A Hwa Chong student taking part in Project Day needs to design a rectangular poster to showcase his project. According to the rules set for the poster design, the poster should contain 1352 cm\textsuperscript{2} of printing with margins of 4 cm each at the top and bottom, and 2 cm each on the left and right sides. By using differentiation, determine the dimensions of the poster to minimise the amount of paper used.

1. Define variables for the printed area and express the dimensions of the poster.

Let the width and height of the printed area be $x$ cm and $y$ cm respectively.
Given the printed area is $1352\text{ cm}^2$, we have $xy = 1352$, which gives $y = \frac{1352}{x}$.
With margins of $4\text{ cm}$ each at the top and bottom, and $2\text{ cm}$ each on the left and right:
Width of poster $= x + 2 + 2 = x + 4$
Height of poster $= y + 4 + 4 = y + 8$

2. Express the total area of the poster, $A$, in terms of $x$.

Area of poster, $A = (x + 4)(y + 8)$
Substitute $y = \frac{1352}{x}$ into the equation:
$A = (x + 4)\left(\frac{1352}{x} + 8\right)$
$A = 1352 + 8x + \frac{5408}{x} + 32$
$A = 1384 + 8x + \frac{5408}{x}$

3. Find the first derivative $\frac{\mathrm{d}A}{\mathrm{d}x}$ and equate it to zero for the minimum area.

Differentiate $A$ with respect to $x$:
$\frac{\mathrm{d}A}{\mathrm{d}x} = 8 - \frac{5408}{x^2}$
For the area to be a minimum, the stationary point condition must be met:
$\frac{\mathrm{d}A}{\mathrm{d}x} = 0$
$8 - \frac{5408}{x^2} = 0$

4. Solve for $x$.

$8 = \frac{5408}{x^2}$
$8x^2 = 5408$
$x^2 = 676$
Since the width must be positive ($x > 0$):
$x = \sqrt{676} = 26$

5. Perform the second derivative test to verify that the area is minimised.

Find the second derivative:
$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = -5408(-2)x^{-3} = \frac{10816}{x^3}$
Substitute $x = 26$:
$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = \frac{10816}{(26)^3} > 0$
Since $\frac{\mathrm{d}^2A}{\mathrm{d}x^2} > 0$, the area $A$ is indeed minimised when $x = 26$.

6. Calculate the dimensions of the poster.

Substitute $x = 26$ back into the expressions for the poster dimensions:
Width $= 26 + 4 = 30\text{ cm}$
Height $= y + 8 = \frac{1352}{26} + 8 = 52 + 8 = 60\text{ cm}$