Modeling Blueprints Slides Calculus III / Applied Analysis
Drafting the Model
Lesson 1: Formulating Objective Functions and Constraints for Industrial Optimization
PROJECT REF: OPT-001
The Cost of a Bad Draft
Consider a shipping container designed with a 10% sub-optimal surface-to-volume ratio.
$0.12
Wasted material cost per unit produced.
If a logistics firm produces 5 million units annually, this simple geometric error costs them $600,000 every year.
Precision Matters
Optimization isn't just a math exercise; it's the financial backbone of industrial engineering.
The Anatomy of an Optimization Problem
Objective Function
"What are we trying to make as large or as small as possible?"
\[ f(x, y) = \dots \]
Profit, Volume, Efficiency
Cost, Surface Area, Time
Constraints
"What limits our choices?"
\[ g(x, y) = k \]
Fixed Budget
Material Limits
Physical Dimensions
The 4-Step Modeling Protocol
01
Variable Audit
Identify every unknown. Sketch the system and label clearly.
02
Goal Definition
Translate "maximize/minimize" into a function \( f \).
03
Constraint Isolation
Use fixed values to link variables (e.g., \( V = 1000 \)).
04
Feasible Domain
Determine the physical boundaries (e.g., \( x > 0 \)).
"A problem well-stated is a problem half-solved." — Charles Kettering
Workshop: The Open Box
PRACTICE 1.1
A manufacturer wants to create an open-topped box from a 24-inch square piece of cardboard by cutting equal squares from the corners and folding up the sides.
Objective: Maximize Volume \( V \)
Variable: Cut size \( x \)
Discuss with your partner:
What is the constraint? What is the domain of \( x \)?
x
x
x
x
24" x 24" SHEET
\[ V(x) = x(24 - 2x)^2 \]
Translation Toolkit Worksheet Translation Toolkit
Applied Calculus III: Optimization Modeling
Student Name:
Date:
Mission Briefing
Your task is to translate industrial requirements into mathematical models. For each scenario, you must define variables, construct the Objective Function , identify Constraints , and determine the Feasible Domain .
01
The Perimeter Fence Conflict
A rancher wants to enclose a rectangular field with a total area of 1,200 square meters. The field is bordered on one side by a straight river, so no fencing is required on that side. The rancher wants to use the minimum amount of fencing material possible.
Variables & Sketch
Objective Function \( f \)
Constraint Equation \( g \)
Feasible Domain
Single Variable Model \( f(x) \)
02
The Dual-Material Cylinder
A chemical storage tank is designed as a right circular cylinder with a volume of 500 \( \text{m}^3 \). The material for the top and bottom costs \( \$15 \) per square meter, while the material for the side costs \( \$10 \) per square meter. The company needs to minimize the total construction cost.
Variables & Sketch
Cost Components (Top/Bottom vs. Side)
Constraint Equation
Total Cost Function \( C(r, h) \)
Domain Constraints
Final Objective Function \( C(r) \)
Self-Verification Protocol
Before proceeding to the calculus phase: Are your units consistent? Does your domain eliminate physical impossibilities (negative dimensions)? Does your final function contain exactly one independent variable?
Modeling Mastery Teacher Guide Teacher Guide
Lesson 1: Drafting the Model (Optimization Foundations)
Level Undergrad
Learning Objectives
Distinguish between objective functions (to be extremized) and constraints (fixed limits).
Model real-world scenarios using multi-variable equations and substitution.
Define the feasible domain based on physical and mathematical restrictions.
Key Vocabulary
Objective Function Constraint Feasible Domain Independent Variable
High-Leverage Discussion Prompts
Prompt 1: The "Why Substitution?" Inquiry
"Why can't we just find the derivative of the cost function \( C(r, h) \) immediately? What mathematical obstacle prevents us from using standard single-variable calculus here?"
Expected Student Insight: We need the function in terms of one variable to apply the Power/Product rules we've learned so far.
Prompt 2: The Domain Criticality
"In the fence problem, why is the domain \( 0 < x < \infty \) mathematically, but perhaps more limited practically? How does the domain change if the rancher only has 500 meters of fencing available?"
