Terrain Surveying & Road Grade Safety Solution

In civil and transportation engineering, route alignment across mountainous topography requires rigorous analysis of surveyed ground elevation models. Before heavy earthmoving equipment and drilling rigs can be deployed to cut roadway grades, geotechnical engineers must establish the exact spatial boundaries of the mathematical elevation model to ensure equipment is never dispatched into unmapped or physically invalid zones.

A civil engineering survey team at UNLV models the localized terrain elevation $z$ (measured in hundreds of meters above a base datum) for a proposed mountain bypass using the multivariable radical function:

A planned construction access corridor requires heavy excavation machinery to traverse the regional grid point $(3, 2)$.

1. State the algebraic domain condition for $f(x, y)$ and determine the mathematical domain $D$ in set-builder notation. Describe its geometric boundary and classify the region as open or closed, and bounded or unbounded.
2. Determine the mathematical range $R$ of the elevation function in interval notation, identifying the maximum mountain peak elevation and the perimeter base elevation in physical units (meters).
3. Evaluate whether the proposed construction access point $(3, 2)$ lies within the valid terrain domain. Explain what this result means for project safety and heavy equipment operations.

Step-by-Step Solution:

1. Determine the Mathematical Domain and Topological Classification: State the Radicand Non-Negativity Condition for Multivariable Radical Functions:

   $$16 - x^2 - y^2 \ge 0$$
   Rearranging terms yields:
   $$x^2 + y^2 \le 16$$
   • Set-Builder Notation:
     $$D = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \le 16\}$$
   • Geometric Description: The domain is a closed circular disk of radius $r = \sqrt{16} = 4\text{ km}$ centered at the base station $(0, 0)$.
   • Classification: Because the region contains its entire circular boundary curve $x^2 + y^2 = 16$ (non-strict inequality $\le$), it is closed. Because the entire disk lies within a finite radius of $4\text{ km}$, it is bounded. (Defined in Domain and Range of Multivariable Functions.)

2. Determine the Elevation Range: State the Extremal Value Analysis for Bounded Quadratic Radicands: The quantity $x^2 + y^2$ attains its minimum value of $0$ at the origin $(0, 0)$:

   $$z_{\max} = f(0, 0) = \sqrt{16 - 0} = 4\text{ (in hundreds of meters)} \implies 400\text{ meters}$$
   The quantity $x^2 + y^2$ attains its maximum value of $16$ along the perimeter boundary $x^2 + y^2 = 16$:
   $$z_{\min} = \sqrt{16 - 16} = 0\text{ meters}$$
   • Range in Interval Notation:
     $$R = [0, 4]$$
     In physical elevation units, the terrain elevation spans from $0\text{ m}$ at the outer survey perimeter to a peak summit of $400\text{ m}$ directly above the base transit station.

3. Evaluate Proposed Access Point $(3, 2)$: Substitute $x = 3$ and $y = 2$ into the domain condition:

   $$x^2 + y^2 = 3^2 + 2^2 = 9 + 4 = 13$$
   Check against the boundary limit:
   $$13 \le 16 \quad \text{(True)}$$
   Because $13 < 16$, the point $(3, 2)$ lies strictly within the interior of the domain disk. Evaluating the actual elevation:
   $$z = f(3, 2) = \sqrt{16 - 3^2 - 2^2} = \sqrt{16 - 13} = \sqrt{3} \approx 1.732\text{ hundreds of meters} \implies 173.2\text{ meters}$$

Meaning of the Answer & Real-Life Application:

Because the coordinate $(3, 2)$ satisfies the domain condition ($x^2 + y^2 = 13 \le 16$), it represents a physically real, accessible location on the hillside at an elevation of $173.2\text{ m}$. Had the coordinate evaluation yielded $x^2 + y^2 > 16$, the radicand would be negative, indicating a point outside the surveyed topography where the mathematical model breaks down (e.g., dropping off a cliff or extending beyond the project survey boundary). Confirming domain containment ensures that earthmoving contractors can safely grade access roads without navigating off-grid drop-offs.