Session 13 - Optimization with Constraints | Tony's Intertwingled Memex

📅 Micro-Lecture

1. Unconstrained Extrema & The Curvature Hessian

Mathematical Foundation: In multivariable modeling, stationary equilibrium states occur where the gradient vector vanishes: $\nabla f = \mathbf{0}$ , creating a completely horizontal tangent plane $z = f(a, b)$ . As established in Critical Points in Multivariable Functions, stationary points represent candidates for local peaks, valleys, or directional inflection saddles. To classify these candidates, The Second Derivative Test inspects the determinant of the Hessian curvature matrix $D = f_{xx}f_{yy} - (f_{xy})^2$ . If $D > 0$ , the surface curves uniformly in all directions (concave up for $f_{xx} > 0 \implies$ local min; concave down for $f_{xx} < 0 \implies$ local max). If $D < 0$ , the surface bends upward in one cross-section and downward in a transverse cross-section, forming a saddle point.
Engineering Context: Aerospace engineers optimize airfoil lift-to-drag performance surfaces $E(\alpha, \beta)$ ; a true local maximum ( $D > 0, E_{\alpha\alpha} < 0$ ) guarantees cruise stability, whereas a saddle point exposes the aircraft to violent aerodynamic pitching oscillations.

2. Compact Domains & The Multi-Tier Boundary Search

Mathematical Foundation: Engineering designs rarely operate across an infinite plane; physical systems are restricted to compact (closed and bounded) operating envelopes. Under the Extreme Value Theorem (explored in Extrema on Closed and Bounded Regions), a continuous multivariable function on a closed and bounded domain is guaranteed to attain an absolute maximum and minimum. Because global extrema frequently reside on perimeter boundaries or structural vertices rather than at interior stationary points, engineers must execute a systematic 3-tier search: interior critical points, 1D parameterized boundary edges, and geometric corner vertices.
Engineering Context: Geotechnical engineers evaluate contact soil bearing pressure $P(x, y)$ under heavy foundation footings to guarantee that peak stresses along foundation edges or corner toes do not exceed the soil's ultimate bearing capacity, preventing structural punch-through failure.

3. Constrained Surfaces & The Method of Lagrange Multipliers

Mathematical Foundation: When optimization variables are strictly bound to an equation constraint $g(x, y, z) = k$ (such as fixed material mass, trajectory pathways, or fluid flow cross-sections), interior gradients do not vanish. Instead, as proven in Lagrange Multipliers, an extreme value along the constraint path occurs when the constraint curve is tangent to a level set of the objective function. Because their tangent lines coincide, their normal gradient vectors are parallel: $\nabla f = \lambda \nabla g$ . The scalar multiplier $\lambda$ represents the marginal sensitivity (shadow price) of the system: $\lambda = \frac{d f_{\text{opt}}}{d k}$ .
Engineering Context: Satellite mission analysts minimize orbit-raising propulsion fuel burn metrics subject to an orbital insertion velocity plane, where $\lambda$ quantifies the exact propellant mass penalty per unit increase in required mission delta-V.

Segment 1 of 8

Skill Block 1

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Unconstrained Local Extrema & The Second Derivative Test

General Overview
Skill Block 1 covers the identification and geometric classification of unconstrained stationary points for functions of two variables. Students learn to compute simultaneous partial derivative systems $\nabla f = \mathbf{0}$ , avoid premature coordinate substitution, construct the Hessian matrix, evaluate the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$ , and classify points as local minima, local maxima, saddle points, or inconclusive.

Conceptual Framework

Critical Point Definition:
A point $(a, b)$ is a critical point of $f(x, y)$ if: $\nabla f(a, b) = \mathbf{0} \iff \frac{\partial f}{\partial x}(a, b) = 0 \quad \text{and} \quad \frac{\partial f}{\partial y}(a, b) = 0$ or if either first partial derivative fails to exist at $(a, b)$ . (Defined in Critical Points in Multivariable Functions.)
The Hessian Matrix & Discriminant:
For a function with continuous second-order partial derivatives, the Hessian matrix and its determinant $D$ are: $H = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}, \quad D(x, y) = f_{xx}f_{yy} - (f_{xy})^2$ (Defined in Second Derivative Test for Multivariable Functions.)
Classification Criteria:
- Local Minimum: $D(a, b) > 0$ and $f_{xx}(a, b) > 0$ .
- Local Maximum: $D(a, b) > 0$ and $f_{xx}(a, b) < 0$ .
- Saddle Point: $D(a, b) < 0$ .
- Inconclusive Test: $D(a, b) = 0$ .

Key Equations & Structural Rules

Differentiate-First Rule: Always compute the symbolic partial derivatives $f_x, f_y, f_{xx}, f_{yy}, f_{xy}$ across the entire domain before substituting numerical coordinates $(a, b)$ .
Clairaut's Symmetry: For smooth surfaces, mixed partial derivatives are equal ( $f_{xy} = f_{yx}$ ), guaranteeing that the Hessian matrix is symmetric.
Same-Sign Rule for $D > 0$ : If $D > 0$ , the product $f_{xx}f_{yy} > (f_{xy})^2 \ge 0$ , which mathematically forces $f_{xx}$ and $f_{yy}$ to have the exact same non-zero algebraic sign.

