Session 14 - Optimizing Areas and Volumes | Tony's Intertwingled Memex

📅 Micro-Lecture

1. The Optimization Modeling Protocol: Bridging Physics to Calculus

Mathematical Foundation: Optimization is the calculus of finding extreme values of an objective function subject to physical constraints. In real-world engineering, functions are rarely handed to us in single-variable form $y = f(x)$ . Instead, physical systems are defined by a primary quantity to optimize (e.g., area $A = xy$ or volume $V = \pi r^2 h$ ) coupled to a secondary constraint equation (e.g., fixed perimeter $2x + 2y = P_0$ or fixed material surface area $S = S_0$ ). Single-variable calculus resolves this via algebraic substitution: we isolate one dependent variable from the constraint and substitute it into the objective function.
Engineering Context: In civil, mechanical, aerospace, and electrical engineering, optimization drives design decisions: sizing solar array perimeters to minimize trenching costs, reducing sheet metal mass on rocket propellant tanks, and optimizing silicon chip floorplans.

2. Context-Driven Feasible Domains: Physics Constrains Mathematics

Mathematical Foundation: Pure mathematical functions frequently have infinite algebraic domains ( $-\infty, \infty$ ). In physical modeling, lengths, radii, and materials cannot be negative ( $x \ge 0, y \ge 0$ ). Applying this non-negativity to the substituted constraint expression ( $y(x) \ge 0$ ) establishes an upper physical boundary $x \le b$ .
Theorem Selection: If the domain is a closed bounded interval $[a, b]$ , the Extreme Value Theorem guarantees that global extrema exist, allowing us to evaluate candidates directly using the Closed Interval Method. If the domain is an open interval $(0, \infty)$ , we prove global optimality using the First or Second Derivative Test at the unique critical point.

3. Geometric Symmetry & Optimal Aspect Ratios

Mathematical Foundation: Unconstrained Euclidean geometry favors symmetry: for a fixed perimeter, the rectangle maximizing area is a square ( $x = y$ ); for a fixed volume, the closed cylinder minimizing surface area has height equal to diameter ( $h = 2r$ ). When engineering constraints introduce asymmetric boundary costs or open tops, calculus predicts exactly how these ideal ratios distort.

Segment 1 of 8

Skill Block 1

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

Constrained Planar Enclosures & Cost-Weighted Boundaries

General Overview
Skill Block 1 focuses on 2D planar optimization. Students learn to extract objective and constraint equations from narrative engineering contexts, apply variable reduction, formulate context-driven domains, and identify global extrema for agricultural pens, partitioned industrial fields, and cost-weighted perimeter boundaries.

Conceptual Framework

Objective Function vs. Constraint Equation:
- The objective function is the quantity being maximized or minimized (e.g., $A = xy$ ).
- The constraint equation is the fixed physical quantity or resource boundary (e.g., $P_0 = ax + by$ ).
Variable Reduction Algorithm:
1. Solve the constraint equation for one variable: $y = g(x)$ .
2. Substitute $g(x)$ into the objective: $A(x) = x \cdot g(x)$ .
3. Simplify before differentiating.
The Physical Non-Negativity Domain:
- Enforce $x \ge 0$ and $g(x) \ge 0$ to find the physical interval $[0, x_{\max}]$ .
- Discard any critical number falling outside this physical interval.

Key Equations & Structural Rules

Applied Optimization Modeling Protocol: $\text{Constraint: } g(x, y) = C \implies y = h(x) \implies \text{Objective: } f(x, h(x)) = F(x)$ (see Applied Optimization Modeling Framework)
Context-Driven Domain Formulation: $x \ge 0 \quad \text{and} \quad y(x) \ge 0 \implies x \in [0, x_{\text{bound}}]$ (see Context-Driven Domain in Geometric Optimization)

Worked Example A: Multi-Pen Enclosure Against a Barn

A farmer constructs three identical adjacent rectangular pens against an existing straight barn wall using 400 feet of fencing. No fencing is needed along the barn wall. Find the dimensions that will maximize the total enclosed area.

💡 Reveal Worked Solution

Name the Governing Formulas: The Applied Optimization Modeling Protocol and Rectangular Area Formula: (These modeling steps are defined in Applied Optimization Modeling Framework.)
Assign Variables and Define Objective Function: Let $x$ be the length of the 4 dividing fence segments perpendicular to the barn (in feet).
Let $y$ be the overall width of the combined pens parallel to the barn (in feet).
Objective: Maximize total area: $A = xy$
Formulate the Constraint Equation and Reduce Variables: The 400 feet of fencing covers 4 vertical segments and 1 horizontal segment: $4x + y = 400$ Isolate $y$ : $y = 400 - 4x$ Substitute into the area function: $A(x) = x(400 - 4x) = 400x - 4x^2$
Establish the Context-Driven Domain: Physical lengths must be non-negative: $x \ge 0$ $y \ge 0 \implies 400 - 4x \ge 0 \implies 4x \le 400 \implies x \le 100$ Feasible closed domain: $x \in [0, 100]\text{ feet}$ .
Differentiate and Locate Critical Numbers: $A'(x) = \frac{d}{dx}[400x - 4x^2] = 400 - 8x$ Set $A'(x) = 0$ : $400 - 8x = 0 \implies 8x = 400 \implies x = 50\text{ feet}$ The critical number $x = 50\text{ ft}$ lies strictly inside $[0, 100]$ .
Execute Closed Interval Evaluation (EVT Guarantee): Evaluate $A(x)$ at critical points and boundaries:
- $A(0) = 400(0) - 4(0)^2 = 0\text{ ft}^2$
- $A(100) = 400(100) - 4(100)^2 = 40,000 - 40,000 = 0\text{ ft}^2$
- $A(50) = 400(50) - 4(50)^2 = 20,000 - 10,000 = 10,000\text{ ft}^2$ The absolute maximum area is $10,000\text{ ft}^2$ .
Find Remaining Dimensions: $y = 400 - 4(50) = 200\text{ feet}$ The dividing fences are $50\text{ feet}$ long and the width parallel to the barn is $200\text{ feet}$ .

