Box Capacity Slides The Maximum Capacity Challenge
Lesson 1: Modeling Volume with Physical Constraints
The Hook
"Who can build the container with the largest possible volume using only a standard sheet of paper and tape?"
UNIT: EFFICIENT PACKAGING
SLIDE 01
The Design Constraints
The Material
One standard sheet of paper (8.5" x 11"). No adding extra material.
The Method
Cut four congruent squares out of the corners of the paper.
The Result
Fold up the flaps to create an open-top box. Secure corners with tape.
x
x
x
x
11"
8.5"
UNIT: EFFICIENT PACKAGING
SLIDE 02
The Big Question
"If we cut out a larger square (x), will the volume increase or decrease?"
Increase?
Because the walls get taller.
Decrease?
Because the base gets smaller.
Sweet Spot?
Is there a specific height that balances base area and height perfectly?
UNIT: EFFICIENT PACKAGING
SLIDE 03
Box Data Worksheet Physical Box Modeling
Data Collection & Observation Worksheet
Name: __________________________
Date: __________________________
The Objective
Using a sheet of 8.5" x 11" cardstock, your team will test different "cut sizes" (x) to determine how the side height affects the total volume of the box.
DATA COLLECTION
Cut Size (x inches) Length of Base (inches) Width of Base (inches) Height of Box (inches) Volume (cubic inches) 0.5 1.0 1.5 2.0 2.5 3.0
1. The Trend
As the cut size (x) increased, what happened to the volume? Did it continuously go up, go down, or change direction?
2. Physical Limits
What is the largest possible square you could theoretically cut out? Why can't you go any larger? (Hint: Look at the 8.5" side of the paper).
3. The Estimation
Based on your table, what is your team's estimate for the "optimal" cut size? What is the maximum volume you achieved?
Packaging Optimization Sequence / Lesson 01 / Worksheet
Modeling Teacher Guide Teacher Guide
Lesson 1: Volume Modeling Activity
Learning Objective
Students will physically model the relationship between a variable (cut size x) and an output (Volume). The goal is to discover that volume does not change linearly and that an "optimal" point exists between the physical extremes.
Materials Needed
8.5" x 11" Paper/Cardstock
Scissors
Transparent Tape
Rulers (Decimal or Fractional)
Instructional Pacing
05m
The Hook & Launch
Present the "Maximum Capacity" prompt. Don't give instructions yet; let them look at the paper and guess.
10m
Method Demo
Show how to cut out corner squares of size x and fold the flaps. Explain that every group will test different x values.
20m
Construction & Data
Students build boxes and calculate volumes. They should notice that as height increases, the base area drops significantly.
15m
Discussion
Compare results on the board. Identify the "peak" in the data. Set the stage for Lesson 2 (Algebraic Modeling).
TEACHER TIP: Physical Domain
Watch for students trying to cut x = 4.25". Since the width of the paper is 8.5", cutting two squares of 4.25" would leave zero width for the base. This is a crucial "aha" moment for identifying the physical domain of our future function: \(0 < x < 4.25\).
Algebraic Modeling Slides The Power of "x"
Lesson 2: From Physical Models to Algebraic Functions
Yesterday we guessed. Today we generalize.
How can we write a single equation that describes every possible box we could build?
Efficiency Sequence
SLIDE 01
The Blueprint Transition
The Width (8.5")
We start with 8.5". We cut x from both sides.
New Width: \(8.5 - 2x\)
The Length (11")
We start with 11". We cut x from both sides.
New Length: \(11 - 2x\)
The Height
When we fold up the flap, the height is exactly...
Height: \(x\)
x
x
x
x
BASE AREA
Fold here
11" sheet
8.5" sheet
Efficiency Sequence
SLIDE 02
The Volume Function
Volume = Length × Width × Height
\(V(x) = x(11 - 2x)(8.5 - 2x)\)
Domain Analysis
For the width to be positive:
Physical Domain: \(0 < x < 4.25\)
Graphing View
This is a cubic function. On our graphing calculators, we'll see a peak within our domain.
Efficiency Sequence
SLIDE 03
Algebraic Modeling Worksheet The Algebra of Form
Lesson 2: Objective Function Derivation
Manufacturing Optimization Unit
Name: __________________________
1
Define Your Variables
We are cutting four congruent squares out of the corners of an 8.5" x 11" piece of cardstock.
Let x = the side length of the square cut-out (inches).
2
Express Dimensions in Terms of x
Height of Box
Length of Base
Width of Base
3
Formulate the Volume Function
Combine your expressions using the formula: Volume = (Length)(Width)(Height)
V(x) =
4
Identify the Realistic Domain
Mathematically, x could be any number. Physically, our box must have positive dimensions. Solve the following inequalities to find the domain of your function:
Length constraint: \(11 - 2x > 0\)
Width constraint: \(8.5 - 2x > 0\)
Conclusion:
The physical domain is: 0 < x < ___________
5
The Polynomial Expansion
Multiply out your volume function into a polynomial in standard form. (You will need this for the calculus tomorrow!)