Worked Example A: Polynomial Gradient and Hessian Point Classification

Find and classify all critical points of the multivariable function:

f(x, y) = 2x^3 - 6x + 2y^2 - 4y

💡 Reveal Worked Solution

Compute First-Order Partial Derivatives: State the Two-Variable Gradient Formula (defined in The Gradient Vector):
$\nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j}$ Differentiate with respect to $x$ treating $y$ as a constant: $f_x(x, y) = \frac{\partial}{\partial x}(2x^3 - 6x + 2y^2 - 4y) = 6x^2 - 6$ Differentiate with respect to $y$ treating $x$ as a constant: $f_y(x, y) = \frac{\partial}{\partial y}(2x^3 - 6x + 2y^2 - 4y) = 4y - 4$
Locate Critical Points: State the Critical Point Condition (defined in Critical Points in Multivariable Functions):
$\nabla f(x, y) = \mathbf{0} \implies f_x(x, y) = 0 \quad \text{and} \quad f_y(x, y) = 0$ Solve the two decoupled equations: $6x^2 - 6 = 0 \implies 6(x^2 - 1) = 0 \implies x = 1 \quad \text{or} \quad x = -1$ $4y - 4 = 0 \implies 4y = 4 \implies y = 1$ Combining $x$ and $y$ solutions yields two distinct critical points: $(1, 1) \quad \text{and} \quad (-1, 1)$
Compute Second-Order Partial Derivatives: State the Second-Order Partial Derivatives (defined in Second-Order Partial Derivatives):
$f_{xx} = \frac{\partial}{\partial x}(6x^2 - 6) = 12x$ $f_{yy} = \frac{\partial}{\partial y}(4y - 4) = 4$ $f_{xy} = \frac{\partial}{\partial y}(6x^2 - 6) = 0$
Construct and Evaluate the Hessian Discriminant: State the Hessian Discriminant Formula (defined in Second Derivative Test for Multivariable Functions):
$D(x, y) = f_{xx}f_{yy} - (f_{xy})^2 = (12x)(4) - (0)^2 = 48x$
- Evaluate at $(1, 1)$ : $D(1, 1) = 48(1) = 48 > 0$ $f_{xx}(1, 1) = 12(1) = 12 > 0$ Because $D > 0$ and $f_{xx} > 0$ , the critical point $(1, 1)$ is a local minimum.
  The local minimum value is: $f(1, 1) = 2(1)^3 - 6(1) + 2(1)^2 - 4(1) = 2 - 6 + 2 - 4 = -6$
- Evaluate at $(-1, 1)$ : $D(-1, 1) = 48(-1) = -48 < 0$ Because $D < 0$ , the critical point $(-1, 1)$ is a saddle point.
  The saddle point value is: $f(-1, 1) = 2(-1)^3 - 6(-1) + 2(1)^2 - 4(1) = -2 + 6 + 2 - 4 = 2$

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) State the algebraic condition on the gradient vector $\nabla f$ required for a point $(a, b)$ to be classified as a critical point, and explain what this means for the surface's tangent plane.
(b) In Worked Example A, explain why knowing $D(1, 1) > 0$ alone is insufficient to conclude that $(1, 1)$ is a local minimum, and state what additional derivative sign must be verified.
(c) In Aerospace Wing Profile Lift-to-Drag Optimization (Aerospace & Mechanical Engineering), explain why an aircraft engineer seeks an operating point with $D > 0$ and $E_{xx} < 0$ rather than a saddle point where $D < 0$ .

Practice Space Solve this problem on paper or tablet.

Review the definition of critical points in Critical Points in Multivariable Functions, inspect the classification criteria in Second Derivative Test for Multivariable Functions, and read the aerodynamic stability discussion in the linked aerospace wing application.

Answer: * (a) $\nabla f(a, b) = \langle 0, 0 \rangle = \mathbf{0}$ ; the tangent plane is completely horizontal ( $z = f(a, b)$ ).

(b) $D > 0$ indicates definite curvature (bowl shape), but it could be concave up (minimum) or concave down (maximum); you must verify $f_{xx} > 0$ (or $f_{yy} > 0$ ).
(c) $D > 0, E_{xx} < 0$ represents a stable local maximum in cruise efficiency; a saddle point ( $D < 0$ ) causes the wing to lose performance under pitch variations.

Step-by-Step Execution

(a) By the Critical Point Definition (defined in Critical Points in Multivariable Functions), all first-order partial derivatives must vanish simultaneously: $\nabla f(a, b) = \mathbf{0} \iff f_x(a, b) = 0$ and $f_y(a, b) = 0$ . Geometrically, the tangent plane equation $z - f(a, b) = f_x(x - a) + f_y(y - b)$ reduces to the constant horizontal plane $z = f(a, b)$ .
(b) By the Hessian Discriminant Criteria (defined in Second Derivative Test for Multivariable Functions), $D > 0$ guarantees that the principal curvatures share the same sign. To distinguish whether the surface forms an upright bowl (local minimum) or an inverted dome (local maximum), one must evaluate the sign of the pure second derivative $f_{xx}$ (or $f_{yy}$ ): $f_{xx} > 0$ confirms upward concavity (local minimum).
(c) In aerospace flight mechanics, an aerodynamic efficiency peak requires negative concavity in all directions ( $D > 0, E_{xx} < 0$ ) so that deviations in angle of attack or camber produce graceful performance boundaries. A saddle point ( $D < 0$ ) represents an unstable equilibrium where slight perturbations induce catastrophic loss of lift.

Practice Problem 2: Decoupled Polynomial Local Extrema

Find and classify all critical points of the function:

f(x, y) = x^3 - 3x + y^2

Practice Space Solve this problem on paper or tablet.

Compute first partial derivatives using the Two-Variable Gradient Formula (see The Gradient Vector). Set $f_x = 0$ and $f_y = 0$ independently, then evaluate the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$ (see Second Derivative Test for Multivariable Functions).

Answer: Critical point $(1, 0)$ is a local minimum ( $f(1, 0) = -2$ ); critical point $(-1, 0)$ is a saddle point ( $f(-1, 0) = 2$ ).

