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.

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.)

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: $C(x, y) = 400x + 300y$
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:

y = \frac{120,000}{x}

Substitute into the cost objective function:

C(x) = 400x + 300\left(\frac{120,000}{x}\right) = 400x + \frac{36,000,000}{x} = 400x + 36,000,000 x^{-1}

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:

C'(x) = \frac{d}{dx}\left[400x + 36,000,000 x^{-1}\right] = 400 - \frac{36,000,000}{x^2}

Set

C'(x) = 0

to identify stationary points:

400 - \frac{36,000,000}{x^2} = 0

400 = \frac{36,000,000}{x^2}

400x^2 = 36,000,000

x^2 = \frac{36,000,000}{400} = 90,000

x = \sqrt{90,000} = 300\text{ meters}

(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:

C''(x) = \frac{d}{dx}\left[400 - 36,000,000 x^{-2}\right] = \frac{72,000,000}{x^3}

Evaluate at

x = 300\text{ m}

C''(300) = \frac{72,000,000}{(300)^3} = \frac{72,000,000}{27,000,000} = \frac{8}{3} > 0

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

Parallel Dimension ( $y$ ): $y = \frac{120,000}{300} = 400\text{ meters}$
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 ( $300 \times 400 = \$120,000$ ).

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