a) To show that \( A(x) = \frac{108}{x} + 2x^2 \):
Volume of the cuboid: \( x \times x \times y = 36 \).
Solve for \( y \):
\( y = \frac{36}{x^2} \)

The container is open-top, so the outside surface area includes the base (\( x \times y \)), two sides parallel to the length (\( x \times y \) each), and two sides parallel to the width (\( x \times x \) each):
\( A(x) = x \times y + x \times y + x \times y + x \times x + x \times x \)
\( = 3 \times (x \times y) + 2 \times (x \times x) \)
\( = 3xy + 2x^2 \)

Substitute \( y = \frac{36}{x^2} \):
\( A(x) = 3 \times x \times \frac{36}{x^2} + 2x^2 \)
\( = \frac{3 \times 36}{x} + 2x^2 \)
\( = \frac{108}{x} + 2x^2 \)

Thus:
\( A(x) = \frac{108}{x} + 2x^2 \) [4]

b) To find \( A'(x) \):
\( A(x) = \frac{108}{x} + 2x^2 = 108x^{-1} + 2x^2 \)
Differentiate:
\( A'(x) = 108 \times (-1) \times x^{-2} + 2 \times 2x \)
\( = -\frac{108}{x^2} + 4x \)

Thus:
\( A'(x) = -\frac{108}{x^2} + 4x \) [2]

c) To find the height when the surface area is a minimum:
Set \( A'(x) = 0 \):
\( -\frac{108}{x^2} + 4x = 0 \)
\( 4x = \frac{108}{x^2} \)
\( 4x^3 = 108 \)
\( x^3 = 27 \)
\( x = 3 \)

Verify minimum using the second derivative:
\( A”(x) = \frac{d}{dx} \left( -\frac{108}{x^2} + 4x \right) = 108 \times 2 \times x^{-3} + 4 = \frac{216}{x^3} + 4 \)
At \( x = 3 \):
\( A”(3) = \frac{216}{3^3} + 4 = \frac{216}{27} + 4 = 8 + 4 = 12 \) (positive, indicating a minimum).

Height of the container:
\( x = 3 \) m [5]

d) To find the total cost of paint tins:
Using \( x = 3 \), calculate the minimum surface area:
\( A(3) = \frac{108}{3} + 2 \times 3^2 = 36 + 2 \times 9 = 36 + 18 = 54 \) m\(^2\)

Number of tins needed (each covers 10 m\(^2\)):
\( \frac{54}{10} = 5.4 \)
Since partial tins cannot be purchased, use 6 tins.

Cost per tin is $20, so total cost:
\( 6 \times 20 = 120 \)

Total cost: $120 [5]