Extrema on Closed and Bounded Regions

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Concept Explanation

The Multi-Tiered Candidate Search Strategy

Because absolute extrema on a compact domain $D$ can reside either strictly inside the interior or along the boundary edges, engineers must execute a structured 3-Tier Search Strategy:

        Boundary Edge 2: y = 4 (0 <= x <= 4)
     (0, 4) +-----------------------+ (4, 4)
            |                       |
            |      * Interior       |
Boundary    |        Critical Point | Boundary Edge 3:
Edge 1:     |        (x_0, y_0)     | x = 4 (0 <= y <= 4)
x = 0       |                       |
            |                       |
     (0, 0) +-----------------------+ (4, 0)
        Boundary Edge 4: y = 0 (0 <= x <= 4)

1. Tier 1: Interior Search

   • Compute first partial derivatives $f_x$ and $f_y$ and set them to zero:
     $$\nabla f(x, y) = \mathbf{0} \implies f_x = 0, \quad f_y = 0$$
   • Domain Filtering Rule: Retain only those critical points that lie strictly inside the open interior of $D$. Any critical point lying outside $D$ must be discarded.
   • Note: You do not need to compute the second derivative test discriminant $D$ here; you only need the function values $f(x, y)$ at the candidate points.

2. Tier 2: Boundary Edge Search

   • Decompose the boundary $\partial D$ into smooth parameterized curves or line segments (e.g., straight sides of a polygon, circular arcs).
   • For each boundary segment, express the boundary mathematically:
     • By direct substitution: Replace one variable with an expression in terms of the other (e.g., $y = mx + b$ or $y^2 = R^2 - x^2$), reducing $f(x, y)$ to a single-variable function $g(x)$.
     • By parameterization: Express the curve as $\vec{r}(t) = \langle x(t), y(t) \rangle$ for $t \in [t_1, t_2]$, giving $g(t) = f(x(t), y(t))$.
   • Differentiate $g'(x) = 0$ (or $g'(t) = 0$) to locate 1D critical points along that boundary segment, verifying they lie within the segment interval.

3. Tier 3: Corner / Vertex Check

   • Identify all intersection points (vertices or corners) where adjacent boundary segments meet (e.g., $(0,0), (4,0), (0,3)$ on a triangle).
   • Boundary derivatives may not be defined across sharp corners, so all vertices are automatically candidate points.

4. Global Evaluation & Comparison

   • Construct a master evaluation table listing every candidate point from Tiers 1, 2, and 3.
   • Evaluate the original function $f(x, y)$ at every candidate.
   • Absolute Maximum: The largest value in the table.
   • Absolute Minimum: The smallest value in the table.

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Engineering Applications

• Civil & Geotechnical Engineering: Foundation footing slabs transmit vertical column loads to underlying subgrade soil. Geotechnical engineers determine the absolute maximum soil bearing pressure $P(x, y)$ across triangular or rectangular footings to ensure values do not exceed the soil's ultimate bearing capacity $q_{\text{ult}}$, preventing catastrophic footing shear failure.
• Mechanical & Materials Engineering: Silicon microprocessor dies operate under high thermal power dissipation. Mechanical engineers calculate temperature fields $T(x, y)$ on closed rectangular dies ($0 \le x \le L, 0 \le y \le W$) to pinpoint maximum hotspot temperatures at interior cores or boundary heat pipe interfaces, preventing thermal throttling or silicon delamination.
• Structural Engineering: Bending stress $\sigma(x, y)$ across structural beam cross-sections subjected to combined axial load and biaxial bending must be evaluated across the perimeter of the cross-section to verify yield safety margins.

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Example Problems

Example 1: Absolute Extrema on a Closed Rectangular Region

Find the absolute maximum and minimum values of:

on the closed square region $D = \{ (x, y) \in \mathbb{R}^2 \mid 0 \le x \le 4, \, 0 \le y \le 4 \}$.

Solution

1. Tier 1: Interior Search:

   $$f_x = 2x - 2 = 0 \implies x = 1$$
   $$f_y = 2y - 4 = 0 \implies y = 2$$
   Critical point: $(1, 2)$. Since $0 < 1 < 4$ and $0 < 2 < 4$, $(1, 2)$ lies strictly inside $D$.
   $$f(1, 2) = 1^2 + 2^2 - 2(1) - 4(2) + 5 = 1 + 4 - 2 - 8 + 5 = 0$$

