x-3)^2}\) is \(\frac{2}{(x-3)^3}\) and the derivative of 1 is 0.

Thus, \(\frac{d^2y}{dx^2} = \frac{2}{(x-3)^3}\).

(ii) To find the critical points, set \(\frac{dy}{dx} = 0\):

\(\frac{-1}{(x-3)^2} + 1 = 0 \Rightarrow (x-3)^2 = 1 \Rightarrow x-3 = \pm 1\).

Solving gives \(x = 4\) and \(x = 2\).

For \(x = 4\), substitute into \(y = \frac{1}{x-3} + x\):

\(y = \frac{1}{4-3} + 4 = 1 + 4 = 5\).

For \(x = 2\), substitute into \(y = \frac{1}{x-3} + x\):

\(y = \frac{1}{2-3} + 2 = -1 + 2 = 1\).

Check the second derivative \(\frac{d^2y}{dx^2}\) to determine concavity:

When \(x = 4\), \(\frac{d^2y}{dx^2} = \frac{2}{(4-3)^3} = 2 > 0\), indicating a minimum.

When \(x = 2\), \(\frac{d^2y}{dx^2} = \frac{2}{(2-3)^3} = -2 < 0\), indicating a maximum.

Thus, the maximum point is \(A(2, 5)\) and the minimum point is \(B(4, 1)\).

Given \(y = \frac{9}{2-x}\), we can rewrite it as \(y = 9(2-x)^{-1}\).

Using the chain rule, \(\frac{dy}{dx} = -9(2-x)^{-2} \times (-1)\).

This simplifies to \(\frac{dy}{dx} = \frac{9}{(2-x)^2}\).

For stationary points, \(\frac{dy}{dx} = 0\). However, \(\frac{9}{(2-x)^2} \neq 0\) for any real \(x\), so there are no stationary points.

(a) Differentiate \(y = 3x + 1 - 4(3x + 1)^{\frac{1}{2}}\).

\(\frac{dy}{dx} = 3 - 4 \times \frac{1}{2} \times (3x+1)^{-\frac{1}{2}} \times 3\)

\(\frac{dy}{dx} = 3 - 6(3x+1)^{-\frac{1}{2}}\)

For \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\):

\(\frac{d^2y}{dx^2} = -6 \times -\frac{1}{2} \times (3x+1)^{-\frac{3}{2}} \times 3\)

\(\frac{d^2y}{dx^2} = 9(3x+1)^{-\frac{3}{2}}\)

(b) Set \(\frac{dy}{dx} = 0\):

\(3 - 6(3x+1)^{-\frac{1}{2}} = 0\)

\((3x+1)^{-\frac{1}{2}} = \frac{1}{2}\)

\(3x+1 = 4\)

\(x = 1\)

Substitute \(x = 1\) into \(y = 3x + 1 - 4(3x + 1)^{\frac{1}{2}}\):

\(y = 3(1) + 1 - 4(4)^{\frac{1}{2}}\)

\(y = 3 + 1 - 4 \times 2\)

\(y = -4\)

Coordinates: (1, -4)

Check nature using \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = 9(3 \times 1 + 1)^{-\frac{3}{2}} = \frac{9}{8} > 0\)

Therefore, the stationary point is a minimum.

(i) To find \(\frac{dy}{dx}\), use the chain rule on \(y = (2x - 3)^3 - 6x\):

\(\frac{dy}{dx} = 3(2x-3)^2 \cdot \frac{d}{dx}(2x-3) - 6\)

\(\frac{dy}{dx} = 3(2x-3)^2 \cdot 2 - 6\)

\(\frac{dy}{dx} = 6(2x-3)^2 - 6\)

For \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\):

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(6(2x-3)^2)\)

\(\frac{d^2y}{dx^2} = 12(2x-3) \cdot \frac{d}{dx}(2x-3)\)

\(\frac{d^2y}{dx^2} = 12(2x-3) \cdot 2\)

\(\frac{d^2y}{dx^2} = 24(2x-3)\)

(ii) Stationary points occur where \(\frac{dy}{dx} = 0\):

\(6(2x-3)^2 - 6 = 0\)

\(6(2x-3)^2 = 6\)

\((2x-3)^2 = 1\)

\(2x-3 = \pm 1\)

\(2x = 4\) or \(2x = 2\)

\(x = 2\) or \(x = 1\)

To determine the nature, use \(\frac{d^2y}{dx^2}\):

For \(x = 2\), \(\frac{d^2y}{dx^2} = 24(2 \cdot 2 - 3) = 24 \cdot 1 = 24\) (positive, minimum)

For \(x = 1\), \(\frac{d^2y}{dx^2} = 24(2 \cdot 1 - 3) = 24 \cdot (-1) = -24\) (negative, maximum)