Blueprint Modeling Slides Unit: Optimal Design
Blueprint Modeling
Translating physical constraints into mathematical objective functions.
CALCULUS L1
The Cardboard Challenge
You have a single sheet of cardboard measuring 24cm by 24cm.
Goal:
Cut squares from the corners and fold up the sides to create the largest possible volume.
How do we turn this "physical" act into a math function?
24 x 24
"Let the cut length be x..."
The Anatomy of Optimization
Objective
The quantity you want to maximize or minimize (Volume, Area, Cost, Time).
Example: V(x) = lwh
Constraint
The limit or boundary. Often a fixed value that links your variables together.
Example: 2x + 2y = 100
Domain
The set of possible values. Determined by physical reality (you can't have negative length!).
Example: x > 0 and x < 12
The Strategy
Moving from multi-variable to single-variable.
STEP-BY-STEP
1
Sketch & Label: Draw the situation. Label unknown variables.
2
Identify Functions: Write an equation for the objective and an equation for the constraint.
3
Substitute: Use the constraint to solve for one variable and plug it into the objective function.
4
Determine Domain: What values of the variable make sense for the real world?
The "Real World" Check
The Math Domain
For a function like \( f(x) = \frac{1}{x-4} \), the domain is all real numbers except \( x = 4 \).
(-\infty, 4) \cup (4, \infty)
The Physical Domain
If \( x \) is the length of a box side, \( x \) must be greater than 0. If we are cutting \( x \) from a 24cm side, then \( 2x \) cannot exceed 24.
0 < x < 12
"Before we find the peak, we must know the edges of the map."
Blueprint Modeling Worksheet Blueprint Modeling
Lesson 1: Translating Reality to Calculus
Student Name:
Date:
Part 1: Defining the Goal
For each scenario below, identify the Objective Function (what needs to be maximized or minimized) and the Constraint (the limiting factor or fixed value).
Scenario A
A farmer wants to fence off a rectangular pasture alongside a straight river. No fence is needed along the river. He has exactly 1,200 meters of fencing available.
Objective
Constraint
Scenario B
An engineer must design a cylindrical soda can that holds exactly 355mL of liquid while using the least amount of aluminum possible.
Objective
Constraint
Part 2: Function Construction
Task 1: The Optimized Window
A window is being designed in the shape of a rectangle surmounted by a semicircle (a Norman Window). The total perimeter of the window is fixed at 16 feet.
1. Sketch & Label
Sketch the window and label variables (e.g., radius r, height h)
2. Write the Constraint Equation
3. Write the Objective Function (Area)
4. Solve for one variable and substitute to create \( A(r) \)
Part 3: The Physical Domain
Consider the Norman Window from Task 1. Mathematical functions can often take any value of \( x \), but physical models have boundaries.
Determine the Physical Domain for the radius \( r \):
Lower Bound: What is the smallest \( r \) can be? Why?
Upper Bound: What is the largest \( r \) can be? Why?
Express the final physical domain in interval notation:
Visualizing Extremes Slides Unit: Optimal Design
Visualizing Extremes
Finding the 'Sweet Spot' through Curve Analysis.
CALCULUS L2
Spotting the Peak
We have our function \( V(x) \).
Without doing any derivative work yet, what does the graph tell us about the "best" choice?
Optimal Point?
Absolute Extremas & Boundaries
The Interior Peak
Where the derivative is zero. These are often our primary targets in optimization.
The Boundaries
Values at the edges of our physical domain. Why might volume be 0 at the boundaries?
The EVT Rule
The Extreme Value Theorem guarantees that if a function is continuous on a closed interval \([a, b]\), it must have an absolute maximum and minimum.
"The search is restricted. The winner is either a peak or an endpoint."
Reading the Curve
Ascending
Slope is positive. As we increase our variable, the objective is still getting better.
f'(x) > 0
Turning
The "Critical Point." The slope flattens out. Momentum is zero.
f'(x) = 0
Descending
Slope is negative. We've passed the peak; more is now less.
f'(x) < 0
Inquiry Mode
Open Your Graphing Tool
Task: Search for the Summit
Input your objective function from Lesson 1.
Adjust your window to match your Physical Domain .
Click the "Peak" to reveal the coordinates.
"What happens to the volume if you cut just a tiny bit more than the optimal x?"