Step-by-Step Execution

Compute First Partial Derivatives: State the Two-Variable Gradient Formula:
$\nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle$ $f_x = \frac{\partial}{\partial x}(x^3 - 3x + y^2) = 3x^2 - 3$ $f_y = \frac{\partial}{\partial y}(x^3 - 3x + y^2) = 2y$
Locate Critical Points: Set both partial derivatives equal to zero:
$3x^2 - 3 = 0 \implies 3(x^2 - 1) = 0 \implies x = 1 \quad \text{or} \quad x = -1$ $2y = 0 \implies y = 0$ This yields two critical points: $(1, 0)$ and $(-1, 0)$ .
Compute Second-Order Partials and Discriminant: State the Hessian Discriminant Formula:
$D(x, y) = f_{xx}f_{yy} - (f_{xy})^2$ $f_{xx} = 6x, \quad f_{yy} = 2, \quad f_{xy} = 0$ $D(x, y) = (6x)(2) - 0^2 = 12x$
Classify Points:
- At $(1, 0)$ : $D(1, 0) = 12(1) = 12 > 0, \quad f_{xx}(1, 0) = 6(1) = 6 > 0$ Since $D > 0$ and $f_{xx} > 0$ , $(1, 0)$ is a local minimum with value $f(1, 0) = 1^3 - 3(1) + 0 = -2$ .
- At $(-1, 0)$ : $D(-1, 0) = 12(-1) = -12 < 0$ Since $D < 0$ , $(-1, 0)$ is a saddle point with value $f(-1, 0) = (-1)^3 - 3(-1) + 0 = 2$ .

Practice Problem 3: Coupled Polynomial Extrema with Cross-Terms

Determine and classify all critical points of the function:

f(x, y) = y^3 - 3xy + 12x

Practice Space Solve this problem on paper or tablet.

Calculate $f_x$ and $f_y$ . Notice that $f_x = 0$ contains only $y$ , allowing you to solve for $y$ immediately. Substitute this into $f_y = 0$ , compute $f_{xx}, f_{yy}, f_{xy}$ , and evaluate $D$ (see Second Derivative Test for Multivariable Functions).

Answer: Critical point $(16, 4)$ is a saddle point ( $f(16, 4) = 64$ ).

Step-by-Step Execution

Compute First Partial Derivatives: State the Two-Variable Gradient Formula:
$\nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle$ $f_x = \frac{\partial}{\partial x}(y^3 - 3xy + 12x) = -3y + 12$ $f_y = \frac{\partial}{\partial y}(y^3 - 3xy + 12x) = 3y^2 - 3x$
Locate Critical Points: Set both equations to zero:
$-3y + 12 = 0 \implies 3y = 12 \implies y = 4$ Substitute $y = 4$ into $f_y = 0$ : $3(4)^2 - 3x = 0 \implies 3(16) - 3x = 0 \implies 48 = 3x \implies x = 16$ The single critical point is $(16, 4)$ .
Compute Second-Order Partials and Discriminant: State the Hessian Discriminant Formula:
$f_{xx} = \frac{\partial}{\partial x}(-3y + 12) = 0$ $f_{yy} = \frac{\partial}{\partial y}(3y^2 - 3x) = 6y$ $f_{xy} = \frac{\partial}{\partial y}(-3y + 12) = -3$ $D(x, y) = f_{xx}f_{yy} - (f_{xy})^2 = (0)(6y) - (-3)^2 = 0 - 9 = -9$
Classify the Point: At $(16, 4)$ , $D = -9 < 0$ .
Since $D < 0$ , the critical point $(16, 4)$ is a saddle point.
Its value is $f(16, 4) = 4^3 - 3(16)(4) + 12(16) = 64 - 192 + 192 = 64$ .

Practice Problem 4: Non-Linear Coordinate Elimination and Local Maxima

Find and classify all critical points of the higher-degree multivariable function:

f(x, y) = 4xy - x^4 - y^4

Practice Space Solve this problem on paper or tablet.

Differentiate to find $f_x$ and $f_y$ . Set them to zero to obtain $y = x^3$ and $x = y^3$ . Substitute $y = x^3$ into $x = y^3$ to get $x(1 - x^8) = 0$ , giving real solutions $x = 0, 1, -1$ . Construct $D(x, y) = 144x^2y^2 - 16$ (see Second Derivative Test for Multivariable Functions).

Answer: Critical point $(0, 0)$ is a saddle point ( $f(0, 0) = 0$ ); critical points $(1, 1)$ and $(-1, -1)$ are local maxima ( $f(1, 1) = 2, f(-1, -1) = 2$ ).

Step-by-Step Execution

Compute First-Order Partials:
$f_x = 4y - 4x^3, \quad f_y = 4x - 4y^3$
Solve the Non-Linear System: Set $f_x = 0 \implies 4y = 4x^3 \implies y = x^3$ . Substitute $y = x^3$ into $f_y = 0$ :
$4x - 4(x^3)^3 = 0 \implies 4x - 4x^9 = 0 \implies 4x(1 - x^8) = 0$
The real solutions for $x$ are $x = 0$ , $x = 1$ , and $x = -1$ .
- If $x = 0 \implies y = 0^3 = 0 \implies (0, 0)$
- If $x = 1 \implies y = 1^3 = 1 \implies (1, 1)$
- If $x = -1 \implies y = (-1)^3 = -1 \implies (-1, -1)$
Compute Second Partials and Discriminant:
$f_{xx} = -12x^2, \quad f_{yy} = -12y^2, \quad f_{xy} = 4$ State the Hessian Discriminant Formula: $D(x, y) = f_{xx}f_{yy} - (f_{xy})^2 = (-12x^2)(-12y^2) - (4)^2 = 144x^2y^2 - 16$
Classify Each Critical Point:
- At $(0, 0)$ : $D(0, 0) = 144(0) - 16 = -16 < 0 \implies \textbf{Saddle point}$ Value: $f(0, 0) = 0$ .
- At $(1, 1)$ : $D(1, 1) = 144(1)^2(1)^2 - 16 = 128 > 0, \quad f_{xx}(1, 1) = -12(1)^2 = -12 < 0 \implies \textbf{Local maximum}$ Value: $f(1, 1) = 4(1)(1) - 1^4 - 1^4 = 2$ .
- At $(-1, -1)$ : $D(-1, -1) = 144(-1)^2(-1)^2 - 16 = 128 > 0, \quad f_{xx}(-1, -1) = -12(-1)^2 = -12 < 0 \implies \textbf{Local maximum}$ Value: $f(-1, -1) = 4(-1)(-1) - (-1)^4 - (-1)^4 = 2$ .

