(i) To find the volume of the cone, we use the formula \(V = \frac{1}{3}\pi r^2 h\). The slant height is 15 cm, and the vertical height is \(h\) cm. By the Pythagorean theorem, \(r^2 = 15^2 - h^2\), so \(r^2 = 225 - h^2\).

Substituting \(r^2\) into the volume formula gives:

\(V = \frac{1}{3}\pi (225 - h^2) h = \frac{1}{3}\pi (225h - h^3)\).

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

\(\frac{dV}{dh} = \frac{\pi}{3} (225 - 3h^2)\).

Set \(\frac{dV}{dh} = 0\) to find the stationary points:

\(225 - 3h^2 = 0\)

\(3h^2 = 225\)

\(h^2 = 75\)

\(h = \sqrt{75}\) or approximately 8.66 cm.

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

\(\frac{d^2V}{dh^2} = \frac{\pi}{3} (-6h)\).

Substituting \(h = \sqrt{75}\) gives a negative value, indicating a maximum.

(i) The volume of the cylinder is given by \(\pi r^2 h = 250\pi\). Solving for \(h\), we get \(h = \frac{250}{r^2}\).

The total surface area \(S\) is the sum of the lateral surface area and the area of the two bases: \(S = 2\pi rh + 2\pi r^2\).

Substitute \(h = \frac{250}{r^2}\) into the surface area formula:

\(S = 2\pi r \left( \frac{250}{r^2} \right) + 2\pi r^2 = \frac{500\pi}{r} + 2\pi r^2\).

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

\(\frac{dS}{dr} = 4\pi r - \frac{500\pi}{r^2}\).

Set \(\frac{dS}{dr} = 0\) to find the critical points:

\(4\pi r = \frac{500\pi}{r^2}\).

Solving for \(r\), we get \(r^3 = 125\), so \(r = 5\).

Substitute \(r = 5\) back into the expression for \(S\):

\(S = 2\pi (5)^2 + \frac{500\pi}{5} = 50\pi + 100\pi = 150\pi\).

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

\(\frac{d^2S}{dr^2} = 4\pi + \frac{1000\pi}{r^3}\).

Since \(\frac{d^2S}{dr^2} > 0\) for \(r = 5\), the stationary value is a minimum.

(i) The area of the playground can be found by subtracting the areas of the triangles and trapezium from the rectangle:

\(Area of rectangle ABCD = 2400 m².\)

Area of triangle AXC = \(\frac{1}{2} \times 40 \times (60 - 2x) = 20(60 - 2x)\).

Area of triangle DYB = \(\frac{1}{2} \times 60 \times (40 - x) = 30(40 - x)\).

Area of trapezium AXYD = \(x \times 30\).

Thus, the area of the playground is:

\(A = 2400 - 20(60 - 2x) - 30(40 - x) - 30x\)

\(A = x^2 - 30x + 1200\).

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

\(\frac{dA}{dx} = 2x - 30\).

Setting \(\frac{dA}{dx} = 0\) gives:

\(2x - 30 = 0\)

\(x = 15\).

Substituting \(x = 15\) into the area formula:

\(A = 15^2 - 30 \times 15 + 1200 = 975\).

Thus, the minimum area of the playground is 975 m².

(i) The perimeter of the L-shaped garden is given by the sum of all its sides: \(y + 3x + 2y + x + 3y + 4x = 48\). Simplifying, we get \(6y + 8x = 48\). Solving for \(y\), we have \(y = \frac{1}{6}(48 - 8x)\).

(ii) The area \(A\) of the L-shaped garden can be expressed as the sum of the areas of the two rectangles: \(A = 4xy + 2xy = 6xy\). Substituting \(y = \frac{1}{6}(48 - 8x)\) into the area equation, we get \(A = x(48 - 8x) = 48x - 8x^2\).

(iii) To find the maximum area, we differentiate \(A\) with respect to \(x\): \(\frac{dA}{dx} = 48 - 16x\). Setting \(\frac{dA}{dx} = 0\) gives \(48 - 16x = 0\), so \(x = 3\). Substituting \(x = 3\) into the area equation, \(A = 48(3) - 8(3)^2 = 144 - 72 = 72\). To confirm it's a maximum, we check the second derivative: \(\frac{d^2A}{dx^2} = -16\), which is less than zero, indicating a maximum.

(i) The volume of the tank is given by \(x \times \frac{1}{2}x \times h = 4\). Thus, \(\frac{1}{2}x^2h = 4\), giving \(h = \frac{8}{x^2}\).

The external surface area \(A\) is the sum of the areas of the sides and the lid:

\(A = x \times \frac{1}{2}x + 2(x \times h) + 2\left(\frac{1}{2}x \times h\right) + \frac{5}{4}x \times \frac{4}{5}x\).

\(A = \frac{1}{2}x^2 + 2xh + xh + \frac{5}{4}x \times \frac{4}{5}x\).

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

Substituting \(h = \frac{8}{x^2}\), we get:

\(A = \frac{3}{2}x^2 + 3x \times \frac{8}{x^2}\).

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

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

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

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

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

\(3x^3 = 24\).

\(x^3 = 8\).

\(x = 2\).

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

\(\frac{d^2A}{dx^2} = 3 + \frac{48}{x^3}\).

\(\frac{d^2A}{dx^2} > 0\) when \(x = 2\), hence \(A\) is a minimum.

(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.

(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.

(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.

(i) To find the area of rectangle OPQR, we need to express y in terms of x. Using similar triangles, we have:

\(\frac{y}{16-x} = \frac{12}{16}\)

Solving for y, we get:

\(y = 12 - \frac{3}{4}x\)

The area A of rectangle OPQR is given by:

\(A = xy = x(12 - \frac{3}{4}x) = 12x - \frac{3}{4}x^2\)

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

\(\frac{dA}{dx} = 12 - \frac{6}{4}x\)

Setting \(\frac{dA}{dx} = 0\), we solve:

\(12 - \frac{6}{4}x = 0\)

\(x = 8\)

Substituting \(x = 8\) back into the area formula:

\(A = 12(8) - \frac{3}{4}(8)^2 = 96 - 48 = 48\)

Since the second derivative \(\frac{d^2A}{dx^2} = -\frac{6}{4}\) is negative, the stationary point is a maximum.

(i) The area A of the track is composed of two rectangles and two semicircles. The area of the rectangles is given by 2xr and the area of the semicircles is \pi r^2. Therefore, the total area is:

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

Given the perimeter is 400 metres, we have:

\(2x + 2\pi r = 400\)

Solving for x gives:

\(x = 200 - \pi r\)

Substitute x into the area equation:

\(A = 2(200 - \pi r)r + \pi r^2\)

\(A = 400r - 2\pi r^2 + \pi r^2\)

\(A = 400r - \pi r^2\)

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

\(\frac{dA}{dr} = 400 - 2\pi r\)

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

\(400 - 2\pi r = 0\)

\(r = \frac{200}{\pi}\)

Substitute back to find x:

\(x = 200 - \pi \left(\frac{200}{\pi}\right) = 0\)

Thus, there are no straight sections.

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

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

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