(i) Using similar triangles, \(\frac{6}{12} = \frac{r}{12-h}\). Solving for \(h\), we get \(h = 12 - 2r\).

The volume of the cylinder is \(V = \pi r^2 h\). Substituting \(h = 12 - 2r\), we have:

\(V = \pi r^2 (12 - 2r) = 12\pi r^2 - 2\pi r^3\).

(ii) Differentiate \(V\) with respect to \(r\):

\(\frac{dV}{dr} = 24\pi r - 6\pi r^2\).

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

\(24\pi r - 6\pi r^2 = 0\).

Factor out \(6\pi r\):

\(6\pi r (4 - r) = 0\).

Thus, \(r = 0\) or \(r = 4\). Since \(r = 0\) is not feasible, \(r = 4\).

Substitute \(r = 4\) into the volume formula:

\(V = 12\pi (4)^2 - 2\pi (4)^3 = 64\pi\).

(i) Given \(V = \frac{1}{3} \pi r^2 h\) and \(h + r = 18\), we can express \(h\) as \(h = 18 - r\). Substituting this into the volume formula gives:

\(V = \frac{1}{3} \pi r^2 (18 - r) = \frac{1}{3} \pi r^2 \times 18 - \frac{1}{3} \pi r^3\)

\(V = 6\pi r^2 - \frac{1}{3} \pi r^3\)

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

\(\frac{dV}{dr} = 12\pi r - \pi r^2\)

Set \(\frac{dV}{dr} = 0\):

\(12\pi r - \pi r^2 = 0\)

\(\pi r (12 - r) = 0\)

\(r = 12\) (since \(r = 0\) is not a valid solution for a cone)

To confirm it's a maximum, find the second derivative:

\(\frac{d^2V}{dr^2} = 12\pi - 2\pi r\)

Substitute \(r = 12\):

\(\frac{d^2V}{dr^2} = 12\pi - 24\pi = -12\pi\)

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

(iii) Substitute \(r = 12\) into \(h = 18 - r\):

\(h = 6\)

Maximum volume \(V = \frac{1}{3} \pi (12)^2 (6) = 288\pi\) or approximately 905 cm³.

(i) The perimeter of the window is given by the sum of the rectangle's width \(2r\), twice the height \(2h\), and the semicircle's circumference \(\pi r\). Thus, the equation is:

\(2h + 2r + \pi r = 8\)

Solving for \(h\):

\(2h = 8 - 2r - \pi r\)

\(h = 4 - r - \frac{1}{2} \pi r\)

(ii) The area \(A\) consists of the rectangle and the semicircle. The rectangle's area is \(2rh\) and the semicircle's area is \(\frac{1}{2} \pi r^2\). Therefore:

\(A = 2rh + \frac{1}{2} \pi r^2\)

Substitute \(h = 4 - r - \frac{1}{2} \pi r\) into the area equation:

\(A = 2r(4 - r - \frac{1}{2} \pi r) + \frac{1}{2} \pi r^2\)

\(A = 8r - 2r^2 - \pi r^2 + \frac{1}{2} \pi r^2\)

\(A = 8r - 2r^2 - \frac{1}{2} \pi r^2\)

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

\(\frac{dA}{dr} = 8 - 4r - \pi r\)

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

\(8 - 4r - \pi r = 0\)

\(8 = r(4 + \pi)\)

\(r = \frac{8}{4 + \pi} \approx 1.12\)

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

\(\frac{d^2A}{dr^2} = -4 - \pi\)

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

(i) The volume of the block is given by \(V = 2x \cdot x \cdot y = 72\). Thus, \(2x^2y = 72\) which implies \(y = \frac{72}{2x^2} = \frac{36}{x^2}\).

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

\(A = 2(2x \cdot x + x \cdot y + 2x \cdot y) = 2(2x^2 + xy + 2xy) = 4x^2 + 6xy\).

Substitute \(y = \frac{36}{x^2}\) into the equation:

\(A = 4x^2 + 6x \cdot \frac{36}{x^2} = 4x^2 + \frac{216}{x}\).

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

\(\frac{dA}{dx} = 8x - \frac{216}{x^2}\).

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

\(8x - \frac{216}{x^2} = 0\).

\(8x^3 = 216\) which gives \(x^3 = 27\) and \(x = 3\).

(iii) Substitute \(x = 3\) back into the expression for \(A\):

\(A = 4(3)^2 + \frac{216}{3} = 36 + 72 = 108\) cm².

To determine if this is a minimum or maximum, find the second derivative:

\(\frac{d^2A}{dx^2} = 8 + \frac{432}{x^3}\).

At \(x = 3\), \(\frac{d^2A}{dx^2} = 8 + \frac{432}{27} = 24\), which is positive, indicating a minimum.

(i) The total surface area of the cylinder is given by \(\pi r^2 + 2\pi rh = 192\pi\). Solving for \(h\), we have:

\(192\pi = \pi r^2 + 2\pi rh\)

\(2\pi rh = 192\pi - \pi r^2\)

\(h = \frac{192\pi - \pi r^2}{2\pi} = \frac{192 - r^2}{2}\)

The volume \(V\) of the cylinder is \(\pi r^2 h\). Substituting for \(h\):

\(V = \pi r^2 \left(\frac{192 - r^2}{2}\right) = \frac{1}{2} \pi (192r - r^3)\)

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

\(\frac{dV}{dr} = \frac{1}{2} \pi (192 - 3r^2)\)

Set \(\frac{dV}{dr} = 0\):

\(192 - 3r^2 = 0\)

\(3r^2 = 192\)

\(r^2 = 64\)

\(r = 8\)

(iii) Substitute \(r = 8\) into the volume formula:

\(V = \frac{1}{2} \pi (192 \times 8 - 8^3) = \frac{1}{2} \pi (1536 - 512) = \frac{1}{2} \pi \times 1024 = 512\pi\)

To determine if this is a maximum or minimum, find the second derivative:

\(\frac{d^2V}{dr^2} = \frac{1}{2} \pi (-6r) = -3\pi r\)

At \(r = 8\), \(\frac{d^2V}{dr^2} = -24\pi\), which is negative, indicating a maximum.