MATH-181 (Calculus I) Curriculum

Session 14 - Optimizing Areas and Volumes

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Translate applied narrative scenarios involving planar boundaries, pastures, and cost-weighted enclosures into single-variable objective functions using the Applied Optimization Modeling Framework and algebraic constraint substitution.
Determine context-driven physical domains ( $x \ge 0, y(x) \ge 0$ ), distinguishing between closed intervals $[a, b]$ evaluated via the Closed Interval Method and open intervals $(0, \infty)$ evaluated via the Second Derivative Test.
Optimize composite 2D geometric configurations involving margin offsets, inscribed rectangles under quadratic curves, and circular arches (Norman windows) by applying symmetry and coordinate modeling.
Model and optimize 3D volumetric containers (folded sheet-metal boxes, cylindrical storage tanks, and open-top bins with differential material costs), verifying optimal aspect ratios ( $h = 2r$ for closed cylinders, $h = r$ for open cylinders) formalized in Optimal Aspect Ratios in Geometric Containers.

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.

Skill Block 1

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.

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?

Solution Key:

(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.

Show your steps and calculations here.

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?

Solution Key:

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 $180,000\text{ ft}^2$ .

Show your steps and calculations here.

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.

Solution Key:

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: $2y = 2(3,000 - 1.5(1,000)) = 2(1,500) = 3,000\text{ feet}$ .

Show your steps and calculations here.

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?

Solution Key:

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: $y = \frac{432}{12\sqrt{2}} = \frac{36}{\sqrt{2}} = 18\sqrt{2}\text{ meters} \approx 25.46\text{ m}$ $\text{Total Fencing } F = 3(12\sqrt{2}) + 2(18\sqrt{2}) = 36\sqrt{2} + 36\sqrt{2} = 72\sqrt{2}\text{ meters} \approx 101.82\text{ m}$

Show your steps and calculations here.

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.

Skill Block 2

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.

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?

Solution Key:

(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.

Show your steps and calculations here.

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.

Solution Key:

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: $\text{Width } W = 2(2) = 4, \quad \text{Height } H = 12 - (2)^2 = 8, \quad \text{Max Area} = 32$

Show your steps and calculations here.

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.

Solution Key:

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: $h = 8 - \left(\frac{2+\pi}{2}\right)\left(\frac{16}{4+\pi}\right) = \frac{16}{4+\pi}\text{ ft} = r$ $A_{\max} = 16\left(\frac{16}{4+\pi}\right) - \left(\frac{4+\pi}{2}\right)\left(\frac{16}{4+\pi}\right)^2 = \frac{256}{4+\pi} - \frac{128}{4+\pi} = \frac{128}{4+\pi}\text{ ft}^2 \approx 17.92\text{ ft}^2$

Show your steps and calculations here.

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.

Skill Block 3

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.

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?

Solution Key:

(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.

Show your steps and calculations here.

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.

Solution Key:

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). $S_{\min} = 2\pi (8)^2 + \frac{2,048\pi}{8} = 128\pi + 256\pi = 384\pi\text{ cm}^2 \approx 1206.37\text{ cm}^2$

Show your steps and calculations here.

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.

Solution Key:

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 ( $x = h$ ). $\text{Minimum Cost } C_{\min} = 6(2\sqrt[3]{4})^2 + \frac{384}{2\sqrt[3]{4}} = 24(4^{2/3}) + \frac{192}{4^{1/3}} = 72(4^{2/3}) \approx \$181.43$

Show your steps and calculations here.

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 ( $x = h$ ).

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.