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Agricultural Runoff Retention Wetland Partitioning (Environmental Engineering) Solution

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Problem Overview

An environmental engineering team optimizes the geometric layout of a constructed treatment wetland basin with active water area A=30,000 m2A = 30,000\text{ m}^2. Bounded on one long side by a natural ridge, the layout requires three transverse berm segments of width xx and one longitudinal berm of length yy. We minimize total berm construction length L=3x+yL = 3x + y.

Step-by-Step Mathematical Solution

Step 1: Formulate Governing Equations

The governing process follows the Applied Optimization Modeling Framework: (Defined in Applied Optimization Modeling Framework.)

  1. Objective Function: Minimize total linear berm length:
    L(x,y)=3x+yL(x, y) = 3x + y
    (3 segments of length xx: 2 outer sidewalls + 1 interior baffle; 1 segment of length yy: outer longitudinal wall).
  2. Hydraulic Surface Area Constraint:
    A=x⋅y=30,000 m2A = x \cdot y = 30,000\text{ m}^2

Step 2: Variable Reduction (Substitution)

Solve the area constraint for yy:

y=30,000xy = \frac{30,000}{x}
Substitute into the berm length objective:
L(x)=3x+30,000x=3x+30,000x−1L(x) = 3x + \frac{30,000}{x} = 3x + 30,000x^{-1}

Step 3: Context-Driven Domain

(Governed by Context-Driven Domain in Geometric Optimization.)

  • Physical dimensions must be positive: x>0x > 0.
  • Feasible domain is the open interval x∈(0,∞)x \in (0, \infty).

Step 4: Differentiate and Find Critical Numbers

Differentiate L(x)L(x) with respect to xx using the Power Rule:

L′(x)=ddx[3x+30,000x−1]=3−30,000x2L'(x) = \frac{d}{dx}\left[3x + 30,000x^{-1}\right] = 3 - \frac{30,000}{x^2}
Set L′(x)=0L'(x) = 0:
3−30,000x2=0  ⟹  3x2=30,000  ⟹  x2=10,0003 - \frac{30,000}{x^2} = 0 \implies 3x^2 = 30,000 \implies x^2 = 10,000
x=10,000=100 metersx = \sqrt{10,000} = 100\text{ meters}
(Reject x=−100 mx = -100\text{ m} as non-physical).

Step 5: Verify Global Minimum via Second Derivative Test

Compute the second derivative:

L′′(x)=ddx[3−30,000x−2]=60,000x3L''(x) = \frac{d}{dx}\left[3 - 30,000x^{-2}\right] = \frac{60,000}{x^3}
Evaluate at x=100 mx = 100\text{ m}:
L′′(100)=60,000(100)3=60,0001,000,000=0.06>0L''(100) = \frac{60,000}{(100)^3} = \frac{60,000}{1,000,000} = 0.06 > 0
Since L′′(100)>0L''(100) > 0 across the entire domain (0,∞)(0, \infty), the curve is concave up, proving that x=100 mx = 100\text{ m} is the unique absolute minimum berm length.

Step 6: Compute Optimal Dimensions and Total Berm Length

  1. Parallel Length (yy):
    y=30,000100=300 metersy = \frac{30,000}{100} = 300\text{ meters}
  2. Total Berm Length Required:
    L=3(100)+300=600 metersL = 3(100) + 300 = 600\text{ meters}

Environmental Design Takeaway: The optimal rectangular aspect ratio is y=3xy = 3x (the wetland is three times as long as it is wide). This elongated aspect ratio is simultaneously hydraulically ideal for preventing short-circuiting in plug-flow wastewater treatment.