(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.

(i) The total surface area of the block is given by the formula for the surface area of a rectangular prism: \(4xh + 2x^2 = 96\).

Solving for \(h\), we have:

\(4xh + 2x^2 = 96\)

\(4xh = 96 - 2x^2\)

\(h = \frac{96 - 2x^2}{4x} = \frac{24}{x} - \frac{x}{2}\)

The volume \(V\) of the block is \(x^2h\):

\(V = x^2 \left( \frac{24}{x} - \frac{x}{2} \right)\)

\(V = 24x - \frac{1}{2}x^3\)

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

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

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

\(24 - \frac{3}{2}x^2 = 0\)

\(\frac{3}{2}x^2 = 24\)

\(x^2 = 16\)

\(x = 4\)

Substitute \(x = 4\) back into the volume equation:

\(V = 24(4) - \frac{1}{2}(4)^3 = 96 - 32 = 64\)

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

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

Substitute \(x = 4\):

\(\frac{d^2V}{dx^2} = -3(4) = -12\)

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

(i) The perimeter of the sector is given by \(2r + r\theta = 50\). Solving for \(\theta\), we have:

\(\theta = \frac{1}{r}(50 - 2r)\).

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

\(A = \frac{1}{2}r^2 \left( \frac{1}{r}(50 - 2r) \right) = 25r - r^2\).

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

\(\frac{dA}{dr} = 25 - 2r\).

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

\(25 - 2r = 0 \Rightarrow r = 12.5\).

Substitute \(r = 12.5\) into \(A = 25r - r^2\):

\(A = 25(12.5) - (12.5)^2 = 156.25\).

The second derivative \(\frac{d^2A}{dr^2} = -2\) is negative, indicating a maximum.

(i) The total length of the wire is 80 cm, so we have the equation:

\(4x + 2\pi r = 80\)

The area of the square is \(x^2\) and the area of the circle is \(\pi r^2\). Therefore, the total area \(A\) is:

\(A = x^2 + \pi r^2\)

From the length equation, solve for \(r\):

\(2\pi r = 80 - 4x\)

\(r = \frac{80 - 4x}{2\pi}\)

Substitute \(r\) into the area equation:

\(A = x^2 + \pi \left(\frac{80 - 4x}{2\pi}\right)^2\)

\(A = x^2 + \frac{(80 - 4x)^2}{4\pi}\)

\(A = x^2 + \frac{6400 - 640x + 16x^2}{4\pi}\)

\(A = x^2 + \frac{16x^2 - 640x + 6400}{4\pi}\)

\(A = \frac{4\pi x^2 + 16x^2 - 640x + 6400}{4\pi}\)

\(A = \frac{(\pi + 4)x^2 - 160x + 1600}{\pi}\)

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

\(\frac{dA}{dx} = \frac{2(\pi + 4)x - 160}{\pi}\)

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

\(2(\pi + 4)x - 160 = 0\)

\(2(\pi + 4)x = 160\)

\(x = \frac{160}{2(\pi + 4)}\)

\(x = 11.2\)

(i) The height of the triangle is 3x. The total area of the cardboard is given by:

\(10y + \frac{1}{2} \times 8x \times 3x = 200\)

Solving for \(y\):

\(10y = 200 - 12x^2\)

\(y = \frac{200 - 24x^2}{10x}\)

(ii) The volume \(V\) is given by the area of the base times the height:

\(V = \frac{1}{2} \times 8x \times 3x \times y = 240x - 28.8x^3\)

(iii) Differentiate \(V\) with respect to \(x\):

\(\frac{dV}{dx} = 240 - 86.4x^2\)

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

\(240 - 86.4x^2 = 0\)

\(x = \frac{1}{3}\)

(iv) Differentiate again to determine the nature of the stationary point:

\(\frac{d^2V}{dx^2} = -172.8x\)

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