Modeling Pitfalls & Misconceptions
Variable Confusion
Students often treat fixed constants (like volume \( V = 1000 \)) as variables during differentiation, leading to unnecessary complexity or zeroed-out terms.
Implicit Constraints
Failure to account for physical constraints (e.g., \( r > 0 \)) results in finding "mathematical" extrema that are physically impossible.
Translation Toolkit Quick Key
Problem Objective Function Constraint Equation 1. Fence P = 2x + y xy = 1200 2. Cylinder C = 15(2\pi r^2) + 10(2\pi rh) \pi r^2 h = 500
Packaging Precision Slides Calculus III / Applied Geometry
Packaging Precision
Lesson 2: Surface Area Minimization and Volume Maximization in Industry
PROJECT REF: OPT-002
The Soda Can Mystery
A standard 355mL soda can has a height of 12.2 cm and a radius of 3.25 cm.
The Challenge:
Does this ratio minimize the surface area for the given volume? Or is there an "engineering ghost" in the machine?
h = 12.2
r = 3.25
Current Market Standard
Verification of Extrema
1st Derivative Test
Finding Critical Points
1. Solve \( f'(x) = 0 \)
2. Analyze sign changes of \( f'(x) \)
Result: Candidate for Min/Max
2nd Derivative Test
Confirming Concavity
If \( f''(c) > 0 \rightarrow \) Local MIN
If \( f''(c) < 0 \rightarrow \) Local MAX
Result: Immediate Verification
Always check Endpoints if the domain is a closed interval!
The "Golden Ratio" of Cylinders
1. Objective: \( SA = 2\pi r^2 + 2\pi rh \)
2. Constraint: \( V = \pi r^2 h \rightarrow h = \frac{V}{\pi r^2} \)
3. Substitute: \( SA(r) = 2\pi r^2 + \frac{2V}{r} \)
4. Differentiate: \( SA'(r) = 4\pi r - \frac{2V}{r^2} \)
5. Solve for 0: \( 4\pi r^3 = 2V \rightarrow r = \sqrt[3]{\frac{V}{2\pi}} \)
Theoretical Optimum
h = 2r
The height should equal the diameter.
Compare this to the soda can (\( h \approx 4r \)). Why the difference?
Why Math Isn't Everything
Ergonomics
A "perfectly optimal" can (h=2r) would be very wide and squat—difficult for human hands to grip comfortably.
Packing Density
Taller cans might allow for more efficient vertical stacking in standard shipping containers or vending machines.
Material Waste
Manufacturing processes (like stamping aluminum) may create more waste for specific ratios, regardless of surface area.
Soda Can Challenge Activity Technical Specification
Soda Can Challenge
Optimization Lab: Surface Area vs. Volume
Engineer ID:
A beverage company is designing a new 500 mL energy drink can. The company wants to minimize the cost of aluminum, which is directly proportional to the total surface area of the can. Use your calculus modeling skills to determine the "mathematically perfect" dimensions.
01
Establish the Model
Define the objective function for Surface Area (\( A \)) and the constraint equation for Volume (\( V \)).
Objective Function \( A(r, h) \):
Constraint Equation (\( V = 500 \)):
02
Reduce and Differentiate
Substitute the constraint into the objective function to get \( A(r) \), then find its derivative \( A'(r) \).
03
Identify Critical Points
Set \( A'(r) = 0 \) and solve for the optimal radius \( r \). Then find the corresponding height \( h \).
Solve for Optimal \( r \):
Solve for Optimal \( h \):
04
Extrema Verification
Apply the Second Derivative Test to prove that your critical point is indeed a minimum.
Industrial Interpretation
The dimensions you found represent the absolute minimum surface area for a 500mL volume. Most energy drink cans are actually 16cm tall and 3.3cm in radius.
The Question:
Calculate the Surface Area of the "Market Standard" can versus your "Optimal" can. What is the percentage increase in material cost for the market version?