Packaging Optimization Sequence / Lesson 02 / Worksheet
Algebraic Modeling Answer Key The Algebra of Form
Answer Key & Teacher Notes
Manufacturing Optimization Unit
For Teacher Use Only
1. The Dimensions
Height (h)
x
Length (l)
11 - 2x
Width (w)
8.5 - 2x
2. The Function
Factored Form:
V(x) = x(11 - 2x)(8.5 - 2x)
Expanded Polynomial Form:
V(x) = 4x³ - 39x² + 93.5x
Note: (11)(8.5) = 93.5. Intermediate: x(93.5 - 22x - 17x + 4x²)
3. The Domain Analysis
Length must be > 0:
11 - 2x > 0
x < 5.5
Width must be > 0:
8.5 - 2x > 0
x < 4.25
Final Domain
0 < x < 4.25
Teacher Insight: The Cubic Nature
This function has three zeros: at x = 0, x = 4.25, and x = 5.5. The local maximum that students are looking for will occur within the domain (0, 4.25). There is another local minimum outside this domain, but it is physically irrelevant.
Packaging Optimization Sequence / Lesson 02 / Key
Optimization Calculus Slides Precision Optimization
Lesson 3: Finding the Exact Maximum
"Data gives us an estimate. Algebra gives us a model.
How do we find the absolute highest point on a curve without graphing?
Efficiency Sequence
SLIDE 01
The Slope of a Peak
At the top of the hill...
The curve is neither going up nor going down. For a split second, it is flat.
The Calculus Definition
A "flat" tangent line has a slope of zero.
V'(x) = 0
MAXIMUM Slope = 0
Efficiency Sequence
SLIDE 02
The Optimization Algorithm
1
Differentiate
Use the Power Rule to find the derivative function, V'(x). This represents the rate of change.
2
Solve for Critical Points
Set V'(x) = 0. Since our derivative is a quadratic, use the Quadratic Formula.
3
Verify Constraints
Check your answers against the physical domain (\(0 < x < 4.25\)). Throw out any impossible answers!
Efficiency Sequence
SLIDE 03
The Calculus Solution Worksheet The Exact Solution
Lesson 3: Finding Critical Points
Calculus Optimization Module
Name: __________________________
STEP 1
The Objective Function
Write your expanded volume function from Lesson 2:
V(x) = 4x³ - 39x² + 93.5x
STEP 2
Apply the Power Rule
Find the derivative, V'(x), by differentiating each term of the polynomial above.
V'(x) =
STEP 3
Set to Zero & Solve
Set your derivative equal to zero. Use the Quadratic Formula to find the values of x:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Show Your Work:
Calculated x-values:
\(x_1 = \) __________________
\(x_2 = \) __________________
STEP 4
The Reality Check
Physical Domain Verification
Recall your physical domain from yesterday: \(0 < x < 4.25\). Which of your critical points is valid?
Optimal Cut Size (x) = ____________ inches
Final Dimensions & Volume
HEIGHT
LENGTH
WIDTH
MAX VOLUME
THE BIG REVEAL
Compare this calculus result to your activity data from Lesson 1. How close was your "guess"? Why do you think calculus is preferred for industrial manufacturing?
Packaging Optimization Sequence / Lesson 03 / Worksheet
The Calculus Solution Key The Exact Solution
Answer Key & Solution Path
Calculus Optimization Module
Teacher Reference
1. The Polynomial
V(x) = 4x³ - 39x² + 93.5x
2. The Derivative
V'(x) = 12x² - 78x + 93.5
Quadratic Solution Path
Values:
a = 12, b = -78, c = 93.5
Discriminant:
b² - 4ac = (-78)² - 4(12)(93.5)
6084 - 4488 = 1596
Solve for x:
x = (78 ± √1596) / 24
x₁ ≈ 4.915 (OUT OF DOMAIN)
x₂ ≈ 1.585 (OPTIMAL)
Optimal Design Specifications
CUT SIZE (x)
1.59 in
BASE LENGTH
7.82 in
BASE WIDTH
5.32 in
MAX VOLUME
66.15 in³
Teaching Key Point
Remind students that x ≈ 4.92 is technically a critical point, but it represents the local minimum of the cubic function if the paper were infinite. Within our physical constraints, any x > 4.25 is impossible because the paper width is only 8.5".
Packaging Optimization Sequence / Lesson 03 / Key
Cost Minimization Slides The Bottom Line
Lesson 4: Minimizing Material Costs
"Volume is what the customer wants.
How do we hold exactly 355ml of soda using the least amount of aluminum?
Efficiency Sequence
SLIDE 01
Constraints & Objectives
The Constraint
What must be true?
V = \(\pi r^2 h = 355\)
The Objective
What are we trying to minimize?
SA = \(2\pi r^2 + 2\pi r h\)
355ml
Fixed Volume
Radius (r) = ?
