Solar Photovoltaic Array Perimeter & Inverter Enclosure (Civil & Construction Engineering) Solution

Problem Overview

A civil engineering team designs an optimized perimeter and partition fencing layout for a $120,000\text{ m}^2$ photovoltaic collection field. Outer security fencing costs $\$150/\text{m}$ while two internal inverter dividing fences cost $\$50/\text{m}$. We minimize total construction cost using single-variable calculus and verify the global minimum using the Second Derivative Test.

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Step-by-Step Mathematical Solution

Step 1: Formulate Governing Objective and Constraint Equations

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

1. Geometry & Cost Allocation:

   • Outer boundary: 2 horizontal sides of length $y$ and 2 vertical sides of length $x$. Unit cost $= \$150/\text{m}$.
   • Interior dividers: 2 vertical lines of length $x$. Unit cost $= \$50/\text{m}$.
   • Total cost of vertical segments ($x$):
     $$C_x = (2 \times 150 \cdot x) + (2 \times 50 \cdot x) = 300x + 100x = 400x$$
   • Total cost of horizontal segments ($y$):
     $$C_y = 2 \times 150 \cdot y = 300y$$
   • Total Cost Objective Function:
     $$C(x, y) = 400x + 300y$$

2. Area Constraint Equation: The total enclosed land footprint is fixed at $120,000\text{ m}^2$:

   $$A = x \cdot y = 120,000$$

Step 2: Variable Reduction (Substitution)

Isolate $y$ from the area constraint:

Substitute into the cost objective function:

Step 3: Establish the Context-Driven Domain

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

• Physical lengths must be positive: $x > 0$.
• As $x \to 0$, $y \to \infty$ and $C(x) \to \infty$.
• As $x \to \infty$, $y \to 0$ and $C(x) \to \infty$.
• Feasible domain is the open interval $x \in (0, \infty)$.

Step 4: Differentiate and Find Critical Numbers

Differentiate $C(x)$ with respect to $x$ using the Power Rule:

Set $C'(x) = 0$ to identify stationary points: (We reject the negative root $x = -300\text{ m}$ because distance cannot be negative).

Step 5: Verify Absolute Global Minimum via Second Derivative Test

Compute the second derivative:

Evaluate at $x = 300\text{ m}$: Since $C''(300) > 0$ and $x = 300\text{ m}$ is the sole critical point in the domain $(0, \infty)$, the Second Derivative Test proves that $x = 300\text{ m}$ yields the unique absolute minimum cost.

Step 6: Compute Optimal Dimensions and Total Construction Cost

1. Parallel Dimension ($y$):
   $$y = \frac{120,000}{300} = 400\text{ meters}$$
2. Minimum Total Construction Expenditure:
   $$C(300) = 400(300) + 300(400) = 120,000 + 120,000 = \$240,000$$

Notice the classic economic equipartition principle: at the cost-optimized optimum, total spending on vertical fencing ($400 \times 300 = \$120,000$) exactly balances total spending on horizontal fencing ($300 \times 400 = \$120,000$).