Packaging Proofs Answer Key Answer Key
Lesson 2: Packaging Precision (Soda Can Lab)
Resource Internal Use Only
Part I: The Derivation (500 mL Can)
Step 1: Models
\[ A = 2\pi r^2 + 2\pi rh \]
\[ \pi r^2 h = 500 \rightarrow h = \frac{500}{\pi r^2} \]
Step 2: Substitution
\[ A(r) = 2\pi r^2 + 2\pi r \left( \frac{500}{\pi r} \right) \]
\[ A(r) = 2\pi r^2 + \frac{1000}{r} \]
Step 3: Differentiation & Critical Point
\[ A'(r) = 4\pi r - \frac{1000}{r^2} \]
Set \( A'(r) = 0 \):
\[ 4\pi r^3 = 1000 \rightarrow r^3 = \frac{250}{\pi} \]
\[ r_{opt} \approx 4.30 \text{ cm} \]
\[ h_{opt} = \frac{500}{\pi(4.30)^2} \approx 8.60 \text{ cm} \]
Note: \( h = 2r \) is verified.
Step 4: Verification (2nd Derivative Test)
\[ A''(r) = 4\pi + \frac{2000}{r^3} \]
Since \( r > 0 \), then \( A''(r) > 0 \) for all \( r \).concave up at the critical point, indicating a Local Minimum .
Part II: Industrial Analysis (Reflections)
Market Standard Comparison
Can Type Dimensions (\( r \), \( h \)) Surface Area (\( A \)) Optimal 4.30 cm, 8.60 cm ~348.7 cm\(^2\) Market (Energy) 3.30 cm, 16.0 cm ~400.2 cm\(^2\)
Efficiency Loss: The market standard can uses 14.8% more aluminum than the mathematically optimal design.
Grading Rubric for Step 4
4 pts Correct objective function and constraint setup.
4 pts Accurate differentiation and critical point calculation.
2 pts Rigorous verification using 2nd Derivative Test or First Derivative sign chart.
Marginal Maximization Slides Calculus III / Applied Economics
Economic Extrema
Lesson 3: Minimizing Cost, Maximizing Profit, and Marginal Analysis
PROJECT REF: OPT-003
The Optimization Trinity
Revenue \( R(x) \)
Total money coming in. Usually Price \(\times\) Quantity.
\( p(x) \cdot x \)
Cost \( C(x) \)
Total expenses. Includes fixed and variable costs.
\( F + V(x) \)
Profit \( P(x) \)
"The Ultimate Objective"
\( R(x) - C(x) \)
Marginal Analysis
In economics, the derivative \( f'(x) \) is called the Marginal function.
Profit Maximization Condition:
\[ R'(x) = C'(x) \]
"Profit is maximized when Marginal Revenue equals Marginal Cost."
The Proof
Maximize \( P(x) = R(x) - C(x) \)
1. Find \( P'(x) = R'(x) - C'(x) \)
2. Set \( P'(x) = 0 \)
3. \( R'(x) - C'(x) = 0 \)
4. \( R'(x) = C'(x) \)
Inventory Management
A factory must produce 10,000 units per year.
Setup Cost: Every time they start a production run, it costs $500.
Holding Cost: Storing a unit for a year costs $2.
How many units should they produce in each run to minimize total annual cost?
Total Cost Function \( C(x) \)
\[ C(x) = 500\left(\frac{10000}{x}\right) + 2\left(\frac{x}{2}\right) \]
Setup Fees Holding Fees
Production Sim
"You are the Operations Manager. If you produce too much at once, your warehouse overflows. If you produce too little, you spend all your time resetting the machines."
Run the Numbers
Maximize Profit
Manufacturing Manager Simulation Worksheet Manufacturing Manager Simulation
Operational Phase: Cost Balance Protocol
Authorization Code
OPS-7742-X
Briefing: Quantum Circuits Inc.
Your facility must fulfill an annual demand of 40,000 circuits. You need to decide how many circuits to produce in each production batch (\( x \)) to minimize the total annual cost.
Setup Cost (\( C_s \))
$2,000 per production run
Cost for cleaning machines, recalibrating lasers, and hiring specialists.
Holding Cost (\( C_h \))
$4 per unit per year
Cost for climate-controlled storage, insurance, and interest on inventory.
1
The Annual Cost Equation
Let \( x \) be the number of units per batch. Express the number of runs per year as a function of \( x \). Then, express the average inventory held throughout the year (assume inventory starts at \( x \) and drops to 0 linearly before the next run).
