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Tony's Intertwingled Memex • Reference Manual

Extrema on Closed and Bounded Regions

Formulas & Definitions
Core Definition

Extreme Value Theorem (EVT) for Functions of Two Variables:
If a real-valued function f(x,y)f(x, y)f(x,y) is continuous on a closed and bounded region D⊂R2D \subset \mathbb{R}^2D⊂R2, then fff attains both an absolute maximum value and an absolute minimum value at some points in DDD.

  • Closed Region: A region that contains all of its boundary points (e.g., defined by non-strict inequalities like ≤\le≤ or ≥\ge≥, such as x2+y2≤R2x^2 + y^2 \le R^2x2+y2≤R2).
  • Bounded Region: A region of finite spatial extent (it can be completely enclosed inside a sufficiently large disk of finite radius centered at the origin).

If DDD fails to be closed (missing boundary points) or fails to be bounded (extending infinitely), the existence of absolute extrema is not guaranteed.


Concept Explanation

The Multi-Tiered Candidate Search Strategy

Because absolute extrema on a compact domain DDD 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 fxf_xfx​ and fyf_yfy​ and set them to zero:
      ∇f(x,y)=0  ⟹  fx=0,fy=0\nabla f(x, y) = \mathbf{0} \implies f_x = 0, \quad f_y = 0∇f(x,y)=0⟹fx​=0,fy​=0
    • Domain Filtering Rule: Retain only those critical points that lie strictly inside the open interior of DDD. Any critical point lying outside DDD must be discarded.
    • Note: You do not need to compute the second derivative test discriminant DDD here; you only need the function values f(x,y)f(x, y)f(x,y) at the candidate points.
  2. Tier 2: Boundary Edge Search

    • Decompose the boundary ∂D\partial D∂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+by = mx + by=mx+b or y2=R2−x2y^2 = R^2 - x^2y2=R2−x2), reducing f(x,y)f(x, y)f(x,y) to a single-variable function g(x)g(x)g(x).
      • By parameterization: Express the curve as r⃗(t)=⟨x(t),y(t)⟩\vec{r}(t) = \langle x(t), y(t) \rangler(t)=⟨x(t),y(t)⟩ for t∈[t1,t2]t \in [t_1, t_2]t∈[t1​,t2​], giving g(t)=f(x(t),y(t))g(t) = f(x(t), y(t))g(t)=f(x(t),y(t)).
    • Differentiate g′(x)=0g'(x) = 0g′(x)=0 (or g′(t)=0g'(t) = 0g′(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)(0,0), (4,0), (0,3)(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)f(x, y)f(x,y) at every candidate.
    • Absolute Maximum: The largest value in the table.
    • Absolute Minimum: The smallest value in the table.

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)P(x, y)P(x,y) across triangular or rectangular footings to ensure values do not exceed the soil's ultimate bearing capacity qultq_{\text{ult}}qult​, 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)T(x, y)T(x,y) on closed rectangular dies (0≤x≤L,0≤y≤W0 \le x \le L, 0 \le y \le W0≤x≤L,0≤y≤W) to pinpoint maximum hotspot temperatures at interior cores or boundary heat pipe interfaces, preventing thermal throttling or silicon delamination.
  • Structural Engineering: Bending stress σ(x,y)\sigma(x, y)σ(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.

Example Problems

Example 1: Absolute Extrema on a Closed Rectangular Region

Find the absolute maximum and minimum values of:

f(x,y)=x2+y2−2x−4y+5f(x, y) = x^2 + y^2 - 2x - 4y + 5f(x,y)=x2+y2−2x−4y+5
on the closed square region D={(x,y)∈R2∣0≤x≤4, 0≤y≤4}D = \{ (x, y) \in \mathbb{R}^2 \mid 0 \le x \le 4, \, 0 \le y \le 4 \}D={(x,y)∈R2∣0≤x≤4,0≤y≤4}.