Practice Problem 1: Concept & Reading Check

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

(a) In the Conceptual Framework, state how the upper boundary of the context-driven domain is derived from the constraint equation.
(b) In Worked Example A, explain why evaluating the function at $x = 0$ and $x = 100$ is sufficient to prove that $x = 50$ is the global maximum without constructing a derivative sign chart.
(c) Review the linked Engineering Application (Solar Photovoltaic Array Perimeter & Inverter Enclosure (Civil & Construction Engineering)). Why does the outer security fence cost three times as much per linear meter as the interior inverter partition fences?

Practice Space Solve this problem on paper or tablet.

Review the variable reduction and domain derivation steps in the Conceptual Framework, state the Extreme Value Theorem hypothesis in Worked Example A, and read the introductory narrative of the Solar Photovoltaic Array application.

Answer: * (a) The upper bound is found by setting the substituted variable non-negative: $y(x) \ge 0 \implies g(x) \ge 0$ and solving for $x$ .

(b) The domain $[0, 100]$ is a closed bounded interval and $A(x)$ is continuous; by the Extreme Value Theorem, comparing critical values to endpoints guarantees the absolute maximum.
(c) Outer security fences require deep concrete trenching, continuous copper grounding loops, and fiber-optic intrusion detection, whereas interior partitions require only simple post holes.

Step-by-Step Execution

(a) In the Conceptual Framework, the upper boundary of a physical domain is established by recognizing that all physical dimensions must be non-negative. After solving the constraint equation for the dependent variable $y = g(x)$ , setting $g(x) \ge 0$ and solving for $x$ identifies the maximum allowable value $x_{\text{bound}}$ before the other dimension is forced to become negative.
(b) In Worked Example A, the domain is the closed bounded interval $[0, 100]$ . Because the objective function $A(x)$ is a polynomial, it is continuous on $[0, 100]$ . Under the Extreme Value Theorem, continuous functions on closed intervals are guaranteed to attain global extrema, and those extrema can only occur at critical points or boundary endpoints. Since $A(0) = 0$ and $A(100) = 0$ while $A(50) = 10,000$ , $x = 50$ is proven to be the absolute global maximum without needing a sign chart.
(c) In utility-scale solar installations, outer perimeter fencing is subjected to high security and electrical code standards, requiring continuous concrete trench footings to deter burrowing, embedded copper ground grids to dissipate high-voltage ground faults, and anti-intrusion monitoring. Internal inverter partitions serve merely to compartmentalize equipment and restrict pedestrian access, requiring only standard posts driven into soil.

Practice Problem 2

A rancher has a pasture bounded on one side by a straight river. The rancher wants to build a rectangular pen for livestock using 1200 feet of fencing material. No fencing is needed along the river boundary. What dimensions will produce the largest pen, and what is the maximum area?

Practice Space Solve this problem on paper or tablet.

Name the Rectangular Area Formula $A = xy$ and Perimeter Constraint $2x + y = 1200$ . Solve for $y$ , substitute into $A(x)$ , determine the closed domain, and locate the critical number where $A'(x) = 0$ .

Answer:

\text{Dimensions: } 300\text{ ft (perpendicular)} \times 600\text{ ft (parallel to river)}, \quad \text{Max Area} = 180,000\text{ ft}^2

Step-by-Step Execution

Name Governing Formulas: Rectangular Area and Perimeter Constraint Equations: (Defined in Applied Optimization Modeling Framework.)
Assign Variables: Let $x$ be the length of each of the two sides perpendicular to the river (in feet).
Let $y$ be the length of the side parallel to the river (in feet).
Formulate Equations:
- Objective: Maximize Area $A = xy$ .
- Constraint: $2x + y = 1200 \implies y = 1200 - 2x$ .
Variable Reduction: $A(x) = x(1200 - 2x) = 1200x - 2x^2$
Establish Domain: $x \ge 0$ and $1200 - 2x \ge 0 \implies x \le 600$ . Feasible closed domain: $x \in [0, 600]$ .
Differentiate and Find Critical Numbers: $A'(x) = 1200 - 4x$ $1200 - 4x = 0 \implies 4x = 1200 \implies x = 300\text{ feet}$
Evaluate Extrema (Closed Interval Method):
- $A(0) = 0\text{ ft}^2$
- $A(600) = 0\text{ ft}^2$
- $A(300) = 300(1200 - 600) = 300(600) = 180,000\text{ ft}^2$
Final Dimensions: $x = 300\text{ ft}, \quad y = 1200 - 2(300) = 600\text{ ft}$ Maximum enclosed area is $180,000\text{ ft}^2$ .