Segment 2 of 8

📅 Regroup 1

Visualizing Curvature: Emphasize why $D < 0$ guarantees a saddle point: the quadratic expansion has eigenvalues of opposite signs, meaning the surface climbs in one direction and falls in another.
Algebraic Pitfall: Highlight that when solving coupled systems like $4y - 4x^3 = 0$ , students must never divide by $x$ or $y$ without accounting for the origin $(0, 0)$ as a solution.

Segment 3 of 8

Skill Block 2

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Global Extrema on Closed and Bounded Regions

General Overview
Skill Block 2 focuses on finding absolute maximum and absolute minimum values for multivariable functions defined over compact sets in $\mathbb{R}^2$ . Students learn how the Extreme Value Theorem guarantees extrema, how to execute the 3-tier candidate search strategy across interiors, edges, and corners, and how to compile a master evaluation table.

Conceptual Framework

Extreme Value Theorem (EVT) for Two Variables:
If $f(x, y)$ is continuous on a closed and bounded set $D \subset \mathbb{R}^2$ , then $f$ attains an absolute maximum value $f(x_1, y_1)$ and an absolute minimum value $f(x_2, y_2)$ at some points in $D$ . (Defined in Extrema on Closed and Bounded Regions.)
Closed vs. Bounded Criteria:
- Closed: Contains its boundary points ( $\le, \ge$ ).
- Bounded: Enclosed completely within a disk of finite radius centered at the origin.
The 3-Tier Search Strategy:
1. Tier 1 (Interior): Solve $\nabla f = \mathbf{0}$ , keeping only points inside $D$ .
2. Tier 2 (Boundary Edges): Parameterize boundary paths, substitute into $f$ , and solve 1D derivative equations $g'(t) = 0$ .
3. Tier 3 (Vertices): Evaluate $f$ at all geometric intersection corners.

Key Equations & Structural Rules

Interior Filtering Rule: Discard any critical point where $(a, b) \notin D$ .
Boundary Substitution Technique: If a boundary is given by $y = mx + b$ or $y^2 = R^2 - x^2$ , substitute directly into $f(x, y)$ to form a single-variable function $g(x)$ .
Global Comparison Standard: You do not need the second derivative test for boundary optimization; simply evaluate $f(x, y)$ numerically across all candidates to find the highest and lowest values.

Worked Example A: Absolute Extrema on a Closed Rectangular Plate

Find the absolute maximum and minimum values of:

f(x, y) = x^2 + y^2 - 2x - 4y + 5

on the closed square domain

D = \{ (x, y) \in \mathbb{R}^2 \mid 0 \le x \le 4, \, 0 \le y \le 4 \}

.

💡 Reveal Worked Solution

Tier 1: Interior Critical Points: Compute the first partial derivatives using the Two-Variable Gradient Formula:
$f_x = 2x - 2 = 0 \implies x = 1$ $f_y = 2y - 4 = 0 \implies y = 2$ The point $(1, 2)$ lies strictly inside $D$ because $0 < 1 < 4$ and $0 < 2 < 4$ .
Evaluate $f$ at $(1, 2)$ : $f(1, 2) = 1^2 + 2^2 - 2(1) - 4(2) + 5 = 1 + 4 - 2 - 8 + 5 = 0$
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: $f(4, 2) = 2^2 - 4(2) + 13 = 4 - 8 + 13 = 9$ .
- 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: $f(0, 2) = 2^2 - 4(2) + 5 = 1$ .
Tier 3: Corner / Vertex Check: Evaluate $f$ at the four square corners:
- At $(0, 0)$ : $f(0, 0) = 5$
- At $(4, 0)$ : $f(4, 0) = 4^2 - 2(4) + 5 = 16 - 8 + 5 = 13$
- At $(4, 4)$ : $f(4, 4) = 4^2 + 4^2 - 2(4) - 4(4) + 5 = 16 + 16 - 8 - 16 + 5 = 13$
- At $(0, 4)$ : $f(0, 4) = 4^2 - 4(4) + 5 = 5$

Master Comparison & Conclusion: Construct the comparison table:

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

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

Practice Problem 5: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) State the two geometric/topological conditions that a domain $D \subset \mathbb{R}^2$ must satisfy for the Extreme Value Theorem to guarantee the existence of absolute extrema.
(b) In Worked Example A, explain why calculating the Hessian discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$ is unnecessary when executing a global boundary-value optimization problem.
(c) In Triangular Foundation Slab Soil Bearing Pressure (Civil & Construction Engineering), explain why civil engineers must evaluate contact pressure at the corner vertices of the footing slab.

Practice Space Solve this problem on paper or tablet.

Check the Extreme Value Theorem conditions in Extrema on Closed and Bounded Regions, review the Global Comparison Standard in the Key Equations, and read the geotechnical structural discussion in the linked foundation application.

Answer: * (a) The domain must be closed (contains all its boundary points) and bounded (enclosed by a disk of finite radius).

(b) The global extrema are determined strictly by comparing the numerical heights $f(x, y)$ across all candidates; local curvature classifications do not dictate absolute maximums or minimums.
(c) Corners represent geometric transitions where boundary derivatives fail to exist, and overturning moments frequently produce peak contact stresses at exterior foundation toes.

Step-by-Step Execution

(a) By the Extreme Value Theorem for Two Variables (defined in Extrema on Closed and Bounded Regions), the domain $D$ must be both closed (including all boundary lines) and bounded (finite in extent). If the domain is open (missing boundary lines) or unbounded (extending infinitely), the function may grow without bound or approach an infimum on a missing edge.
(b) The second derivative test classifies local behavior in an open neighborhood. For global optimization on compact domains, the Extreme Value Theorem guarantees that the highest and lowest points must be among the interior critical points, boundary critical points, or corner vertices. A direct numerical evaluation of $f(x, y)$ across this finite list automatically identifies the global extrema without computing $D$ .
(c) In geotechnical foundation engineering, eccentric overturning loads induce linear and quadratic stress gradients. Because edges intersect sharply at vertices, mathematical boundary parameterizations terminate at corners. Physical foundation toe blowout frequently occurs at these sharp vertices.