Solution
  1. Tier 1: Interior Search:

    fx=2x−2=0  ⟹  x=1f_x = 2x - 2 = 0 \implies x = 1fx​=2x−2=0⟹x=1
    fy=2y−4=0  ⟹  y=2f_y = 2y - 4 = 0 \implies y = 2fy​=2y−4=0⟹y=2
    Critical point: (1,2)(1, 2)(1,2). Since 0<1<40 < 1 < 40<1<4 and 0<2<40 < 2 < 40<2<4, (1,2)(1, 2)(1,2) lies strictly inside DDD.
    f(1,2)=12+22−2(1)−4(2)+5=1+4−2−8+5=0f(1, 2) = 1^2 + 2^2 - 2(1) - 4(2) + 5 = 1 + 4 - 2 - 8 + 5 = 0f(1,2)=12+22−2(1)−4(2)+5=1+4−2−8+5=0

  2. Tier 2: Boundary Edge Search:

    • Bottom Edge (y=0,0≤x≤4y = 0, 0 \le x \le 4y=0,0≤x≤4):
      g1(x)=f(x,0)=x2−2x+5  ⟹  g1′(x)=2x−2=0  ⟹  x=1g_1(x) = f(x, 0) = x^2 - 2x + 5 \implies g_1'(x) = 2x - 2 = 0 \implies x = 1g1​(x)=f(x,0)=x2−2x+5⟹g1′​(x)=2x−2=0⟹x=1
      Candidate: (1,0)(1, 0)(1,0). Value: f(1,0)=12−2(1)+5=4f(1, 0) = 1^2 - 2(1) + 5 = 4f(1,0)=12−2(1)+5=4.
    • Right Edge (x=4,0≤y≤4x = 4, 0 \le y \le 4x=4,0≤y≤4):
      g2(y)=f(4,y)=16+y2−8−4y+5=y2−4y+13  ⟹  g2′(y)=2y−4=0  ⟹  y=2g_2(y) = f(4, y) = 16 + y^2 - 8 - 4y + 5 = y^2 - 4y + 13 \implies g_2'(y) = 2y - 4 = 0 \implies y = 2g2​(y)=f(4,y)=16+y2−8−4y+5=y2−4y+13⟹g2′​(y)=2y−4=0⟹y=2
      Candidate: (4,2)(4, 2)(4,2). Value: f(4,2)=22−4(2)+13=9f(4, 2) = 2^2 - 4(2) + 13 = 9f(4,2)=22−4(2)+13=9.
    • Top Edge (y=4,0≤x≤4y = 4, 0 \le x \le 4y=4,0≤x≤4):
      g3(x)=f(x,4)=x2+16−2x−16+5=x2−2x+5  ⟹  g3′(x)=2x−2=0  ⟹  x=1g_3(x) = f(x, 4) = x^2 + 16 - 2x - 16 + 5 = x^2 - 2x + 5 \implies g_3'(x) = 2x - 2 = 0 \implies x = 1g3​(x)=f(x,4)=x2+16−2x−16+5=x2−2x+5⟹g3′​(x)=2x−2=0⟹x=1
      Candidate: (1,4)(1, 4)(1,4). Value: f(1,4)=12−2(1)+5=4f(1, 4) = 1^2 - 2(1) + 5 = 4f(1,4)=12−2(1)+5=4.
    • Left Edge (x=0,0≤y≤4x = 0, 0 \le y \le 4x=0,0≤y≤4):
      g4(y)=f(0,y)=y2−4y+5  ⟹  g4′(y)=2y−4=0  ⟹  y=2g_4(y) = f(0, y) = y^2 - 4y + 5 \implies g_4'(y) = 2y - 4 = 0 \implies y = 2g4​(y)=f(0,y)=y2−4y+5⟹g4′​(y)=2y−4=0⟹y=2
      Candidate: (0,2)(0, 2)(0,2). Value: f(0,2)=22−4(2)+5=1f(0, 2) = 2^2 - 4(2) + 5 = 1f(0,2)=22−4(2)+5=1.
  3. Tier 3: Corner / Vertex Check:

    • (0,0)(0, 0)(0,0): f(0,0)=5f(0, 0) = 5f(0,0)=5
    • (4,0)(4, 0)(4,0): f(4,0)=42−2(4)+5=16−8+5=13f(4, 0) = 4^2 - 2(4) + 5 = 16 - 8 + 5 = 13f(4,0)=42−2(4)+5=16−8+5=13
    • (4,4)(4, 4)(4,4): f(4,4)=42+42−2(4)−4(4)+5=16+16−8−16+5=13f(4, 4) = 4^2 + 4^2 - 2(4) - 4(4) + 5 = 16 + 16 - 8 - 16 + 5 = 13f(4,4)=42+42−2(4)−4(4)+5=16+16−8−16+5=13
    • (0,4)(0, 4)(0,4): f(0,4)=42−4(4)+5=5f(0, 4) = 4^2 - 4(4) + 5 = 5f(0,4)=42−4(4)+5=5
  4. Master Comparison:

    Candidate (x,y)(x, y)(x,y) Source f(x,y)f(x, y)f(x,y) Value
    (1,2)(1, 2)(1,2) Interior Critical Point 000 (Absolute Minimum)
    (0,2)(0, 2)(0,2) Left Edge 111
    (1,0)(1, 0)(1,0) Bottom Edge 444
    (1,4)(1, 4)(1,4) Top Edge 444
    (0,0)(0, 0)(0,0) Bottom-Left Corner 555
    (0,4)(0, 4)(0,4) Top-Left Corner 555
    (4,2)(4, 2)(4,2) Right Edge 999
    (4,0)(4, 0)(4,0) Bottom-Right Corner 131313 (Absolute Maximum)
    (4,4)(4, 4)(4,4) Top-Right Corner 131313 (Absolute Maximum)

    Conclusion:

    • Absolute Minimum: 000, attained at (1,2)(1, 2)(1,2).
    • Absolute Maximum: 131313, attained at (4,0)(4, 0)(4,0) and (4,4)(4, 4)(4,4).

Example 2: Absolute Extrema on a Triangular Region

Find the absolute extrema of f(x,y)=x2+y2f(x, y) = x^2 + y^2f(x,y)=x2+y2 on the closed triangular region bounded by the vertices (0,0)(0, 0)(0,0), (4,0)(4, 0)(4,0), and (0,3)(0, 3)(0,3).

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

  2. Boundary Edges:

    • Bottom Edge (y=0,0≤x≤4y = 0, 0 \le x \le 4y=0,0≤x≤4): f(x,0)=x2f(x, 0) = x^2f(x,0)=x2. Minimum at x=0x = 0x=0 (f=0f = 0f=0); maximum at x=4x = 4x=4 (f=16f = 16f=16).
    • Left Edge (x=0,0≤y≤3x = 0, 0 \le y \le 3x=0,0≤y≤3): f(0,y)=y2f(0, y) = y^2f(0,y)=y2. Minimum at y=0y = 0y=0 (f=0f = 0f=0); maximum at y=3y = 3y=3 (f=9f = 9f=9).
    • Hypotenuse (y=−34x+3,0≤x≤4y = -\frac{3}{4}x + 3, 0 \le x \le 4y=−43​x+3,0≤x≤4):
      g(x)=x2+(−34x+3)2=x2+916x2−92x+9=2516x2−92x+9g(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 + 9g(x)=x2+(−43​x+3)2=x2+169​x2−29​x+9=1625​x2−29​x+9
      Set g′(x)=0g'(x) = 0g′(x)=0:
      g′(x)=5016x−92=258x−92=0  ⟹  x=9/225/8=92⋅825=3625=1.44g'(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} = 1.44g′(x)=1650​x−29​=825​x−29​=0⟹x=25/89/2​=29​⋅258​=2536​=1.44
      Corresponding yyy:
      y=−34(3625)+3=−2725+7525=4825=1.92y = -\frac{3}{4}\left(\frac{36}{25}\right) + 3 = -\frac{27}{25} + \frac{75}{25} = \frac{48}{25} = 1.92y=−43​(2536​)+3=−2527​+2575​=2548​=1.92
      Evaluate function:
      f(1.44,1.92)=(1.44)2+(1.92)2=2.0736+3.6864=5.76f(1.44, 1.92) = (1.44)^2 + (1.92)^2 = 2.0736 + 3.6864 = 5.76f(1.44,1.92)=(1.44)2+(1.92)2=2.0736+3.6864=5.76
  3. Vertices:

    • (0,0)  ⟹  f(0,0)=0(0, 0) \implies f(0, 0) = 0(0,0)⟹f(0,0)=0
    • (4,0)  ⟹  f(4,0)=16(4, 0) \implies f(4, 0) = 16(4,0)⟹f(4,0)=16
    • (0,3)  ⟹  f(0,3)=9(0, 3) \implies f(0, 3) = 9(0,3)⟹f(0,3)=9
  4. Conclusion:

    • Absolute Minimum: 000 at the origin (0,0)(0, 0)(0,0).
    • Absolute Maximum: 161616 at vertex (4,0)(4, 0)(4,0).

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