Practice Problem 3

A rectangular industrial storage yard is to be enclosed using two different types of fencing. Two opposite sides will use heavy-duty security fencing costing $\$4.50$ per foot. The remaining two sides will use standard security fencing costing $\$3.00$ per foot. If the total available construction budget is $\$18,000$ , find the total length of heavy-duty and standard fencing that should be purchased to maximize the enclosed area.

Practice Space Solve this problem on paper or tablet.

Assign $x$ as the length of the heavy-duty sides and $y$ as the length of the standard sides. State the Cost Constraint $2(4.50)x + 2(3.00)y = 18,000 \implies 9x + 6y = 18,000$ . Solve for $y$ , substitute into $A = xy$ , and differentiate.

Answer:

\text{Heavy-Duty Fencing: } 2,000\text{ ft}, \quad \text{Standard Fencing: } 3,000\text{ ft}, \quad \text{Max Area} = 1,500,000\text{ ft}^2

Step-by-Step Execution

Name Governing Formulas: Budget-Weighted Perimeter Constraint and Rectangular Area Formula: (Defined in Applied Optimization Modeling Framework.)
Assign Variables: Let $x$ be the length of each heavy-duty fence side (in feet).
Let $y$ be the length of each standard fence side (in feet).
Formulate Equations:
- Objective: Maximize Area $A = xy$ .
- Cost Constraint: $2(4.50)x + 2(3.00)y = 18,000 \implies 9x + 6y = 18,000$
Variable Reduction: $6y = 18,000 - 9x \implies y = 3,000 - 1.5x$ $A(x) = x(3,000 - 1.5x) = 3,000x - 1.5x^2$
Establish Domain: $x \ge 0$ and $3,000 - 1.5x \ge 0 \implies 1.5x \le 3,000 \implies x \le 2,000$ . Feasible domain: $x \in [0, 2000]$ .
Differentiate and Find Critical Numbers: $A'(x) = 3,000 - 3x$ $3,000 - 3x = 0 \implies 3x = 3,000 \implies x = 1,000\text{ feet}$
Evaluate Extrema (Closed Interval Method):
- $A(0) = 0$
- $A(2,000) = 0$
- $A(1,000) = 1,000(3,000 - 1,500) = 1,000(1,500) = 1,500,000\text{ ft}^2$
Compute Total Purchased Fencing:
- Heavy-duty fencing: $2x = 2(1,000) = 2,000\text{ feet}$ .
- Standard fencing: $2y = 2(3,000 - 1.5(1,000)) = 2(1,500) = 3,000\text{ feet}$ .

Practice Problem 4

A rectangular agricultural test plot covering an area of $432\text{ m}^2$ is to be enclosed by an outer fence and divided into two equal halves by an interior dividing fence parallel to one of the sides. What dimensions will require the smallest total length of fencing material, and how much fencing is needed?

Practice Space Solve this problem on paper or tablet.

Let $x$ be the length of the three parallel fence segments and $y$ be the length of the remaining two outer sides. State the Area Constraint $xy = 432$ and Fencing Objective $F = 3x + 2y$ . Substitute $y = 432/x$ and find the critical point on $(0, \infty)$ .

Answer:

\text{Dimensions: } 12\sqrt{2}\text{ m} \approx 16.97\text{ m} \times 18\sqrt{2}\text{ m} \approx 25.46\text{ m}, \quad \text{Min Fencing} = 72\sqrt{2}\text{ m} \approx 101.82\text{ m}

Step-by-Step Execution

Name Governing Formulas: Partitioned Enclosure Fencing Formula and Area Constraint: (Defined in Applied Optimization Modeling Framework.)
Assign Variables: Let $x$ be the length of the 3 parallel fence segments (in meters).
Let $y$ be the length of the 2 outer boundary sides perpendicular to $x$ (in meters).
Formulate Equations:
- Objective: Minimize Fencing $F = 3x + 2y$ .
- Constraint: Area $xy = 432 \implies y = \frac{432}{x}$ .
Variable Reduction: $F(x) = 3x + 2\left(\frac{432}{x}\right) = 3x + \frac{864}{x} = 3x + 864x^{-1}$
Establish Domain: Physical dimensions require $x > 0$ . Feasible domain: $x \in (0, \infty)$ .
Differentiate and Find Critical Numbers: $F'(x) = 3 - \frac{864}{x^2}$ $3 - \frac{864}{x^2} = 0 \implies 3x^2 = 864 \implies x^2 = 288$ $x = \sqrt{288} = \sqrt{144 \cdot 2} = 12\sqrt{2}\text{ meters} \approx 16.97\text{ m}$
Verify Global Minimum (Second Derivative Test): $F''(x) = \frac{1728}{x^3} \implies F''(12\sqrt{2}) > 0$ Concave up everywhere on $(0, \infty)$ ; hence $x = 12\sqrt{2}\text{ m}$ is the absolute minimum.
Compute Dimensions and Total Material: $y = \frac{432}{12\sqrt{2}} = \frac{36}{\sqrt{2}} = 18\sqrt{2}\text{ meters} \approx 25.46\text{ m}$ $\text{Total Fencing } F = 3(12\sqrt{2}) + 2(18\sqrt{2}) = 36\sqrt{2} + 36\sqrt{2} = 72\sqrt{2}\text{ meters} \approx 101.82\text{ m}$

Segment 2 of 8

📅 Regroup 1

Constraint Substitution Traps: When isolating a variable, always verify that the isolated variable has been completely cleared from the objective function before taking derivatives.
Cost Equipartition: In linear cost problems with two orthogonal dimensions, notice that the total dollars spent on horizontal elements ( $300 \times 400 = \$120,000$ ) exactly equals the dollars spent on vertical elements ( $400 \times 300 = \$120,000$ ).
Physical vs. Algebraic Extrema: Why do we discard negative critical numbers? Always cite the physical non-negativity principle.

