Stage Truss Projection Screen Aspect Ratio (Entertainment Engineering & Design) Solution

Problem Overview

An entertainment engineering team inscribes a maximum-area rectangular LED video display wall under a parabolic lighting truss arch governed by $y = 9 - \frac{1}{12}x^2$ resting on the stage floor $y = 0$. We express area as a single-variable cubic function, establish the closed physical domain, and evaluate extrema via the Closed Interval Method.

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Step-by-Step Mathematical Solution

Step 1: Geometric Modeling and Dimension Formulas

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

Let $(x, y)$ denote the coordinates of the upper right corner of the inscribed rectangular video display in the first quadrant ($x > 0, y > 0$).

1. By bilateral symmetry across the centerline ($x = 0$):
   • Total Display Width:
     $$W = 2x$$
   • Total Display Height:
     $$H = y = 9 - \frac{1}{12}x^2$$
2. Objective Function: Maximize Total Screen Area:
   $$A(x) = W \cdot H = (2x)\left(9 - \frac{1}{12}x^2\right) = 18x - \frac{1}{6}x^3$$

Step 2: Establish the Context-Driven Domain

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

1. The half-width must be non-negative: $x \ge 0$.
2. The height must be non-negative:
   $$y \ge 0 \implies 9 - \frac{1}{12}x^2 \ge 0 \implies \frac{1}{12}x^2 \le 9 \implies x^2 \le 108$$
   $$x \le \sqrt{108} = 6\sqrt{3} \approx 10.39\text{ meters}$$
   (The truss anchors into the stage floor at $x = \pm 6\sqrt{3}\text{ m}$).
3. Feasible domain is the closed interval:
   $$x \in [0, \sqrt{108}]$$

Step 3: Differentiate and Locate Critical Numbers

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

Set $A'(x) = 0$ to identify stationary critical points: (Reject $x = -6\text{ m}$ because $x \ge 0$). The critical number $x = 6\text{ m}$ lies strictly within the interior of the domain $[0, \sqrt{108}]$ since $6 < \sqrt{108} \approx 10.39$.

Step 4: Execute Closed Interval Method (EVT Verification)

The area function $A(x)$ is a polynomial (continuous everywhere) on the closed, bounded interval $[0, \sqrt{108}]$. By the Extreme Value Theorem (defined in Extreme Value Theorem), an absolute maximum is guaranteed to exist.

Evaluate $A(x)$ at both boundary endpoints and the interior critical point:

1. Left Boundary ($x = 0\text{ m}$ - zero width line):
   $$A(0) = 18(0) - \frac{1}{6}(0)^3 = 0\text{ m}^2$$
2. Right Boundary ($x = \sqrt{108}\text{ m}$ - zero height strip on floor):
   $$A(\sqrt{108}) = 18(\sqrt{108}) - \frac{1}{6}(\sqrt{108})^3 = 18(6\sqrt{3}) - \frac{1}{6}(108 \cdot 6\sqrt{3}) = 108\sqrt{3} - 108\sqrt{3} = 0\text{ m}^2$$
3. Interior Critical Point ($x = 6\text{ m}$):
   $$A(6) = 18(6) - \frac{1}{6}(6)^3 = 108 - \frac{216}{6} = 108 - 36 = 72\text{ m}^2$$

Comparing candidate values $\{0, 0, 72\}$, the absolute maximum display area is $72\text{ m}^2$, occurring at $x = 6\text{ meters}$.

Step 5: Final Screen Dimensions and Aspect Ratio

1. Total Width:
   $$W = 2x = 2(6) = 12\text{ meters}$$
2. Total Height:
   $$H = 9 - \frac{1}{12}(6)^2 = 9 - \frac{36}{12} = 9 - 3 = 6\text{ meters}$$
3. Optimized Aspect Ratio:
   $$\text{Aspect Ratio} = \frac{W}{H} = \frac{12\text{ m}}{6\text{ m}} = 2:1 \quad (\text{Univisium widescreen panoramic format})$$

Theatrical Rigging Insight: Notice that at the optimum, the screen height ($6\text{ m}$) is exactly two-thirds of the total arch apex height ($9\text{ m}$), leaving the upper $3\text{ meters}$ of the truss arch open for overhead theatrical spotlights and line-array audio speakers.