Runs per year:
Average Inventory:
Construct Total Annual Cost \( C(x) \) = (Setup Cost) + (Holding Cost)
2
Optimization Execution
Find the optimal batch size \( x^* \) that minimizes total cost. Show your derivative and verification steps.
3
Managerial Analysis
Based on your optimal \( x^* \), how many production runs will occur per year? What is the minimum total annual cost?
Sensitivity Inquiry:
Suppose the warehouse rent (Holding Cost) doubles due to a local tax increase. Without re-calculating everything, intuitively explain: Should your batch size \( x \) increase or decrease? Why?
Economic Extrema Teacher Notes Teacher Notes
Lesson 3: Economic Extrema & Marginal Theory
Lesson ID OPT-003
Instructional Context
This lesson bridges the gap between abstract optimization and business logic. The "Manufacturing Simulation" uses a simplified version of the Economic Order Quantity (EOQ) model. Students often find this more intuitive than geometric problems because the trade-offs (Setup vs. Holding) are distinct and grounded in common sense.
Key Bridge Concept
"Marginal Cost is the cost of producing one more unit. Profit stops growing when the cost to make the next unit is higher than the money it brings in."
Discrete vs. Continuous
Remind students that while calculus treats \( x \) as continuous, in reality, a manager can't produce \( 200.4 \) batches. We use the continuous model as an approximation for the optimal integer solution.
Simulation Solution Key (Quantum Circuits Inc.)
The Equation
\[ C(x) = 2000\left(\frac{40000}{x}\right) + 4\left(\frac{x}{2}\right) \]
\[ C(x) = \frac{80,000,000}{x} + 2x \]
The Solution
\[ C'(x) = -\frac{80,000,000}{x^2} + 2 \]
\[ 2 = \frac{80,000,000}{x^2} \rightarrow x^2 = 40,000,000 \]
\[ x^* \approx 6,325 \text{ units} \]
Annual Runs: \( 40,000 / 6,325 \approx 6.3 \) runs per year.
Min Annual Cost: \( C(6325) \approx \$25,298.22 \)
Discussion & Debrief
"How does a change in fixed cost (Setup Fee) affect the optimal batch size?"
Answer: Increased fixed costs drive \( x \) higher (economies of scale); we want to run the machine fewer times.
"Is there a scenario where the Second Derivative Test fails in an economic context?"
Answer: If the cost function is linear, the derivative is constant and no extrema exist within the interval (boundary solution).
Pathfinder Optimization Slides Calculus III / Physical Systems
Optimal Paths
Lesson 4: Physical Efficiency and Trigonometric Optimization
PROJECT REF: OPT-004
The Lifeguard Dilemma
A lifeguard at point A needs to reach a swimmer at point B.
The Variables:
Speed on Sand (\( v_1 \)) = 8 m/s
Speed in Water (\( v_2 \)) = 1.5 m/s
The shortest distance (straight line) is rarely the shortest time. Why?
Water
Sand
B
A
Angle-Based Models
Optimal Viewing Angle
A painting is hung high on a wall. At what distance from the wall is the viewing angle \(\theta\) maximized?
\[ \theta(x) = \arctan\left(\frac{b}{x}\right) - \arctan\left(\frac{a}{x}\right) \]
Trig Toolset
Required Derivatives:
\[ \frac{d}{dx}[\sin(x)] = \cos(x) \]
\[ \frac{d}{dx}[\cos(x)] = -\sin(x) \]
\[ \frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2} \]
Snell's Law via Calculus
1. Time \( T = \frac{d_1}{v_1} + \frac{d_2}{v_2} \)
2. \( d_1 = \sqrt{x^2 + a^2}, d_2 = \sqrt{(L-x)^2 + b^2} \)
3. Differentiate \( T \) w.r.t \( x \)
4. Set \( T'(x) = 0 \)
The Optic Result
\[ \frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2} \]
Nature always minimizes time, not distance.
Beyond Snell
Energy Efficiency
Optimization in biology: Birds adjusting wing-beat frequency and angle to minimize oxygen consumption during long-haul migration.
Structural Integrity
Finding the optimal angle for support trusses in a bridge to minimize total material volume while maintaining a required load capacity.