Segment 3 of 8

Skill Block 2

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

Geometric Embedding, Margin Offsets & Composite Shapes

General Overview
Skill Block 2 expands optimization into embedded geometry and composite shapes. Students learn to handle margin offsets (active area vs. total material area in posters and silicon wafers), inscribed rectangles under coordinate curves, and composite geometric perimeters (Norman windows).

Conceptual Framework

Margin Offset Translation:
- Let $x$ and $y$ represent the inner active dimensions (e.g., printed text or active silicon logic).
- Adding margins of width $m_x$ on each side and $m_y$ on top/bottom expands the total dimensions to: $W = x + 2m_x, \quad H = y + 2m_y$
- Total Area to minimize: $A(x) = (x + 2m_x)(y(x) + 2m_y)$ .
Coordinate Symmetry in Inscribed Geometry:
- For a rectangle inscribed under an even function $y = f(x)$ with base on the $x$ -axis, the vertices are $(-x, 0), (x, 0), (x, f(x)), (-x, f(x))$ .
- Width is $2x$ , height is $f(x)$ , and area is $A(x) = 2x f(x)$ .
Composite Boundary Curves (Norman Windows):
- A Norman window combines a rectangular base (width $2r$ , height $h$ ) surmounted by a semicircular arch of radius $r$ .
- Perimeter includes 3 straight sides plus 1 semicircular arc: $P = 2r + 2h + \pi r$ .

Key Equations & Structural Rules

Margin Expansion Formula: $A_{\text{total}}(x) = (x + 2m_x)\left(\frac{A_{\text{active}}}{x} + 2m_y\right)$ (see Applied Optimization Modeling Framework)
Inscribed Symmetric Area Formula: $A(x) = 2x \cdot f(x) \quad \text{for } x \in [0, x_{\text{intercept}}]$ (see Context-Driven Domain in Geometric Optimization)

Worked Example A: Poster Layout with Margins

A graphic designer is laying out a rectangular poster that must contain $60\text{ in}^2$ of printed text. The design specifications require margins of 3 inches at the top and bottom, and 5 inches on each side. Find the overall dimensions of the poster that will minimize the total amount of paper used.

💡 Reveal Worked Solution

Name the Governing Formulas: Margin Offset Expansion and Rectangular Area Formula: (Defined in Applied Optimization Modeling Framework.)
Assign Variables: Let $x$ be the width of the printed text region (in inches).
Let $y$ be the height of the printed text region (in inches).
Formulate Constraint and Dimensions:
- Printed Text Area Constraint: $xy = 60 \implies y = \frac{60}{x}$
- Overall Poster Width: $W = x + 2(5) = x + 10\text{ inches}$
- Overall Poster Height: $H = y + 2(3) = y + 6\text{ inches}$
Formulate Objective Function: Minimize total paper area $A = W \cdot H$ : $A(x) = (x + 10)(y + 6) = (x + 10)\left(\frac{60}{x} + 6\right)$ Expand the expression: $A(x) = x\left(\frac{60}{x}\right) + 6x + 10\left(\frac{60}{x}\right) + 60$ $A(x) = 60 + 6x + \frac{600}{x} + 60 = 6x + 600x^{-1} + 120$
Establish Context-Driven Domain: Physical dimensions must be positive: $x > 0$ . Feasible domain: $x \in (0, \infty)$ .
Differentiate and Find Critical Numbers: $A'(x) = 6 - \frac{600}{x^2}$ $6 - \frac{600}{x^2} = 0 \implies 6x^2 = 600 \implies x^2 = 100 \implies x = 10\text{ inches}$ (Reject $x = -10\text{ in}$ as non-physical).
Verify Global Minimum (Second Derivative Test): $A''(x) = \frac{1200}{x^3} \implies A''(10) = \frac{1200}{1000} = 1.2 > 0$ Since $A''(10) > 0$ and $x = 10$ is the unique critical point on $(0, \infty)$ , it yields the absolute minimum area.
Compute Final Overall Dimensions:
- Text height: $y = \frac{60}{10} = 6\text{ inches}$ .
- Overall Width: $W = 10 + 10 = 20\text{ inches}$ .
- Overall Height: $H = 6 + 6 = 12\text{ inches}$ . Minimum total paper area is $20 \times 12 = 240\text{ in}^2$ .

Practice Problem 5: Concept & Reading Check

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

(a) In the Conceptual Framework, why does an inscribed rectangle centered on the $y$ -axis have width $2x$ rather than $x$ ?
(b) Deconstruct Worked Example A: When expanding $A(x) = (x + 10)\left(\frac{60}{x} + 6\right)$ , where does the constant term $120$ originate?
(c) Review the linked Engineering Application (Stage Truss Projection Screen Aspect Ratio (Entertainment Engineering & Design)). At the maximum display area, what fraction of the total truss apex height does the screen occupy?

Practice Space Solve this problem on paper or tablet.

Review the symmetry rules in the Conceptual Framework, multiply out the terms in Worked Example A, and read the Theatrical Rigging Insight in the Stage Truss Screen application.

Answer: * (a) The rectangle extends from $-x$ to $+x$ across the $y$ -axis, giving total width $x - (-x) = 2x$ .