Practice Problem 6: Absolute Extrema on a Closed Triangular Region

Find the absolute maximum and minimum values of:

f(x, y) = x^2 + y^2

on the closed triangular region bounded by the vertices

(0, 0)

,

(4, 0)

, and

(0, 3)

.

Practice Space Solve this problem on paper or tablet.

Check the interior for critical points (is $(0, 0)$ strictly in the interior?). Parameterize the three boundary segments: $y = 0$ ( $0 \le x \le 4$ ), $x = 0$ ( $0 \le y \le 3$ ), and the hypotenuse line $y = -\frac{3}{4}x + 3$ . Test all edge critical points and corner vertices (see Extrema on Closed and Bounded Regions).

Answer: Absolute minimum is $0$ at $(0, 0)$ ; absolute maximum is $16$ at $(4, 0)$ .

Step-by-Step Execution

Tier 1: Interior Search:
$f_x = 2x = 0 \implies x = 0, \quad f_y = 2y = 0 \implies y = 0$ The point $(0, 0)$ is on the boundary (it is a vertex); thus there are no critical points strictly in the open interior.
Tier 2: Boundary Edge Search:
- Bottom Edge ( $y = 0, 0 \le x \le 4$ ): $f(x, 0) = x^2 \implies \frac{d}{dx}[x^2] = 2x = 0 \implies x = 0$ This reproduces vertex $(0, 0)$ ( $f = 0$ ); endpoint $x = 4$ yields $(4, 0)$ ( $f = 16$ ).
- Left Edge ( $x = 0, 0 \le y \le 3$ ): $f(0, y) = y^2 \implies \frac{d}{dy}[y^2] = 2y = 0 \implies y = 0$ This reproduces vertex $(0, 0)$ ( $f = 0$ ); endpoint $y = 3$ yields $(0, 3)$ ( $f = 9$ ).
- Hypotenuse Line ( $y = -\frac{3}{4}x + 3, 0 \le x \le 4$ ): Substitute $y = -\frac{3}{4}x + 3$ into $f(x, y)$ : $g(x) = x^2 + \left(-\frac{3}{4}x + 3\right)^2 = x^2 + \frac{9}{16}x^2 - \frac{9}{2}x + 9 = \frac{25}{16}x^2 - \frac{9}{2}x + 9$ Differentiate and set to zero: $g'(x) = \frac{50}{16}x - \frac{9}{2} = \frac{25}{8}x - \frac{9}{2} = 0 \implies x = \frac{9/2}{25/8} = \frac{9}{2} \cdot \frac{8}{25} = \frac{36}{25} = 1.44$ Since $1.44 \in [0, 4]$ , calculate $y$ : $y = -\frac{3}{4}\left(\frac{36}{25}\right) + 3 = -\frac{27}{25} + \frac{75}{25} = \frac{48}{25} = 1.92$ Evaluate function at $(1.44, 1.92)$ : $f(1.44, 1.92) = (1.44)^2 + (1.92)^2 = 2.0736 + 3.6864 = 5.76$
Tier 3: Corner Vertices:
- At $(0, 0)$ : $f(0, 0) = 0$
- At $(4, 0)$ : $f(4, 0) = 16$
- At $(0, 3)$ : $f(0, 3) = 9$
Comparison & Conclusion:
- Candidate values: $0$ (at $(0, 0)$ ), $5.76$ (at $(1.44, 1.92)$ ), $9$ (at $(0, 3)$ ), $16$ (at $(4, 0)$ ).
- Absolute Minimum: $0$ at $(0, 0)$ .
- Absolute Maximum: $16$ at $(4, 0)$ .

Practice Problem 7: Absolute Extrema on a Closed Circular Disk

Find the absolute maximum and minimum values of:

f(x, y) = 2x^2 + 3y^2 - 4x - 5

on the closed circular disk

D = \{ (x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \le 4 \}

.

Practice Space Solve this problem on paper or tablet.

Find interior critical points satisfying $\nabla f = \mathbf{0}$ and check if $x^2 + y^2 < 4$ . For the boundary $x^2 + y^2 = 4$ , express $y^2 = 4 - x^2$ for $x \in [-2, 2]$ . Substitute this into $f(x, y)$ to create a single-variable function $g(x)$ , find critical points of $g(x)$ , and check interval endpoints $x = \pm 2$ (see Extrema on Closed and Bounded Regions).

Answer: Absolute minimum is $-7$ at $(1, 0)$ ; absolute maximum is $11$ at $(-2, 0)$ .

Step-by-Step Execution

Tier 1: Interior Search: Compute partial derivatives:
$f_x = 4x - 4 = 0 \implies x = 1$ $f_y = 6y = 0 \implies y = 0$ Check boundary inequality: $1^2 + 0^2 = 1 < 4$ , so $(1, 0)$ lies strictly inside $D$ .
Evaluate $f$ at $(1, 0)$ : $f(1, 0) = 2(1)^2 + 3(0)^2 - 4(1) - 5 = 2 - 4 - 5 = -7$
Tier 2: Circular Boundary Search ( $x^2 + y^2 = 4$ for $-2 \le x \le 2$ ): Solve for $y^2 = 4 - x^2$ and substitute directly into $f(x, y)$ :
$g(x) = 2x^2 + 3(4 - x^2) - 4x - 5 = 2x^2 + 12 - 3x^2 - 4x - 5 = -x^2 - 4x + 7$ Differentiate $g(x)$ with respect to $x$ and set to zero: $g'(x) = -2x - 4 = 0 \implies 2x = -4 \implies x = -2$ Since $x = -2$ lies in the interval $[-2, 2]$ , find $y$ : $y^2 = 4 - (-2)^2 = 4 - 4 = 0 \implies y = 0$ Candidate boundary point: $(-2, 0)$ .
Evaluate $f$ at $(-2, 0)$ : $f(-2, 0) = 2(-2)^2 + 3(0)^2 - 4(-2) - 5 = 2(4) + 8 - 5 = 8 + 8 - 5 = 11$
Check Endpoints of the Interval $x \in [-2, 2]$ :
- At $x = -2$ : $y = 0$ , giving $(-2, 0)$ already evaluated ( $f = 11$ ).
- At $x = 2$ : $y^2 = 4 - 2^2 = 0 \implies y = 0$ , giving point $(2, 0)$ .
  Evaluate $f$ at $(2, 0)$ : $f(2, 0) = 2(2)^2 + 3(0)^2 - 4(2) - 5 = 8 - 8 - 5 = -5$

Master Comparison & Conclusion:

Candidate Point	Region Location	Function Value $f(x, y)$
$(1, 0)$	Interior Critical Point	$-7$ (Absolute Minimum)
$(2, 0)$	Boundary Point	$-5$
$(-2, 0)$	Boundary Point	$11$ (Absolute Maximum)

Absolute Minimum: $-7$ at $(1, 0)$ .
Absolute Maximum: $11$ at $(-2, 0)$ .

