Regional Wastewater Effluent Pipeline Routing (Civil & Environmental Engineering) Solution

Problem Overview

A civil and environmental engineering team designs an optimal utility pipeline trajectory connecting a wastewater treatment plant across a $600\text{-m}$ wide wetland corridor to a discharge terminal $2000\text{ m}$ downstream. Wetland directional drilling costs $\$400/\text{m}$ while dry utility easement trenching costs $\$240/\text{m}$. We formulate the cost objective, establish the physical domain, find the optimal crossing point using the Closed Interval Method, and demonstrate Snell's Law equivalence.

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Step-by-Step Mathematical Solution

Step 1: Formulate the Total Construction Cost Objective Function

The total project capital expenditure is the sum of the wetland installation cost and the dry overland easement installation cost:

(Defined in Applied Optimization Modeling Framework.)

1. Wetland Segment: The pipeline originates at $(0, 600)$ and drills diagonally under the wetland to point $P(x, 0)$ on the southern easement bank. By the Pythagorean Theorem:
   $$d_{\text{wetland}}(x) = \sqrt{600^2 + x^2} = \sqrt{360,000 + x^2}\text{ meters}$$
   Cost: $C_{\text{wetland}}(x) = 400\sqrt{360,000 + x^2}$ dollars.
2. Dry Easement Segment: From $P(x, 0)$, the pipe runs along the dry riverbank to terminal $T(2000, 0)$:
   $$d_{\text{land}}(x) = 2000 - x\text{ meters}$$
   Cost: $C_{\text{land}}(x) = 240(2000 - x)$ dollars.
3. Combined Objective Function:
   $$C(x) = 400\sqrt{360,000 + x^2} + 240(2000 - x)$$

Step 2: Establish the Context-Driven Physical Domain

The downstream crossing point $P(x, 0)$ must physically lie between the direct perpendicular alignment ($x = 0$) and the terminal outfall location ($x = 2000$):

Because the domain is a closed bounded interval $[a, b]$, the Extreme Value Theorem guarantees the existence of an absolute minimum, which can be identified using the Closed Interval Method.

Step 3: Differentiate and Locate Critical Numbers

Differentiate $C(x)$ with respect to $x$:

Set $C'(x) = 0$ to find the stationary point:

Square both sides to eliminate the radical:

Cross-multiply and isolate $x^2$: Taking the positive physical square root: Because $450 \in [0, 2000]$, $x = 450\text{ m}$ is a valid physical critical number.

Step 4: Deconstructing Snell's Law Equivalence

Notice the structure of the critical point condition:

Let $\theta$ represent the angle between the diagonal wetland pipeline and the perpendicular bank normal. Then $\sin \theta = \frac{x}{\sqrt{600^2 + x^2}} = \frac{3}{5}$.
The ratio of unit construction costs directly mirrors Snell's Law in optics: (Formalized in Fastest Path Across Media and Snell's Law.)

Step 5: Execute the Closed Interval Method

Evaluate $C(x)$ at the domain boundary endpoints and the interior critical number:

1. Perpendicular Crossing ($x = 0\text{ m}$):
   • Wetland length: $600\text{ m}$; Easement length: $2000\text{ m}$.
     $$C(0) = 400(600) + 240(2000) = 240,000 + 480,000 = \$720,000$$
2. Direct Straight-Line Diagonal ($x = 2000\text{ m}$):
   • Wetland length: $\sqrt{600^2 + 2000^2} = \sqrt{4,360,000} \approx 2088.06\text{ m}$; Easement length: $0\text{ m}$.
     $$C(2000) = 400(2088.06) \approx \$835,224$$
3. Optimal Refracted Alignment ($x = 450\text{ m}$):
   • Wetland length: $\sqrt{600^2 + 450^2} = \sqrt{360,000 + 202,500} = \sqrt{562,500} = 750\text{ m}$ (a scaled $3-4-5$ right triangle).
   • Easement length: $2000 - 450 = 1550\text{ m}$.
     $$C(450) = 400(750) + 240(1550) = 300,000 + 372,000 = \$672,000$$

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Engineering Significance & Conclusion

• Optimal Route: The effluent pipeline should cross the wetland along a $750\text{-meter}$ diagonal path reaching the southern riverbank exactly $450\text{ meters}$ downstream, then follow the dry utility easement for the remaining $1,550\text{ meters}$ to outfall terminal $T$.
• Capital Cost Savings: This mathematically optimized trajectory yields an absolute minimum capital cost of $\$672,000$, generating:
  • $\$48,000$ in savings ($6.7\%$) compared to a naive perpendicular crossing ($\$720,000$).
  • $\$163,224$ in savings ($19.5\%$) compared to a direct diagonal line ($\$835,224$).