(i) The volume of the prism is given by \(\frac{1}{2} \times \frac{\sqrt{3}}{2} x^2 \times h = 2000\). Solving for \(h\), we have:

\(h = \frac{8000}{\sqrt{3}x^2}.\)

The total surface area \(A\) is the sum of the areas of the three rectangles and the two triangular bases:

\(A = 3xh + 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} x^2.\)

Substituting for \(h\), we get:

\(A = \frac{\sqrt{3}}{2}x^2 + \frac{24000}{\sqrt{3}}x^{-1}.\)

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

\(\frac{dA}{dx} = \frac{\sqrt{3}}{2} \times 2x - \frac{24000}{\sqrt{3}}x^{-2}.\)

Set \(\frac{dA}{dx} = 0\):

\(\sqrt{3}x - \frac{24000}{\sqrt{3}}x^{-2} = 0.\)

Solving for \(x\), we find:

\(x^3 = 8000 \Rightarrow x = 20.\)

(iii) To determine the nature of the stationary value, find the second derivative:

\(\frac{d^2A}{dx^2} = \sqrt{3} + \frac{48000}{\sqrt{3}}x^{-3}.\)

Since \(\frac{d^2A}{dx^2} > 0\), the stationary value is a minimum.

(i) The total area \(A\) is given by \(A = 2y \times 4x = 8xy\).

The total fencing used is given by the perimeter equation: \(10y + 12x = 480\).

Solving for \(y\) in terms of \(x\):

\(y = \frac{480 - 12x}{10} = 48 - 1.2x\).

Substitute \(y\) into the area equation:

\(A = 8x(48 - 1.2x) = 384x - 9.6x^2\).

(ii) To find the maximum area, differentiate \(A\) with respect to \(x\):

\(\frac{dA}{dx} = 384 - 19.2x\).

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

\(384 - 19.2x = 0\).

Solving gives \(x = 20\).

Substitute \(x = 20\) back into the equation for \(y\):

\(y = 48 - 1.2 \times 20 = 24\).

Thus, the dimensions for maximum area are \(x = 20\) m and \(y = 24\) m.

(i) The surface area \(A\) of a closed cylinder is given by \(A = 2\pi r^2 + 2\pi rh\).

The volume \(V\) of the cylinder is \(\pi r^2 h = 1000\), so \(h = \frac{1000}{\pi r^2}\).

Substitute \(h\) into the surface area formula:

\(A = 2\pi r^2 + 2\pi r \left( \frac{1000}{\pi r^2} \right) = 2\pi r^2 + \frac{2000}{r}\).

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

\(\frac{dA}{dr} = 4\pi r - \frac{2000}{r^2}\).

Set \(\frac{dA}{dr} = 0\) to find the stationary point:

\(4\pi r - \frac{2000}{r^2} = 0\).

Solving for \(r\), we get \(r^3 = \frac{2000}{4\pi}\).

\(r = \sqrt[3]{\frac{2000}{4\pi}} \approx 5.4\) cm.

To verify it's a minimum, check the second derivative:

\(\frac{d^2A}{dr^2} = 4\pi + \frac{4000}{r^3}\).

Since \(\frac{d^2A}{dr^2} > 0\), \(A\) is minimized, confirming the flask is most efficient at \(r = 5.4\) cm.

(i) The perimeter of the sector is given by \(r + r + r\theta = 24\), which simplifies to \(2r + r\theta = 24\). Solving for \(\theta\), we get \(\theta = \frac{24 - 2r}{r}\).

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

\(A = \frac{1}{2} r^2 \left( \frac{24 - 2r}{r} \right) = \frac{24r}{2} - \frac{r^2}{2} = 12r - r^2\).

(ii) To express \(A\) in the form \(a - (r - b)^2\), we complete the square:

\(A = 12r - r^2 = -(r^2 - 12r)\).

Completing the square gives \(-(r^2 - 12r) = -( (r - 6)^2 - 36 ) = 36 - (r - 6)^2\).

Thus, \(A = 36 - (r - 6)^2\).

(iii) The expression \(A = 36 - (r - 6)^2\) is maximized when \((r - 6)^2 = 0\), which occurs when \(r = 6\).

Substituting \(r = 6\) into the perimeter equation \(2r + r\theta = 24\), we get:

\(12 + 6\theta = 24\) which simplifies to \(\theta = 2\).

Therefore, the greatest value of \(A\) is 36 when \(r = 6\) and \(\theta = 2\).

(i) Let the height of the cuboid be \(y\). The volume of the cuboid is given by \(3x^2y = 288\), so \(y = \frac{288}{3x^2} = \frac{96}{x^2}\).

The surface area \(A\) is given by:

\(A = 2(x \cdot 3x + x \cdot y + 3x \cdot y) = 2(3x^2 + xy + 3xy).\)

Substitute \(y = \frac{96}{x^2}\) into the equation:

\(A = 2(3x^2 + x \cdot \frac{96}{x^2} + 3x \cdot \frac{96}{x^2}) = 2(3x^2 + \frac{96}{x} + \frac{288}{x}).\)

\(A = 6x^2 + \frac{768}{x}.\)

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

\(\frac{dA}{dx} = 12x - \frac{768}{x^2}.\)

Set \(\frac{dA}{dx} = 0\):

\(12x = \frac{768}{x^2}.\)

\(12x^3 = 768.\)

\(x^3 = 64.\)

\(x = 4.\)

Substitute \(x = 4\) back into the expression for \(A\):

\(A = 6(4)^2 + \frac{768}{4} = 96 + 192 = 288.\)

To determine the nature, find the second derivative:

\(\frac{d^2A}{dx^2} = 12 + \frac{1536}{x^3}.\)

At \(x = 4\), \(\frac{d^2A}{dx^2} = 12 + \frac{1536}{64} = 36 > 0\), indicating a minimum.