一名参加“项目日”（Project Day）的华中（Hwa Chong）学生需要设计一张矩形海报来展示他的项目。根据海报设计的规则，海报应包含 1352 cm\textsuperscript{2} 的印刷区域，上下边距各为 4 cm，左右边距各为 2 cm。利用微分法，确定海报的尺寸以使所使用的纸张面积最小。

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 $= y + 8 = \frac{1352}{26} + 8 = 52 + 8 = 60\text{ cm}$