环境管理局希望建造一个围栏，围住下图所示的一个公园。$BCEF$ 是一个边长为 2 千米的正方形。$ECD$ 和 $ABF$ 是两个完全相同的三角形，其中 $CD = CE$，且角 $DCE = \theta$ 弧度。建造围栏的费用为 28000 美元。已知 $\theta$ 足够小，以至于 $\theta^2$ 及更高次幂的 $\theta$ 可以忽略不计，证明建造围栏的单位成本大约为每千米 $\char36{}(a + b\theta)$，其中 $a$ 和 $b$ 是待定的常数。

1. Find the length of DE in terms of $\theta$ using the cosine rule and small angle approximations.

In $\triangle ECD$, $CE = 2$ (side of the square) and $CD = 2$. By the cosine rule,
$DE^2 = CD^2 + CE^2 - 2(CD)(CE)\cos \theta$
$DE^2 = 2^2 + 2^2 - 2(2)(2)\cos \theta = 8(1 - \cos \theta)$
Using the small angle approximation $\cos \theta \approx 1 - \frac{\theta^2}{2}$,
$DE^2 \approx 8\left(1 - \left(1 - \frac{\theta^2}{2}\right)\right) = 4\theta^2$
$DE \approx 2\theta$
Since $\triangle ABF$ is identical to $\triangle ECD$, we also have $FA = DE \approx 2\theta$.

2. Determine the total perimeter of the park.

The boundary of the park passes through $D, E, F, A, B, C, D$.
The lengths of the segments are:
$DE \approx 2\theta$
$EF = 2$ (top side of the square)
$FA \approx 2\theta$
$AB = 2$ (since $CD=2$ and triangles are identical)
$BC = 2$ (bottom side of the square)
$CD = 2$
$\text{Perimeter} = DE + EF + FA + AB + BC + CD$
$\text{Perimeter} \approx 2\theta + 2 + 2\theta + 2 + 2 + 2 = 8 + 4\theta \text{ km}$

3. Formulate an expression for the unit cost of building the fence.

The total cost of building the fence is $\char36{}28000$.
$\text{Unit Cost} = \frac{\text{Total Cost}}{\text{Perimeter}}$
$\text{Unit Cost} \approx \frac{28000}{8 + 4\theta} = \frac{28000}{8(1 + 0.5\theta)} = 3500(1 + 0.5\theta)^{-1}$

4. Use the binomial expansion to approximate the unit cost in the required form.

Using the standard series expansion $(1 + x)^n \approx 1 + nx$ for small $x$, where higher powers of $\theta$ are neglected:
$(1 + 0.5\theta)^{-1} \approx 1 + (-1)(0.5\theta) = 1 - 0.5\theta$
$\text{Unit Cost} \approx 3500(1 - 0.5\theta) = 3500 - 1750\theta$
This matches the form $\char36{}(a + b\theta)$, so $a = 3500$ and $b = -1750$.