(b) The constant $120$ is the sum of $x(60/x) = 60$ and $10(6) = 60$ .
(c) The screen height ( $6\text{ m}$ ) occupies exactly two-thirds ( $2/3$ ) of the total arch apex height ( $9\text{ m}$ ).

Step-by-Step Execution

(a) When a geometric figure is symmetric with respect to the $y$ -axis, the right half of the figure extends from the centerline $x = 0$ to coordinate $+x$ , while the left half extends symmetrically from $x = 0$ to $-x$ . The total horizontal span is the Euclidean distance from $-x$ to $x$ : $\Delta x = x - (-x) = 2x$ .
(b) In Worked Example A, expanding the binomial product $(x + 10)\left(\frac{60}{x} + 6\right)$ yields four terms: $x \cdot \frac{60}{x} = 60$ , $x \cdot 6 = 6x$ , $10 \cdot \frac{60}{x} = \frac{600}{x}$ , and $10 \cdot 6 = 60$ . Combining the two numerical constants $60 + 60$ gives $120$ .
(c) In the UNLV Entertainment Engineering application, the parabolic truss has an apex height of $9\text{ meters}$ . The calculus optimization determines that the maximum viewing area occurs when the display height is $H = 6\text{ meters}$ . Thus, the screen occupies $\frac{6}{9} = \frac{2}{3}$ of the total arch elevation, deliberately preserving the top $3\text{ meters}$ for rigging lighting fixtures and line arrays.

Practice Problem 6

A rectangular banner is to be inscribed beneath the parabola $y = 12 - x^2$ in the upper half-plane ( $y \ge 0$ ), with its base resting on the $x$ -axis and its upper two vertices lying on the parabola. Determine the dimensions of the rectangle that will maximize its area, and find the maximum area.

Practice Space Solve this problem on paper or tablet.

Let the upper right corner be $(x, 12 - x^2)$ . State the Inscribed Area Formula $A(x) = 2x(12 - x^2) = 24x - 2x^3$ . Determine the closed interval $[0, \sqrt{12}]$ from the $x$ -intercepts and evaluate extrema via the Closed Interval Method.

Answer:

\text{Width} = 4, \quad \text{Height} = 8, \quad \text{Max Area} = 32

Step-by-Step Execution

Name Governing Formulas: Symmetric Inscribed Area Formula: (Defined in Applied Optimization Modeling Framework.)
Assign Coordinates and Dimensions: By symmetry across the $y$ -axis, the upper vertices are $(-x, y)$ and $(x, y)$ where $y = 12 - x^2$ .
- Width: $W = 2x$ .
- Height: $H = 12 - x^2$ .
Formulate Objective Function: $A(x) = 2x(12 - x^2) = 24x - 2x^3$
Establish Context-Driven Domain: Height must be non-negative: $12 - x^2 \ge 0 \implies x^2 \le 12 \implies x \in [0, \sqrt{12}]$ .
Differentiate and Find Critical Numbers: $A'(x) = 24 - 6x^2$ $24 - 6x^2 = 0 \implies 6x^2 = 24 \implies x^2 = 4 \implies x = 2$ (Reject $x = -2$ as $x \ge 0$ ).
Evaluate Extrema (Closed Interval Method):
- $A(0) = 24(0) - 2(0)^3 = 0$
- $A(\sqrt{12}) = 24\sqrt{12} - 2(\sqrt{12})^3 = 24(2\sqrt{3}) - 2(24\sqrt{3}) = 0$
- $A(2) = 24(2) - 2(2)^3 = 48 - 16 = 32$
Final Dimensions: $\text{Width } W = 2(2) = 4, \quad \text{Height } H = 12 - (2)^2 = 8, \quad \text{Max Area} = 32$

Practice Problem 7

A Norman window consists of a rectangle surmounted by a semicircle. If the total perimeter of the outer window frame is fixed at $16\text{ feet}$ , determine the radius $r$ of the semicircle that maximizes the total light-admitting area, and compute the maximum area.

Practice Space Solve this problem on paper or tablet.

Let $r$ be the radius of the semicircle (so window width is $2r$ ) and let $h$ be the height of the rectangular section. State the Perimeter Constraint $P = 2r + 2h + \pi r = 16$ . Solve for $h$ , substitute into Total Area $A = 2rh + \frac{1}{2}\pi r^2$ , and differentiate with respect to $r$ .

Answer:

r = \frac{16}{4 + \pi}\text{ ft} \approx 2.24\text{ ft}, \quad \text{Max Area} = \frac{128}{4 + \pi}\text{ ft}^2 \approx 17.92\text{ ft}^2