Teacher: Demonstrate Desmos/GeoGebra here
Curve Analysis Lab Activity Curve Analysis Lab
Lesson 2: Predicting Extremes through Visualization
Student Name:
Date:
Lab Objective
Use a graphing utility (Desmos or GeoGebra) to analyze your objective function within its physical domain. Identify trends, estimate the optimal point, and verify boundary conditions.
1. Setting the Window
Enter your function \( V(x) \) into the graphing tool. Adjust the window to match the Physical Domain you calculated in Lesson 1.
My Viewport Settings:
x-min: 0
x-max: ________
y-min: 0
y-max: ________
Sketch the curve as it appears in your window:
2. Trend & Endpoint Analysis
A. The "Sweet Spot" Estimation
Click the highest point on your graph. Record the coordinates and interpret them in the context of the problem.
Optimal Input (\( x \))
Maximized Output (\( V \))
What do these numbers physically represent for your design?
B. Endpoint Evaluation
Check the value of the function at the boundaries of your domain.
Why does the volume/area often go to zero at the boundaries?
Is it possible for the optimal solution to be an endpoint in a real-world project? Why/why not?
3. Slope Intuition
Pick a point on the graph to the left of your estimated peak and a point to the right .
Region Chosen x Visual Slope (Pos/Neg/Zero) Physical Interpretation Left of Peak Adding to x ________ the volume. Right of Peak Adding to x ________ the volume.
Summit Search Slides Unit: Optimal Design
Summit Search
Using Derivatives to Find Exact Certainty.
CALCULUS L3
Precision Matters
Last lesson, we found the "peak" by clicking a graph.
The Problem with Zoom:
Graphs are made of pixels. They are approximations. In engineering, a 0.01% error could mean a structural failure or millions in wasted material.
Calculus gives us the exact coordinate of the summit.
Visual Estimate
x ≈ 3.46
Analytical Exact
x = 2\sqrt{3}
The "Flat Line" Condition
At the exact moment a function switches from rising to falling...
\( f'(x) = 0 \)
Step 1
Differentiate
Step 2
Set to Zero
Step 3
Solve for \( x \)
Verification: Don't Assume!
1st Derivative Test
Analyze the sign of \( f'(x) \) on both sides of your critical point.
+ to - = Absolute Maximum
- to + = Absolute Minimum
2nd Derivative Test
Analyze the concavity at your critical point.
f''(c) < 0 = Concave Down (Max)
f''(c) > 0 = Concave Up (Min)
The Optimization Workflow
1
Identify \( f(x) \) and \( \text{Domain} \)
From Lesson 1: Build the single-variable model.
2
Find \( f'(x) \) and Critical Points
Set the derivative to zero and solve.
3
Verify & Compare
Test the critical points AND the endpoints (EVT).
Remember: You are designing a real product. Dimensions must be positive!
Analytical Verification Worksheet The Summit Search
Lesson 3: Analytical Verification of Extrema
Student Name:
Date:
Part 1: The Critical Point Hunt
The Project Function:
Use the function you built in Lesson 1 (e.g., the Norman Window or the Cardboard Box).
Objective Function \( f(x) \)
Physical Domain
1. Find the First Derivative \( f'(x) \)
Differentiate your function with respect to your variable.
2. Solve for Critical Points
Set \( f'(x) = 0 \) and solve for \( x \). List only the values within your physical domain.
Part 2: Verification Tests
Method A: First Derivative Test
Create a sign chart for \( f'(x) \) around your critical point \( c \).
x = c
Test point (left):
Test point (right):
Method B: Second Derivative Test
Find \( f''(x) \) and evaluate it at your critical point \( c \).
\( f''(x) = \)
Evaluate \( f''(c) \):
Part 3: The Optimized Result
Final Design Specs
Based on your calculus, what are the exact dimensions of your optimized design?
Calculate the Absolute Maximum value of your objective function.
Verification: Compare this analytical value with your graphical estimate from Lesson 2. How close were they?
Concavity Context Slides Unit: Optimal Design
Concavity Context
Analyzing the "Shape of Change" and Marginal Acceleration.
CALCULUS L4
Acceleration of Costs
Imagine you are managing a factory. Costs are rising.
Scenario:
Which is worse: costs increasing at a steady pace, or costs increasing at an accelerating pace?
One is a steady hill; the other is a rocket ship taking off.
Linear
f''(x) = 0
Accelerating
f''(x) > 0
The Marginal Meaning
Concave Up
f''(x) > 0
The "marginal" rate of change is increasing.
