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.

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:

C(x) = C_{\text{wetland}} \cdot d_{\text{wetland}}(x) + C_{\text{land}} \cdot d_{\text{land}}(x)

(Defined in Applied Optimization Modeling Framework.)

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.
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.
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 ( $x = 2000$ ):

\text{Domain: } x \in [0, 2000]\text{ meters}

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 $x$ :

C'(x) = 400 \cdot \frac{1}{2\sqrt{360,000 + x^2}} \cdot (2x) + 240(-1) = \frac{400x}{\sqrt{360,000 + x^2}} - 240

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

\frac{400x}{\sqrt{360,000 + x^2}} = 240 \implies \frac{x}{\sqrt{360,000 + x^2}} = \frac{240}{400} = \frac{3}{5}

Square both sides to eliminate the radical:

\frac{x^2}{360,000 + x^2} = \frac{9}{25}

Cross-multiply and isolate

x^2

25x^2 = 9(360,000 + x^2) = 3,240,000 + 9x^2

16x^2 = 3,240,000 \implies x^2 = \frac{3,240,000}{16} = 202,500

Taking the positive physical square root:

x = \sqrt{202,500} = 450\text{ meters}

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:

\frac{x}{\sqrt{600^2 + x^2}} = \frac{240}{400} = 0.6

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:

\sin \theta = \frac{C_{\text{land}}}{C_{\text{wetland}}} = \frac{240}{400} = 0.6

(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:

Perpendicular Crossing ( $x = 0\text{ m}$ ):
- Wetland length: $600\text{ m}$ ; Easement length: $2000\text{ m}$ . $C(0) = 400(600) + 240(2000) = 240,000 + 480,000 = \$720,000$
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: $0\text{ m}$ . $C(2000) = 400(2088.06) \approx \$835,224$
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: $2000 - 450 = 1550\text{ m}$ . $C(450) = 400(750) + 240(1550) = 300,000 + 372,000 = \$672,000$

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 ( $\$835,224$ ).

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