2. Tier 2: Boundary Edge Search:

   • Bottom Edge ($y = 0, 0 \le x \le 4$):
     $$g_1(x) = f(x, 0) = x^2 - 2x + 5 \implies g_1'(x) = 2x - 2 = 0 \implies x = 1$$
     Candidate: $(1, 0)$. Value: $f(1, 0) = 1^2 - 2(1) + 5 = 4$.
   • Right Edge ($x = 4, 0 \le y \le 4$):
     $$g_2(y) = f(4, y) = 16 + y^2 - 8 - 4y + 5 = y^2 - 4y + 13 \implies g_2'(y) = 2y - 4 = 0 \implies y = 2$$
     Candidate: $(4, 2)$. Value: $f(4, 2) = 2^2 - 4(2) + 13 = 9$.
   • Top Edge ($y = 4, 0 \le x \le 4$):
     $$g_3(x) = f(x, 4) = x^2 + 16 - 2x - 16 + 5 = x^2 - 2x + 5 \implies g_3'(x) = 2x - 2 = 0 \implies x = 1$$
     Candidate: $(1, 4)$. Value: $f(1, 4) = 1^2 - 2(1) + 5 = 4$.
   • Left Edge ($x = 0, 0 \le y \le 4$):
     $$g_4(y) = f(0, y) = y^2 - 4y + 5 \implies g_4'(y) = 2y - 4 = 0 \implies y = 2$$
     Candidate: $(0, 2)$. Value: $f(0, 2) = 2^2 - 4(2) + 5 = 1$.

3. Tier 3: Corner / Vertex Check:

   • $(0, 0)$: $f(0, 0) = 5$
   • $(4, 0)$: $f(4, 0) = 4^2 - 2(4) + 5 = 16 - 8 + 5 = 13$
   • $(4, 4)$: $f(4, 4) = 4^2 + 4^2 - 2(4) - 4(4) + 5 = 16 + 16 - 8 - 16 + 5 = 13$
   • $(0, 4)$: $f(0, 4) = 4^2 - 4(4) + 5 = 5$

4. Master Comparison:

   Candidate $(x, y)$	Source	$f(x, y)$ Value
   $(1, 2)$	Interior Critical Point	$0$ (Absolute Minimum)
   $(0, 2)$	Left Edge	$1$
   $(1, 0)$	Bottom Edge	$4$
   $(1, 4)$	Top Edge	$4$
   $(0, 0)$	Bottom-Left Corner	$5$
   $(0, 4)$	Top-Left Corner	$5$
   $(4, 2)$	Right Edge	$9$
   $(4, 0)$	Bottom-Right Corner	$13$ (Absolute Maximum)
   $(4, 4)$	Top-Right Corner	$13$ (Absolute Maximum)

   Conclusion:

   • Absolute Minimum: $0$, attained at $(1, 2)$.
   • Absolute Maximum: $13$, attained at $(4, 0)$ and $(4, 4)$.

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Example 2: Absolute Extrema on a Triangular Region

Find the absolute extrema of $f(x, y) = x^2 + y^2$ on the closed triangular region bounded by the vertices $(0, 0)$, $(4, 0)$, and $(0, 3)$.

Solution

1. Interior: $f_x = 2x = 0 \implies x = 0$, $f_y = 2y = 0 \implies y = 0$. The point $(0, 0)$ lies on the boundary (vertex), so there are no strictly interior critical points.

2. Boundary Edges:

   • Bottom Edge ($y = 0, 0 \le x \le 4$): $f(x, 0) = x^2$. Minimum at $x = 0$ ($f = 0$); maximum at $x = 4$ ($f = 16$).
   • Left Edge ($x = 0, 0 \le y \le 3$): $f(0, y) = y^2$. Minimum at $y = 0$ ($f = 0$); maximum at $y = 3$ ($f = 9$).
   • Hypotenuse ($y = -\frac{3}{4}x + 3, 0 \le x \le 4$):
     $$g(x) = x^2 + \left(-\frac{3}{4}x + 3\right)^2 = x^2 + \frac{9}{16}x^2 - \frac{9}{2}x + 9 = \frac{25}{16}x^2 - \frac{9}{2}x + 9$$
     Set $g'(x) = 0$:
     $$g'(x) = \frac{50}{16}x - \frac{9}{2} = \frac{25}{8}x - \frac{9}{2} = 0 \implies x = \frac{9/2}{25/8} = \frac{9}{2} \cdot \frac{8}{25} = \frac{36}{25} = 1.44$$
     Corresponding $y$:
     $$y = -\frac{3}{4}\left(\frac{36}{25}\right) + 3 = -\frac{27}{25} + \frac{75}{25} = \frac{48}{25} = 1.92$$
     Evaluate function:
     $$f(1.44, 1.92) = (1.44)^2 + (1.92)^2 = 2.0736 + 3.6864 = 5.76$$

3. Vertices:

   • $(0, 0) \implies f(0, 0) = 0$
   • $(4, 0) \implies f(4, 0) = 16$
   • $(0, 3) \implies f(0, 3) = 9$

4. Conclusion:

   • Absolute Minimum: $0$ at the origin $(0, 0)$.
   • Absolute Maximum: $16$ at vertex $(4, 0)$.

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Connections & References

• Prerequisites:
  • Critical Points in Multivariable Functions — Finding interior stationary points.
  • Partial Derivatives — Differentiating multi-input functions.
• Direct Successors:
  • Lagrange Multipliers — Systematic method for finding extrema along curved or implicit boundaries.
• Course Sessions:
  • Session 13 - Optimization with Constraints