"Every extra unit we produce is becoming MORE expensive than the last one."
Concave Down
f''(x) < 0
The "marginal" rate of change is decreasing.
"Efficiency is kicking in. Every extra unit is adding less to the total cost than the previous one."
The Inflection Point
In business, this is the point of Diminishing Returns.
It's where you stop gaining efficiency and start losing it (or vice versa). It's where the 2nd derivative switches signs.
Calculus Rule:
f''(x) = 0
Inflection Point Concave Up (Increasing rate) Concave Down (Decreasing rate)
The Efficiency Defense
In your project presentation, you won't just say "This is the peak."
You must use concavity to explain if your design is stable, or if a small change in dimensions would cause an explosion in costs.
NEXT: DESIGN DEFENSE
Efficiency Case Study Worksheet Efficiency Case Study
Lesson 4: Marginal Analysis & Concavity
Student Name:
Date:
The Situation: Solar Panel Production
A manufacturing firm models its weekly production cost \( C(x) \) (in thousands of dollars) as a function of the number of panels produced \( x \). Through regression analysis, they've found the model:
\( C(x) = \frac{1}{3}x^3 - 6x^2 + 40x + 50 \)
1. Marginal Cost Analysis
A. Find the Marginal Cost function, \( C'(x) \).
B. Explain what \( C'(10) \) physically represents to the production manager.
2. Cost Acceleration
A. Find the second derivative, \( C''(x) \).
B. Find the Inflection Point of the cost curve.
Concave Down Interval:
In this region, efficiency is _________.
Concave Up Interval:
In this region, efficiency is _________.
3. Design Defense Prep
Apply this thinking to your own project design. Look at the concavity of your objective function at your optimal point.
Question:
If you increase your optimized dimension by just 1%, does the output (Volume/Area) drop off sharply or gradually? How does the 2nd derivative help you explain this "stability" to your client?
Design Defense Slides The Final Review
Design Defense
Proving Efficiency through Calculus.
Executive Session
The Boardroom Pitch
Today, you aren't just students. You are Optimization Consultants.
Your Mission:
Convince the "Board of Directors" (your peers and teacher) that your dimensions will save the company the most money while meeting all constraints.
"Numbers lie. Calculus speaks the truth."
The "Evidence Package"
Visual Blueprint: Sketch of the function & labeled dimensions.
The Model: Objective function and constraint equations.
The Calculus: 1st and 2nd derivative tests for verification.
Domain Validity: Proof that your solution is physically possible.
Pitch Structure
Introduction of the Problem
Visual Model Representation
Mathematical Derivation
Concavity & Stability Analysis
Q&A Defense Session
The "Review Board" Protocol
Validate
"The calculus you used to verify the maximum was clear because..."
Probe
"How did you account for the physical domain boundaries in your final answer?"
Challenge
"I noticed your second derivative test result was positive. Does that truly prove a maximum?"
Goal: Every design must be bulletproof before it leaves the boardroom.
Floor is Open
Who will present their Optimal Design first?
Start Presentations
Optimization Project Rubric Optimization Rubric
Board of Directors Assessment Suite
Lead Designer:
Assessor:
Criteria Exceptional (4) Proficient (3) Developing (2) Emerging (1) Mathematical Modeling Objective function and constraints are perfectly derived; variables are clearly defined with units. Models are mostly correct; minor errors in notation or variable definition. Fundamental error in the relationship between constraint and objective. Equations are incorrect or missing. Derivative Analysis Differentiation is flawless. Critical points are found analytically and checked against domain. Differentiation is correct; minor arithmetic error in solving for critical points. Conceptual error in differentiation or solving process. Derivatives are missing or used incorrectly. Verification Testing Rigorous 1st or 2nd derivative test confirms absolute extremum; endpoints are checked. Verification test is present and mostly correct; logic is clear. Extremum is identified but the test is flawed or incomplete. Assumption of extremum without analytical proof. Contextual Interpretation Deep insight into concavity (marginal stability) and inflection points; domain boundaries are defended. Correct interpretation of the optimal point in real-world terms. Surface-level explanation of what the numbers mean. No connection between math and the design problem.
Board Commendations
Highlight specific moments of technical clarity or design innovation.
Structural Concerns
Identify areas where the mathematical defense was vulnerable to critique.
Final Efficiency Score:
/ 16
Approved
Requires Revision