Segment 4 of 8

📅 Regroup 2

Boundary Substitution Standard: Emphasize how substituting $y^2 = 4 - x^2$ reduces a 2D boundary search into single-variable Calculus 1 optimization.
Filtering Discarded Candidates: Remind students to always verify that interior critical points satisfy the domain inequalities before adding them to the candidate table.

Segment 5 of 8

Skill Block 3

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Constrained Optimization via Lagrange Multipliers

General Overview
Skill Block 3 presents the method of Lagrange multipliers for optimizing objective functions subject to smooth constraint equations. Students learn the geometric foundation of gradient tangency ( $\nabla f = \lambda \nabla g$ ), formulate component equation systems, avoid illegal division by zero when eliminating variables, and interpret $\lambda$ as a measure of system sensitivity and shadow pricing.

Conceptual Framework

Lagrange Multiplier Vector System:
To find the extrema of $f(x, y, z)$ subject to $g(x, y, z) = k$ : $\nabla f = \lambda \nabla g \quad \text{and} \quad g(x, y, z) = k$ (Defined in Lagrange Multipliers.)
Geometric Tangency Condition:
At a constrained extremum, the level surface of $f$ and the constraint surface $g = k$ share a common tangent plane. Because gradient vectors are orthogonal to level sets, $\nabla f$ and $\nabla g$ must be collinear (parallel).
Physical Meaning of the Multiplier $\lambda$ :
$\lambda = \frac{d f_{\text{opt}}}{d k}$ $\lambda$ represents the marginal change in the optimal objective value per unit relaxation of the constraint threshold $k$ .

Key Equations & Structural Rules

Vector Component Equations:
$\hat{i}: \frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x}, \quad \hat{j}: \frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}, \quad \hat{k}: \frac{\partial f}{\partial z} = \lambda \frac{\partial g}{\partial z}$
Zero-Division Ban: Never divide by a variable without verifying whether that variable can equal zero.
Candidate Testing Rule: After solving for coordinate candidates, substitute all points back into the objective function $f$ to determine the maximum and minimum values.

Worked Example A: Constrained Distance Minimization in 3D Space

Find the point on the plane $x + y + z = 12$ that is closest to the origin, and evaluate the minimum squared distance and minimum Euclidean distance.

💡 Reveal Worked Solution

Formulate Objective and Constraint Functions: Minimizing Euclidean distance $\sqrt{x^2 + y^2 + z^2}$ is mathematically equivalent to minimizing the squared distance:
$f(x, y, z) = x^2 + y^2 + z^2$ subject to the linear plane constraint: $g(x, y, z) = x + y + z = 12$
Compute Gradient Vectors: State the Three-Variable Gradient Formula (defined in The Gradient Vector):
$\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle = \langle 2x, 2y, 2z \rangle = 2x\hat{i} + 2y\hat{j} + 2z\hat{k}$ $\nabla g = \left\langle \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right\rangle = \langle 1, 1, 1 \rangle = \hat{i} + \hat{j} + \hat{k}$
Set Up the Lagrange Multiplier System: State the Lagrange Multiplier Vector System (defined in Lagrange Multipliers):
$\nabla f = \lambda \nabla g \quad \text{and} \quad x + y + z = 12$ Equate the scalar components: $\hat{i}: 2x = \lambda(1) \implies x = \frac{\lambda}{2}$ $\hat{j}: 2y = \lambda(1) \implies y = \frac{\lambda}{2}$ $\hat{k}: 2z = \lambda(1) \implies z = \frac{\lambda}{2}$
Eliminate $\lambda$ and Relate Variables: Since all three variables equal $\frac{\lambda}{2}$ , we obtain:
$x = y = z$
Substitute into the Constraint Equation:
$x + x + x = 12 \implies 3x = 12 \implies x = 4$ Since $x = y = z$ , our candidate optimal point is $(4, 4, 4)$ , with multiplier: $\lambda = 2x = 2(4) = 8$
Evaluate Minimum Distances:
- Minimum squared distance: $f(4, 4, 4) = 4^2 + 4^2 + 4^2 = 16 + 16 + 16 = 48$
- Minimum Euclidean distance: $d_{\min} = \sqrt{f(4, 4, 4)} = \sqrt{48} = 4\sqrt{3} \approx 6.93$ The point on the plane closest to the origin is $(4, 4, 4)$ .

Practice Problem 8: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain why the gradient vectors $\nabla f$ and $\nabla g$ must be parallel ( $\nabla f = \lambda \nabla g$ ) at a constrained optimal point, using the geometry of level curves.
(b) In Worked Example A, state the value of the Lagrange multiplier $\lambda$ , and explain what this number predicts if the plane constant is relaxed from $12$ to $13$ .
(c) In Orbital Insertion Satellite Delta-V Thruster Burn (Aerospace Engineering), explain how the Lagrange multiplier $\lambda$ informs mission control regarding fuel mass penalties.

Practice Space Solve this problem on paper or tablet.

Review the geometric tangency derivation in Lagrange Multipliers, inspect the sensitivity formula $\lambda = d f_{\text{opt}} / dk$ , and read the mission delta-V analysis in the linked aerospace application.

