(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.