Step-by-Step Execution

Name Governing Formulas: Norman Window Perimeter and Composite Area Formulas: (Defined in Geometric Volume and Surface Area Formulas.)
Assign Variables: Let $r$ be the radius of the semicircular top (in feet).
The width of the window is $2r$ .
Let $h$ be the height of the rectangular lower portion (in feet).
Formulate Constraint Equation: Perimeter includes the bottom side ( $2r$ ), two vertical sides ( $2h$ ), and semicircular arc ( $\pi r$ ): $2r + 2h + \pi r = 16 \implies 2h = 16 - (2 + \pi)r \implies h = 8 - \left(1 + \frac{\pi}{2}\right)r$
Formulate Area Objective Function: $A = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{semicircle}} = (2r)h + \frac{1}{2}\pi r^2$ Substitute $h$ : $A(r) = 2r\left[8 - \left(1 + \frac{\pi}{2}\right)r\right] + \frac{1}{2}\pi r^2 = 16r - (2 + \pi)r^2 + \frac{1}{2}\pi r^2$ $A(r) = 16r - \left(2 + \frac{\pi}{2}\right)r^2 = 16r - \left(\frac{4 + \pi}{2}\right)r^2$
Establish Domain: $r \ge 0$ and $h \ge 0 \implies r \le \frac{16}{2 + \pi}$ .
Differentiate and Find Critical Numbers: $A'(r) = 16 - (4 + \pi)r$ $16 - (4 + \pi)r = 0 \implies r = \frac{16}{4 + \pi}\text{ feet} \approx 2.24\text{ ft}$
Verify Global Maximum: $A''(r) = -(4 + \pi) < 0$ Concave down everywhere, guaranteeing an absolute maximum.
Compute Rectangular Height and Maximum Area: $h = 8 - \left(\frac{2+\pi}{2}\right)\left(\frac{16}{4+\pi}\right) = \frac{16}{4+\pi}\text{ ft} = r$ $A_{\max} = 16\left(\frac{16}{4+\pi}\right) - \left(\frac{4+\pi}{2}\right)\left(\frac{16}{4+\pi}\right)^2 = \frac{256}{4+\pi} - \frac{128}{4+\pi} = \frac{128}{4+\pi}\text{ ft}^2 \approx 17.92\text{ ft}^2$

Segment 4 of 8

📅 Regroup 2

The Role of Symmetry: In parabolic and coordinate curve problems, recognizing symmetry across the axis reduces a multivariable coordinate problem $(x_1, x_2, y)$ to a single independent variable $x$ .
Algebraic Pre-simplification: In margin problems like Worked Example A, fully expanding the factored product before differentiating avoids cumbersome product rules and isolates simple power-rule terms.
Why Norman Windows Optimize at $h = r$ : Notice that the optimal rectangular height equals the radius of the arch! This geometric harmony is a classic property of constrained circular-rectangular composites.

Segment 5 of 8

Skill Block 3

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

Volumetric Containers, Material Minimization & Optimal Ratios

General Overview
Skill Block 3 extends optimization into 3D containers. Students learn to optimize volumes and surface areas of folded sheet-metal boxes, closed and open-top cylinders, and cost-weighted storage bins, validating the universal $h = 2r$ and $h = r$ aspect ratio theorems.

Conceptual Framework

The Cut-and-Fold Box Algorithm:
- Given a flat rectangular sheet of dimensions $W \times L$ (with $W \le L$ ).
- Cutting square corners of size $x$ produces: $\text{Height } h = x, \quad \text{Width } w = W - 2x, \quad \text{Length } l = L - 2x$
- Volume: $V(x) = x(W - 2x)(L - 2x)$ .
- Bounded Domain: $x \in \left[0, \frac{W}{2}\right]$ .
Cylindrical Enclosures (The $h = 2r$ Rule):
- For a closed cylinder of fixed volume $V_0 = \pi r^2 h$ , surface area is minimized when height equals diameter: $h = 2r$ .
- For an open-top cylinder (no top lid), surface area is minimized when height equals radius: $h = r$ .
Cost-Weighted Asymmetry:
- When container materials have unequal costs (e.g., base material costs $c_b$ while side material costs $c_s$ ), the optimal ratio shifts: $h = 2\left(\frac{c_b}{c_s}\right)r$ .

Key Equations & Structural Rules

Cylindrical Volume & Surface Area: $V = \pi r^2 h, \quad S_{\text{closed}} = 2\pi r^2 + 2\pi rh, \quad S_{\text{open}} = \pi r^2 + 2\pi rh$ (see Geometric Volume and Surface Area Formulas)
Optimal Aspect Ratio Theorems: $\text{Closed Cylinder: } h = 2r, \quad \text{Open Cylinder: } h = r$ (see Optimal Aspect Ratios in Geometric Containers)

Worked Example A: The Classic Cut-and-Fold Cardboard Box

An open-top rectangular shipping box is constructed from a 15-inch by 24-inch sheet of cardboard by cutting equal square notches of side length $x$ from each corner and folding up the flaps. Find the cut size $x$ that maximizes box volume, and state the final dimensions.

💡 Reveal Worked Solution

Name Governing Formulas: Prismatic Volume Formula: (Defined in Geometric Volume and Surface Area Formulas.)
Assign Variables and Dimensions: Let $x$ be the side length of the square cut from each corner (in inches).
When the flaps are folded upward:
- Height: $h = x$ .
- Width: $w = 15 - 2x$ .
- Length: $l = 24 - 2x$ .
Formulate Objective Function: $V(x) = l \cdot w \cdot h = (24 - 2x)(15 - 2x)x$ Expand the polynomial before differentiating: $(24 - 2x)(15 - 2x) = 360 - 48x - 30x + 4x^2 = 4x^2 - 78x + 360$ $V(x) = x(4x^2 - 78x + 360) = 4x^3 - 78x^2 + 360x$
Establish Context-Driven Domain: Physical dimensions must be non-negative: $x \ge 0, \quad 15 - 2x \ge 0 \implies x \le 7.5, \quad 24 - 2x \ge 0 \implies x \le 12$ The most restrictive upper bound is $x \le 7.5$ . Feasible closed domain: $x \in [0, 7.5]\text{ inches}$
Differentiate and Find Critical Numbers: $V'(x) = 12x^2 - 156x + 360$ Divide by 12: $12(x^2 - 13x + 30) = 0$ Factor the quadratic: $12(x - 3)(x - 10) = 0 \implies x = 3\text{ in} \quad \text{or} \quad x = 10\text{ in}$ Filter Through Domain: $x = 10$ lies outside the physical domain $[0, 7.5]$ and is discarded. The only viable critical number is $x = 3\text{ inches}$ .
Evaluate Extrema (Closed Interval Method):
- $V(0) = 0\text{ in}^3$
- $V(7.5) = (24 - 15)(15 - 15)(7.5) = 0\text{ in}^3$
- $V(3) = (24 - 6)(15 - 6)(3) = 18 \times 9 \times 3 = 486\text{ in}^3$ The absolute maximum volume is $486\text{ in}^3$ .
Final Dimensions:
- Length: $l = 24 - 2(3) = 18\text{ inches}$ .
- Width: $w = 15 - 2(3) = 9\text{ inches}$ .
- Height: $h = 3\text{ inches}$ .

