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 2 - 2014 - Paper 1

The total surface area S of a prism can be calculated using the areas of the two bases along with the area of the side walls:

$S = 2 \cdot \text{Base Area} + \text{Lateral Surface Area}$

The lateral area can be obtained from the perimeter of the base multiplied by the height (length of the prism, y):

$\text{Perimeter} = 4 + 5 + 6 + 9 = 24 \text{ cm}$

So:

$\text{Lateral Surface Area} = \text{Perimeter} \cdot y = 24 \cdot y = 24 \cdot \frac{320}{x^2}$

Therefore, substituting y into the surface area equation:

$S = 2(27) + 24(320/x^2)$

After simplifying:

$S = 60x + 7680/x$

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

$S' = 60 - \frac{7680}{x^2}$

Setting the derivative equal to zero to find critical points:

$60 - \frac{7680}{x^2} = 0$

This gives:

$\frac{7680}{x^2} = 60$

Solving for x:

$x^2 = \frac{7680}{60} = 128 \Rightarrow x = 8$

To verify it's a minimum, we can check the second derivative:

$S'' = \frac{15360}{x^3}$

Since S'' is positive when x > 0, S has a local minimum at this point.

Taking the second derivative confirmed that:

$S'' = \frac{15360}{x^3} > 0$

for all x > 0. Since the second derivative is positive, the function S is concave up at the point x = 8, confirming that this is indeed a minimum value for S.