作为艺术项目的一部分，一个学生设计了一个如图所示的八面体灯笼。灯笼的框架由十二根直铁丝组成，端点相连，形成两个相同的正四棱锥（其底面为边长为 $4x\text{ m}$ 的正方形，侧棱长为 $y\text{ m}$）。铁丝框架用彩色纸紧紧包住。 铁丝的价格为每米 8 美元，彩色纸的价格为每平方米 15 美元。学生需要在该项目中列出灯笼的预算。已知使用的铁丝总长度为 $3.2\text{ m}$，利用求导来求其最大预算金额，结果四舍五入到最接近的元。

1. Express the slant edge $y$ in terms of $x$ using the total length of the wire.

The wire frame consists of the square base (4 edges of length $4x$) and the slant edges of the two pyramids (8 edges of length $y$).
Total length of wire $= 4(4x) + 8y = 16x + 8y$
Given that the total length of the wire used is $3.2\text{ m}$:
$16x + 8y = 3.2$
$8y = 3.2 - 16x$
$y = 0.4 - 2x$

2. Find an expression for the total surface area of the lantern in terms of $x$ and $y$.

The surface of the lantern is formed by 8 identical isosceles triangular faces.
For one triangular face with base $4x$ and two slant edges $y$, the height $h$ is given by Pythagoras' theorem:
$h = \sqrt{y^2 - (2x)^2} = \sqrt{y^2 - 4x^2}$
Area of one triangular face $= \frac{1}{2}(4x)h = 2x\sqrt{y^2 - 4x^2}$
Total surface area $A = 8 \times 2x\sqrt{y^2 - 4x^2} = 16x\sqrt{y^2 - 4x^2}$

3. Substitute $y$ to express the total budget $C$ entirely in terms of $x$.

Substitute $y = 0.4 - 2x$ into $A$:
$A = 16x\sqrt{(0.4 - 2x)^2 - 4x^2}$
$A = 16x\sqrt{0.16 - 1.6x + 4x^2 - 4x^2} = 16x\sqrt{0.16 - 1.6x}$
The total budget $C$ is the sum of the wire cost and the paper cost.
Cost of wire $= 8 \times 3.2 = 25.6$
Cost of paper $= 15 \times A = 15 \times 16x\sqrt{0.16 - 1.6x} = 240x\sqrt{0.16 - 1.6x}$
$C = 25.6 + 240x\sqrt{0.16 - 1.6x}$

4. Differentiate $C$ with respect to $x$ and equate to $0$ to find the stationary point.

Using the product rule and chain rule to find $\frac{\mathrm{d}C}{\mathrm{d}x}$:
$\frac{\mathrm{d}C}{\mathrm{d}x} = 240 \left[ 1 \cdot \sqrt{0.16 - 1.6x} + x \cdot \frac{1}{2\sqrt{0.16 - 1.6x}}(-1.6) \right]$
$\frac{\mathrm{d}C}{\mathrm{d}x} = 240 \left[ \sqrt{0.16 - 1.6x} - \frac{0.8x}{\sqrt{0.16 - 1.6x}} \right]$
To maximize the budget, let $\frac{\mathrm{d}C}{\mathrm{d}x} = 0$:
$\sqrt{0.16 - 1.6x} - \frac{0.8x}{\sqrt{0.16 - 1.6x}} = 0$
Multiply through by $\sqrt{0.16 - 1.6x}$:
$0.16 - 1.6x - 0.8x = 0$
$0.16 - 2.4x = 0$
$x = \frac{0.16}{2.4} = \frac{1}{15}$

5. Verify that this value of $x$ gives a maximum budget using the First Derivative Test.

Testing the sign of $\frac{\mathrm{d}C}{\mathrm{d}x}$ around $x = \frac{1}{15} \approx 0.0667$:

• When $x = 0.06$, $\frac{\mathrm{d}C}{\mathrm{d}x} \propto 0.16 - 2.4(0.06) = 0.016 > 0$
• When $x = 0.07$, $\frac{\mathrm{d}C}{\mathrm{d}x} \propto 0.16 - 2.4(0.07) = -0.008 < 0$
  Since the gradient changes from positive to negative, $x = \frac{1}{15}$ yields a maximum point.

6. Calculate the maximum budget and round it to the nearest dollar.

Substitute $x = \frac{1}{15}$ into the budget equation:
$C = 25.6 + 240\left(\frac{1}{15}\right)\sqrt{0.16 - 1.6\left(\frac{1}{15}\right)}$
$C = 25.6 + 16\sqrt{\frac{4}{25} - \frac{16}{150}}$
$C = 25.6 + 16\sqrt{\frac{24 - 16}{150}} = 25.6 + 16\sqrt{\frac{8}{150}}$
$C = 25.6 + 16\sqrt{\frac{4}{75}} = 25.6 + \frac{32}{5\sqrt{3}}$
$C \approx 25.6 + 3.6950 = 29.295$
Rounding to the nearest dollar, the maximum budget is $\char36{}29$.