"Calculus is the language in which God has written the laws of the universe." — Galileo
Trig Tactic Worksheet Trig Tactic Worksheet
Pathfinding & Efficiency in Non-Uniform Mediums
Problem Set
OPT-PATH-04
Navigator:
Timestamp:
01
The Shoreline Crossing
A lifeguard is 40 meters from the shoreline. A swimmer is 30 meters offshore, and 100 meters down the beach (horizontally) from the lifeguard's current position. The lifeguard can run at 6 m/s on sand and swim at 2 m/s in water.
1. Variable Definition & Diagram
Sketch the path and label the crossing point distance 'x' from the start.
2. Objective: Total Time \( T(x) \)
3. The Derivative \( T'(x) \)
4. Optimal Crossing Distance \( x \)
02
The Gallery Angle
A 10-foot tall painting is mounted on a gallery wall. The bottom edge of the painting is 7 feet above your eye level. At what distance \( x \) from the wall is the angle \( \theta \) subtended by the painting at your eye maximized?
x
SUBTENDED ANGLE MODEL
1. Construct \( \theta(x) \) using Arctan
θ(x) = arctan(...) - arctan(...)
2. Solve for \( \theta'(x) = 0 \)
The Challenge:
Find the numerical value of x and the resulting maximum angle in degrees.
Convergence Protocol
Verify that your optimal value is not a boundary solution. In Problem 2, what happens as \( x \rightarrow 0 \) and \( x \rightarrow \infty \)? How does this confirm your solution is a global maximum?
Lifeguard Logic Teacher Key Answer Key
Lesson 4: Optimal Paths & Trig Systems
Security Level High
Part I: Shoreline Crossing (Lifeguard)
The Model
\[ T(x) = \frac{\sqrt{x^2 + 40^2}}{6} + \frac{\sqrt{(100-x)^2 + 30^2}}{2} \]
The Derivative
\[ T'(x) = \frac{x}{6\sqrt{x^2 + 1600}} - \frac{100-x}{2\sqrt{(100-x)^2 + 900}} \]
Conceptual Solution:
At the minimum, \( \frac{\sin \theta_1}{6} = \frac{\sin \theta_2}{2} \). The ratio of the sines of the angles of incidence must equal the ratio of the speeds.
Numerical Result:
x ≈ 90.7 meters
The lifeguard should run most of the way down the beach before entering the water.
Part II: Gallery Angle Maximization
Modeling Angle Subtension
\[ \theta(x) = \arctan\left(\frac{17}{x}\right) - \arctan\left(\frac{7}{x}\right) \]
Differentiation
\[ \theta'(x) = \frac{-17/x^2}{1+(17/x)^2} - \frac{-7/x^2}{1+(7/x)^2} \]
\[ \theta'(x) = \frac{-17}{x^2+289} + \frac{7}{x^2+49} \]
Critical Point Logic
Set \( \theta'(x) = 0 \rightarrow 17(x^2+49) = 7(x^2+289) \)
\[ 17x^2 + 833 = 7x^2 + 2023 \]
\[ 10x^2 = 1190 \rightarrow x^2 = 119 \]
x ≈ 10.91 feet
θ ≈ 24.6°
Rubric Highlights
Modeling (40%)
Correctly identifying Pythagorean components in T(x) or Arctan components in \(\theta(x)\).
Chain Rule (40%)
Accuracy in differentiating composite functions, specifically handling negative signs and \(x^2\) denominators.
Final Logic (20%)
Interpreting the mathematical result back into physical context (e.g., feet, meters, degrees).
Infrastructure RFP Slides Calculus III / Capstone Project
The Final Blueprint
Lesson 5: Synthesizing Optimization for Industrial Infrastructure
PHASE 05: MASTERY
Request for Proposal (RFP)
Reference: INFRA-OPT-X
The Objective:
"Design a pipeline route connecting a refinery to a distribution hub. The route must cross two distinct terrain types: Wetlands and Hardrock Ridge."