Practice Problem 8: Concept & Reading Check

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

(a) In Worked Example A, explain why the mathematical critical point $x = 10$ was discarded from the analysis.
(b) In the Conceptual Framework, state the optimal ratio of height to diameter ( $h/d$ ) for a closed cylindrical container of fixed volume minimizing surface area.
(c) Review the linked Engineering Application (Cryogenic Propellant Tank Bulkhead Surface Area Minimization (Mechanical & Aerospace Engineering)). Why does minimizing the surface area of a rocket propellant tank directly increase orbital payload capacity?

Practice Space Solve this problem on paper or tablet.

Review the domain derivation in Worked Example A, check the closed cylinder rule in the Conceptual Framework, and read the introduction of the Cryogenic Propellant Tank application.

Answer: * (a) $x = 10$ exceeds the physical domain upper limit $x \le 7.5$ ; cutting 10-inch corners from a 15-inch sheet would produce negative width ( $-5\text{ in}$ ).

(b) For a closed cylinder, the optimal ratio is $h = 2r = d$ , meaning $h/d = 1$ (square profile).
(c) Propellant tanks are pressurized Al-Li metal structures; reducing surface area reduces dry structural mass, allowing the rocket to carry more payload under the rocket equation.

Step-by-Step Execution

(a) In Worked Example A, the shorter dimension of the cardboard sheet is 15 inches. Cutting two corner notches of size $x$ leaves a remaining box width of $w = 15 - 2x$ . If $x = 10$ , the calculated width would be $15 - 2(10) = -5\text{ inches}$ , which is physically impossible. Therefore, the physical non-negativity constraint restricts $x$ to $[0, 7.5]$ , disqualifying $x = 10$ .
(b) In the Conceptual Framework and Optimal Aspect Ratios in Geometric Containers, unconstrained surface-area minimization for a closed right circular cylinder yields an optimal height $h = 2r$ . Because the diameter is $d = 2r$ , the ratio of height to diameter is $\frac{h}{d} = \frac{2r}{2r} = 1$ . Its vertical cross-section forms a square.
(c) In orbital rocketry, dry tank mass is dead weight that must be accelerated to orbital velocity. Because propellant tank walls and insulation have a fixed mass per square meter ( $\rho_A = 15\text{ kg/m}^2$ ), every square meter of surface area eliminated directly removes $15\text{ kg}$ of structural dead weight, converting that mass capacity directly into revenue-generating satellite payload.

Practice Problem 9

An engineering packaging designer must create a can shaped like a closed right circular cylinder that can hold an internal fluid volume of $1,024\pi\text{ cm}^3$ . Determine the radius $r$ and height $h$ that will minimize the total surface area of sheet aluminum required to manufacture the can.

Practice Space Solve this problem on paper or tablet.

State the Cylindrical Volume Formula $V = \pi r^2 h = 1024\pi \implies h = 1024/r^2$ and Surface Area Formula $S = 2\pi r^2 + 2\pi rh$ . Substitute $h$ , differentiate $S(r)$ , and locate the critical number on $(0, \infty)$ .

Answer:

r = 8\text{ cm}, \quad h = 16\text{ cm}, \quad \text{Min Surface Area} = 384\pi\text{ cm}^2 \approx 1206.37\text{ cm}^2

Step-by-Step Execution

Name Governing Formulas: Cylindrical Volume and Surface Area Formulas: (Defined in Geometric Volume and Surface Area Formulas.)
Assign Variables: Let $r$ be the radius of the circular base (in cm).
Let $h$ be the height of the cylinder (in cm).
Formulate Equations:
- Objective: Minimize Surface Area $S = 2\pi r^2 + 2\pi rh$ .
- Constraint: Volume $V = \pi r^2 h = 1,024\pi \implies r^2 h = 1,024 \implies h = \frac{1,024}{r^2}$ .
Variable Reduction: $S(r) = 2\pi r^2 + 2\pi r\left(\frac{1,024}{r^2}\right) = 2\pi r^2 + \frac{2,048\pi}{r} = 2\pi r^2 + 2,048\pi r^{-1}$
Establish Domain: Physical dimensions require $r > 0$ . Feasible domain: $r \in (0, \infty)$ .
Differentiate and Find Critical Numbers: $S'(r) = 4\pi r - \frac{2,048\pi}{r^2}$ $4\pi r - \frac{2,048\pi}{r^2} = 0 \implies 4\pi r^3 = 2,048\pi \implies r^3 = 512$ $r = \sqrt[3]{512} = 8\text{ cm}$
Verify Global Minimum (Second Derivative Test): $S''(r) = 4\pi + \frac{4,096\pi}{r^3} \implies S''(8) = 4\pi + \frac{4,096\pi}{512} = 4\pi + 8\pi = 12\pi > 0$ Concave up everywhere on $(0, \infty)$ ; hence $r = 8\text{ cm}$ is the unique absolute minimum.
Compute Height and Minimum Surface Area: $h = \frac{1,024}{(8)^2} = \frac{1,024}{64} = 16\text{ cm}$ Notice that $h = 2r = 16\text{ cm}$ (height equals diameter). $S_{\min} = 2\pi (8)^2 + \frac{2,048\pi}{8} = 128\pi + 256\pi = 384\pi\text{ cm}^2 \approx 1206.37\text{ cm}^2$