Answer: * (a) The constraint curve is tangent to a level curve of $f$ ; because gradients are perpendicular to level sets, the two normal gradients must point along the same line ( $\nabla f = \lambda \nabla g$ ).

(b) $\lambda = 8$ ; it predicts that increasing the plane constraint constant by $+1$ will increase the minimum squared distance by approximately $\Delta f \approx 8(1) = 8$ units.
(c) $\lambda$ represents the marginal fuel burn penalty per $\text{km/s}$ of required velocity change, guiding whether to invest in trajectory correction burns.

Step-by-Step Execution

(a) By the Geometric Tangency Principle (defined in Lagrange Multipliers), if a constraint curve $g = k$ crosses a level curve $f = c$ at an angle, the value of $f$ can be increased or decreased by sliding along the constraint. An extremum occurs only where the curves are tangent. Because $\nabla f$ is orthogonal to the level curve of $f$ and $\nabla g$ is orthogonal to the constraint curve, the normal vectors must be collinear, yielding $\nabla f = \lambda \nabla g$ .
(b) In Worked Example A, $\lambda = 8$ . By the Sensitivity Theorem $\lambda = \frac{d f_{\text{opt}}}{d k}$ , relaxing the plane equation from $x + y + z = 12$ to $x + y + z = 13$ ( $\Delta k = +1$ ) increases the minimum squared distance by $\Delta f \approx \lambda \Delta k = 8(1) = 8$ units (exact new value: $3 \times (13/3)^2 = 169/3 \approx 56.33$ , $\Delta f = 8.33$ ).
(c) In aerospace propulsion, propellant is mass-constrained. The multiplier $\lambda = 12$ gives the direct marginal cost in fuel metric units per unit increase in target delta-V, allowing systems engineers to evaluate aerodynamic gravity assist trade-offs.

Practice Problem 9: Constrained Distance to a 2D Line

Minimize the objective function $f(x, y) = x^2 + y^2$ subject to the linear constraint $4x + 3y = 12$ .

Practice Space Solve this problem on paper or tablet.

Calculate $\nabla f = \langle 2x, 2y \rangle$ and $\nabla g = \langle 4, 3 \rangle$ . Formulate component equations $2x = 4\lambda$ and $2y = 3\lambda$ . Express $x$ and $y$ in terms of $\lambda$ , substitute into $4x + 3y = 12$ , and solve for $\lambda$ (see Lagrange Multipliers).

Answer: Minimum value is $5.76$ (or $144/25$ ), occurring at coordinates $(1.92, 1.44) = (48/25, 36/25)$ .

Step-by-Step Execution

Compute Gradients:
$\nabla f = \langle 2x, 2y \rangle, \quad \nabla g = \langle 4, 3 \rangle$
Formulate Component Equations ( $\nabla f = \lambda \nabla g$ ):
$\hat{i}: 2x = 4\lambda \implies x = 2\lambda$ $\hat{j}: 2y = 3\lambda \implies y = 1.5\lambda = \frac{3\lambda}{2}$
Substitute into Constraint:
$4(2\lambda) + 3(1.5\lambda) = 12 \implies 8\lambda + 4.5\lambda = 12 \implies 12.5\lambda = 12$ $\lambda = \frac{12}{12.5} = \frac{24}{25} = 0.96$
Compute Optimal Coordinates and Minimum Value:
$x = 2(0.96) = 1.92 = \frac{48}{25}$ $y = 1.5(0.96) = 1.44 = \frac{36}{25}$ Evaluate $f(x, y)$ : $f(1.92, 1.44) = (1.92)^2 + (1.44)^2 = 3.6864 + 2.0736 = 5.76 = \frac{144}{25}$ (The shortest Euclidean distance from the origin to the line is $\sqrt{5.76} = 2.4$ ).

Practice Problem 10: Constrained Extrema on an Elliptical/Circular Boundary

Find the maximum and minimum values of $f(x, y) = xy$ subject to the circular constraint $x^2 + y^2 = 8$ .

Practice Space Solve this problem on paper or tablet.

Compute $\nabla f = \langle y, x \rangle$ and $\nabla g = \langle 2x, 2y \rangle$ . Set up $y = 2\lambda x$ and $x = 2\lambda y$ . Multiply the first by $y$ and the second by $x$ to show $y^2 = x^2 \implies y = \pm x$ . Substitute into $x^2 + y^2 = 8$ (see Lagrange Multipliers).

Answer: Maximum value is $4$ at $(2, 2)$ and $(-2, -2)$ ; minimum value is $-4$ at $(2, -2)$ and $(-2, 2)$ .

Step-by-Step Execution

Compute Gradients:
$\nabla f = \langle y, x \rangle, \quad \nabla g = \langle 2x, 2y \rangle$
Lagrange System:
1. $y = 2\lambda x$
2. $x = 2\lambda y$
3. $x^2 + y^2 = 8$
Eliminate $\lambda$ : Multiply equation (1) by $y$ and equation (2) by $x$ :
$y^2 = 2\lambda xy \quad \text{and} \quad x^2 = 2\lambda xy \implies x^2 = y^2 \implies y = \pm x$
Substitute into Constraint:
$x^2 + x^2 = 8 \implies 2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$
- For $y = x$ , candidates are $(2, 2)$ and $(-2, -2)$ .
- For $y = -x$ , candidates are $(2, -2)$ and $(-2, 2)$ .
Evaluate Objective $f(x, y) = xy$ :
- At $(2, 2)$ : $f(2, 2) = (2)(2) = 4$
- At $(-2, -2)$ : $f(-2, -2) = (-2)(-2) = 4$
- At $(2, -2)$ : $f(2, -2) = (2)(-2) = -4$
- At $(-2, 2)$ : $f(-2, 2) = (-2)(2) = -4$
Conclusion:
- Maximum value: $4$ at $(2, 2)$ and $(-2, -2)$ .
- Minimum value: $-4$ at $(2, -2)$ and $(-2, 2)$ .