Economic Constraint:
$12M / mile (Wetlands)
$18M / mile (Hardrock Ridge)
Wetlands
Hardrock
Proposed Site: Sector 7G
Multi-Faceted Synthesis
Geometric Model
Determine the path distances \( d_1 \) and \( d_2 \) based on crossing points and terrain boundaries.
\[ \sqrt{x^2 + a^2} + \sqrt{(L-x)^2 + b^2} \]
Cost Scaling
Apply cost coefficients to the geometric model to create the Objective Cost Function \( C(x) \).
\[ C(x) = k_1 d_1 + k_2 d_2 \]
Physical Limits
Account for boundary conditions where crossing at an angle might exceed legal wetland encroachment limits.
\[ 0 \leq x \leq L_{limit} \]
Deliverable Protocol
1
Mathematical Derivation
Full algebraic and calculus proof of the optimal crossing point \( x^* \).
2
Boundary Justification
Explanation of why your solution is the absolute minimum (2nd Derivative or Domain Endpoints).
3
Visual Blueprint
A scale drawing of the proposed route with all dimensions and angles clearly labeled.
4
Financial Summary
Comparison of the optimal cost against the straight-line "traditional" route cost.
Optimization is Mastery
"Engineering is the art of modeling materials we do not wholly understand, into shapes we cannot precisely analyze, so as to withstand forces we cannot properly assess, in such a way that the public has no reason to suspect the extent of our ignorance." — A.R. Dykes
Proceed to Project Documents
Pipeline Proposal Project Project Proposal
Sector 7G: Pipeline Infrastructure Optimization
Project Number
2026-CALC-SYN
The Challenge
A petroleum logistics firm needs to connect a coastal refinery (Point A) to an inland processing plant (Point B). Point A is at \( (0, 0) \). Point B is located 10 miles inland and 20 miles along the coast at \( (20, 10) \). The first 4 miles of the inland journey (from the coast) are through protected wetlands , while the remaining 6 miles are through stable plains .
Wetland Cost: $15 Million / Mile
Plains Cost: $8 Million / Mile
I
The Objective Function
Construct a function \( C(x) \) representing the total cost of the pipeline as a function of the coastal distance \( x \) where the pipeline leaves the shore and enters the wetlands.
II
Differentiation & Critical Point Analysis
Show all calculus steps to find the optimal crossing point. Use the 2nd Derivative Test or a sign chart to prove your result is a minimum.
III
Comparative Analysis
Route Type Calculation Path Total Projected Cost A. Coastal-First 20 miles along coast + 10 miles straight inland B. Straight-Line Direct hypotenuse from (0,0) to (20,10) C. Your Optimal Calculated crossing at point \( x^* \)
Executive Summary:
Based on your analysis, why should the board of directors approve your proposal over the "Straight-Line" approach? Quantify the savings and address any potential environmental risks of choosing a longer coastal path.
Optimization Rubric Teacher Resource Assessment Rubric
Lesson 5: Capstone Pipeline Project
Mastery Level 4.0 / Synthesis
Criteria Mastery (5 pts) Proficiency (3 pts) Developing (1 pt) Modeling Strategy Correctly identifies \( C(x) \) using Pythagorean distances for two distinct terrain types with appropriate cost coefficients. Creates a cost function but misses one coefficient or uses a simplified distance model. Fails to connect cost to distance or geometric variables correctly. Calculus Execution Flawless differentiation using Chain Rule. Correctly solves for \( x^* \) and verifies extrema using 2nd Derivative. Minor algebraic or differentiation errors. Solves for \( x^* \) but skips rigorous verification. Significant errors in differentiation or unable to solve for critical points. Comparative Analysis Calculates costs for all three routes. Demonstrates clear financial advantage of the optimal solution. Calculates optimal cost but fails to compare it effectively against the Straight-Line or Coastal routes. Incorrect financial calculations or no comparative reasoning provided. Executive Argument Synthesizes math and business logic. Addresses environmental/practical trade-offs of the chosen route. Reasoning is based purely on cost without considering physical or environmental context. Brief or superficial justification; lacks professional tone.
Project Solution Key (Internal)
Optimal Result
x* ≈ 14.15 miles
Distance from refinery before turning inland.
Financial Comparison
Straight-Line: ~$268.3M
Coastal-First: ~$430.0M
Optimal Path: ~$212.1M
Teaching Note: Encourage students to use Snell's Law intuition here. \( \frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2} \) translates here to \( \text{Cost}_1 \sin \theta_1 = \text{Cost}_2 \sin \theta_2 \) when minimizing cost.
