Figure 4 shows a closed letter box ABFEHGCD, which is made to be attached to a wall of a house - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 1

The total surface area $S$ of the prism consists of the area of the two bases and the lateral surface area:

1. Area of two bases (rectangles) = $2 \cdot 45 = 90$ cm²
2. Lateral surface area = perimeter of the base $ABCD \times \text{Height} = (4 + 6 + 5 + 9) \cdot y = 24y$ cm²

Thus, the complete surface area is: $S = 90 + 24y$

Substituting the expression for $y$ we found: $S = 90 + 24 \cdot \left(\frac{320}{x^2}\right) = 90 + \frac{7680}{x^2}$

To find the minimum value of the surface area $S$, we first differentiate it with respect to $x$:

$\frac{dS}{dx} = \frac{d}{dx}(60x + \frac{7680}{x}) = 60 - \frac{7680}{x^2}$

Setting the derivative to zero for critical points: $60 - \frac{7680}{x^2} = 0\Rightarrow 60x^2 = 7680\Rightarrow x^2 = 128\Rightarrow x = 8\text{ cm}$

To confirm it's a minimum, we check the second derivative: $\frac{d^2S}{dx^2} = \frac{15360}{x^3}$ Since this is positive when $x = 8$, $S$ has a local minimum there.

To confirm that the critical point we found is indeed a minimum, we use the second derivative test. As we calculated earlier:

$\frac{d^2S}{dx^2} = \frac{15360}{x^3}$

For $x = 8$: $\frac{d^2S}{dx^2} = \frac{15360}{8^3} > 0$ This indicates that the surface area $S$ is concave up at this point, confirming that $S$ achieves a minimum at $x = 8$ cm.