Practice Problem 11: Geometric Optimization of a Rectangular Channel

An environmental engineer is sizing a rectangular open water channel with cross-sectional area $A = xy = 9\text{ m}^2$ , where $x$ and $y$ are positive dimensions. Using Lagrange multipliers, find the dimensions $(x, y)$ that minimize the total perimeter $P(x, y) = 2x + 2y$ .

Practice Space Solve this problem on paper or tablet.

Objective is $P(x, y) = 2x + 2y$ , constraint is $g(x, y) = xy = 9$ . Compute $\nabla P = \langle 2, 2 \rangle$ and $\nabla g = \langle y, x \rangle$ . Equate $2 = \lambda y$ and $2 = \lambda x$ to show $x = y$ (see Lagrange Multipliers).

Answer: Optimal dimensions are $x = 3\text{ m}$ and $y = 3\text{ m}$ ; minimum perimeter is $12\text{ m}$ .

Step-by-Step Execution

Formulate Functions:
- Objective: $P(x, y) = 2x + 2y$
- Constraint: $g(x, y) = xy = 9$ (with $x > 0, y > 0$ )
Compute Gradients:
$\nabla P = \langle 2, 2 \rangle, \quad \nabla g = \langle y, x \rangle$
Set Up System:
$\hat{i}: 2 = \lambda y \implies \lambda = \frac{2}{y}$ $\hat{j}: 2 = \lambda x \implies \lambda = \frac{2}{x}$
Eliminate $\lambda$ and Solve:
$\frac{2}{y} = \frac{2}{x} \implies x = y$ Substitute $x = y$ into $xy = 9$ : $x^2 = 9 \implies x = 3\text{ m}$ Since $x = y$ , $y = 3\text{ m}$ . Minimum perimeter: $P(3, 3) = 2(3) + 2(3) = 6 + 6 = 12\text{ m}$

Practice Problem 12: Three-Variable Box Surface Area Minimization

An open-top rectangular shipping crate must have a specified volume of $V = xyz = 32\text{ m}^3$ . Using the method of Lagrange multipliers, determine the dimensions $x, y, z$ that minimize the external surface area $S(x, y, z) = xy + 2xz + 2yz$ .

Practice Space Solve this problem on paper or tablet.

Objective: $S(x, y, z) = xy + 2xz + 2yz$ ; constraint: $g(x, y, z) = xyz = 32$ . Compute $\nabla S = \langle y + 2z, x + 2z, 2x + 2y \rangle$ and $\nabla g = \langle yz, xz, xy \rangle$ . Show that $x = y$ and $x = 2z$ , then substitute into $xyz = 32$ (see Lagrange Multipliers).

Answer: Base dimensions are $x = 4\text{ m}$ and $y = 4\text{ m}$ , height is $z = 2\text{ m}$ ; minimum surface area is $48\text{ m}^2$ .

Step-by-Step Execution

Formulate Functions:
- Objective: $S(x, y, z) = xy + 2xz + 2yz$ (base plus four vertical faces)
- Constraint: $g(x, y, z) = xyz = 32$ , with $x, y, z > 0$
Compute Gradients:
$\nabla S = \langle y + 2z, \, x + 2z, \, 2x + 2y \rangle$ $\nabla g = \langle yz, \, xz, \, xy \rangle$
Lagrange Component Equations ( $\nabla S = \lambda \nabla g$ ):
1. $y + 2z = \lambda yz$
2. $x + 2z = \lambda xz$
3. $2x + 2y = \lambda xy$
Eliminate $\lambda$ and Relate Dimensions: Multiply equation (1) by $x$ , (2) by $y$ , and (3) by $z$ :
$xy + 2xz = \lambda xyz$ $xy + 2yz = \lambda xyz$ $2xz + 2yz = \lambda xyz$ Equating the first two expressions: $xy + 2xz = xy + 2yz \implies 2xz = 2yz \implies x = y$ Equating the first and third expressions with $y = x$ : $x^2 + 2xz = 2xz + 2xz \implies x^2 + 2xz = 4xz \implies x^2 = 2xz \implies x = 2z \implies z = \frac{x}{2}$
Substitute into Volume Constraint:
$xyz = 32 \implies (x)(x)\left(\frac{x}{2}\right) = 32 \implies \frac{x^3}{2} = 32 \implies x^3 = 64 \implies x = 4\text{ m}$ Now find $y$ and $z$ : $y = x = 4\text{ m}$ $z = \frac{4}{2} = 2\text{ m}$
Evaluate Minimum Surface Area:
$S(4, 4, 2) = (4)(4) + 2(4)(2) + 2(4)(2) = 16 + 16 + 16 = 48\text{ m}^2$ The optimal crate has a square base of $4\text{ m} \times 4\text{ m}$ and a height of $2\text{ m}$ , achieving a minimum surface area of $48\text{ m}^2$ .

Segment 6 of 8

📅 Regroup 3

Component Equation Setup: Reiterate how multiplying by $x$ , $y$ , and $z$ cleanly aligns terms to $\lambda xyz$ in symmetric 3D problems.
Physical Meaning of Constraints: Discuss how $\lambda$ serves as an economic shadow price in industrial fabrication when evaluating the cost of material limits.

Segment 7 of 8

📅 Synthesis Wrap-up

Core Takeaways:
1. Unconstrained Extrema: Use $\nabla f = \mathbf{0}$ and classify with the Hessian discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$ .
2. Closed & Bounded Extrema: Use the 3-tier strategy (interior, boundary curves, vertices) and evaluate directly in a global comparison table.
3. Constrained Extrema: Use Lagrange multipliers $\nabla f = \lambda \nabla g$ when restricted to smooth curves or surfaces, and interpret $\lambda = \partial f_{\text{opt}}/\partial k$ as sensitivity.
Homework Assignment (Knewton Alta):
- Critical Points and the Second Derivative Test for Functions of Several Variables
- Maximization and Minimization Problems for Functions of Several Variables
- Lagrange Multipliers
Next Session Preview: Session 14 introduces Double Integrals over Rectangular and General Regions—accumulating volume and mass under multivariable surfaces.

Segment 8 of 8