Practice Problem 10

A heavy-duty open-top rectangular industrial storage bin with a square base must have an internal volume of $32\text{ m}^3$ . Heavy wear-resistant steel plate for the base costs $\$6/\text{m}^2$ , while standard steel sheet for the four vertical side walls costs $\$3/\text{m}^2$ . Determine the base width $x$ and height $h$ that will minimize the total material fabrication cost.

Practice Space Solve this problem on paper or tablet.

Let $x$ be the side length of the square base and $h$ be the height. State the Volume Constraint $x^2 h = 32 \implies h = 32/x^2$ . State the Cost Objective $C = 6(x^2) + 3(4xh) = 6x^2 + 12xh$ . Substitute $h$ and differentiate.

Answer:

x = 2\sqrt[3]{4}\text{ m} \approx 3.17\text{ m}, \quad h = 2\sqrt[3]{4}\text{ m} \approx 3.17\text{ m}, \quad \text{Min Cost} \approx \$181.43

Step-by-Step Execution

Name Governing Formulas: Open Rectangular Container Cost Formula and Volume Constraint: (Defined in Applied Optimization Modeling Framework.)
Assign Variables: Let $x$ be the width and length of the square base (in meters).
Let $h$ be the height of the bin (in meters).
Formulate Equations:
- Volume Constraint: $V = x^2 h = 32 \implies h = \frac{32}{x^2}$ .
- Cost Objective Function: $C = (\text{Base Cost}) + (\text{Sides Cost}) = 6(x^2) + 3(4xh) = 6x^2 + 12xh$
Variable Reduction: $C(x) = 6x^2 + 12x\left(\frac{32}{x^2}\right) = 6x^2 + \frac{384}{x} = 6x^2 + 384x^{-1}$
Establish Domain: $x > 0$ . Feasible domain: $x \in (0, \infty)$ .
Differentiate and Find Critical Numbers: $C'(x) = 12x - \frac{384}{x^2}$ $12x - \frac{384}{x^2} = 0 \implies 12x^3 = 384 \implies x^3 = 32$ $x = \sqrt[3]{32} = \sqrt[3]{8 \cdot 4} = 2\sqrt[3]{4}\text{ meters} \approx 3.1748\text{ m}$
Verify Global Minimum (Second Derivative Test): $C''(x) = 12 + \frac{768}{x^3} \implies C''(2\sqrt[3]{4}) = 12 + \frac{768}{32} = 12 + 24 = 36 > 0$ Concave up everywhere; $x = 2\sqrt[3]{4}\text{ m}$ is the absolute minimum.
Compute Height and Total Cost: $h = \frac{32}{(2\sqrt[3]{4})^2} = \frac{32}{4 \cdot 4^{2/3}} = \frac{8}{4^{2/3}} = 2 \cdot 4^{1/3} = 2\sqrt[3]{4}\text{ meters} \approx 3.1748\text{ m}$ Notice that for this 2-to-1 material cost weighting ( $\$6/\$3 = 2$ ), the optimal container geometry is a perfect cube ( $x = h$ ). $\text{Minimum Cost } C_{\min} = 6(2\sqrt[3]{4})^2 + \frac{384}{2\sqrt[3]{4}} = 24(4^{2/3}) + \frac{192}{4^{1/3}} = 72(4^{2/3}) \approx \$181.43$

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📅 Regroup 3

Closed vs. Open Cylinders: For a closed cylinder, $h = 2r$ . For an open cylinder, $h = r$ . Recognizing this pattern gives you an instant sanity check on exam and homework problems.
Handling Non-Closed Domains: When optimizing open tanks on $(0, \infty)$ , we cannot evaluate endpoints because $x = 0$ is undefined and $x \to \infty$ is unbounded. Always execute the Second Derivative Test ( $f''(c) > 0$ ) to prove an absolute minimum.
Cost-Weighted Symmetry: When base plate steel is twice as expensive as side plate steel, the open container naturally balances out to a cube ( $x = h$ ).

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📅 Synthesis Wrap-up

Core Takeaways from Session 14:
1. Optimization translates physical words into calculus: identify objective and constraints, substitute to $f(x)$ , and enforce context-driven domains.
2. The nature of the domain dictates your proof: use the Closed Interval Method on $[a, b]$ ; use the Second Derivative Test on $(0, \infty)$ .
3. Geometric symmetry is optimal unless cost weighting or boundary barriers intentionally break that symmetry.
Homework Transition: Complete Knewton Alta assignment: Optimization Problems in a Geometric Context. Prepare for Session 15: Optimizing Distance and Time.

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