Height (h) = ?
Efficiency Sequence
SLIDE 02
The Substitution Play
Problem: We can't differentiate the Surface Area because it has two variables (r and h).
1. Solve Constraint for h
h = \(\frac{355}{\pi r^2}\)
2. Substitute into SA
SA(r) = \(2\pi r^2 + 2\pi r (\frac{355}{\pi r^2})\)
Ready for Calculus!
Now we have a function of only one variable. Just simplify, differentiate, and solve.
Efficiency Sequence
SLIDE 03
The 355ml Challenge Worksheet The 355ml Challenge
Lesson 4: Surface Area Minimization
Manufacturing Optimization
Name: __________________________
PROJECT BRIEF: Aluminum Savings
A standard soda can holds 355ml (355 cm³). Aluminum costs money. Your task is to calculate the ideal dimensions (radius and height) that use the absolute minimum surface area to hold that volume.
Constraint Equation
The Volume must be exactly 355.
V = \(\pi r^2 h = 355\)
Solve for h:
Objective Function
The Surface Area we want to minimize.
SA = \(2\pi r^2 + 2\pi r h\)
Substitute h into SA:
Simplified Objective Function
After substituting and simplifying, your function should look like: \(SA(r) = 2\pi r^2 + \frac{710}{r}\)
Apply the Calculus:
1. Differentiate SA(r)
2. Solve for Critical Radius (r)
Final Manufacturing Specs
Optimal Radius
centimeters
Optimal Height
centimeters
Minimum Surface Area
sq. centimeters
INDUSTRY ANALYSIS
Go find a real soda can and measure it. Are the dimensions close to your "mathematical ideal"? If they are different, why might a company choose a tall, thin can even if it uses slightly more aluminum?
Packaging Optimization Sequence / Lesson 04 / Activity
Substitution Optimization Guide Substitution Mastery
Teacher Resource: Optimization Guide
Manufacturing Optimization Unit
Instructional Strategies
The Substitution Logic
The biggest hurdle for students in Lesson 4 is the transition from two variables to one. Emphasize that the Constraint is our "rulebook" and the Objective is our "goal."
The Substitution Step
Substitute \(h = \frac{355}{\pi r^2}\) into \(SA = 2\pi r^2 + 2\pi r h\)
SA = \(2\pi r^2 + 2\pi r (\frac{355}{\pi r^2})\)
SA = \(2\pi r^2 + \frac{710}{r}\)
The Derivative Step
Differentiate with respect to r:
SA'(r) = \(4\pi r - \frac{710}{r^2}\)
Setting \(SA'(r) = 0\):
\(4\pi r^3 = 710 \Rightarrow r = \sqrt[3]{\frac{710}{4\pi}}\)
Ideal Target Values
Optimal Radius
~ 3.84 cm
Optimal Height
~ 7.67 cm
Height/Radius Ratio
h = 2r
Insight: In an ideal scenario, the height should equal the diameter (\(h = 2r\)) for a cylinder with a closed top and bottom.
Discussion Guide: Math vs. Reality
1
Human Factors: Why are soda cans taller and thinner than our optimal result? (Answer: Easier to grip with a human hand, fits in car cup holders).
2
Retail Constraints: How does shelf space in a grocery store affect packaging design? (Answer: Thinner cans allow more units to be displayed in a narrow facing).
3
Branding: Does the shape of the container change the customer's perception of value? (Answer: Tall containers are often perceived as "more" even if the volume is the same).
Packaging Optimization Sequence / Lesson 04 / Resource
The Design Defense Slides The Design Defense
Lesson 5: Pitching Your Optimization
"Numbers don't lie, but they do need
How do you convince a stakeholder that your calculus is the key to their profit?
Efficiency Sequence
SLIDE 01
The Three Pillars of Your Defense
The Math
Show your objective function and your derivative . Prove where the zero occurs and why it is the absolute maximum/minimum.
The Model
Present your physical dimensions . How does it look? Is it functional for the consumer as well as the manufacturer?
The Justification
Address the trade-offs . If your design isn't the "mathematical perfect," explain why (grip, shelf space, aesthetics).
Efficiency Sequence
SLIDE 02
Preparing for Cross-Examination
Expect these questions:
"What happens to the cost if the volume requirement increases by 10%?"
"How did you use the first derivative test to guarantee this wasn't a minimum?"
"Can a human hand comfortably hold this 'ideal' container?"
"How much material are we wasting per 1,000 units?"
Efficiency Sequence
SLIDE 03
Design Defense Organizer Design Defense Organizer
Lesson 5: Pitch Preparation
Packaging Optimization Sequence
Name: __________________________
Product Selected
Open-Top Box
Cylindrical Can
Other: __________
I. The Calculus Argument
The Objective Function & Derivative
Show the function and its derivative here...
The Critical Point Verification
Show the math proving this is a Max/Min...
II. Final Specifications
Dimension Value (with Units) Economic Significance Length/Radius Minimal footprint? Width/Height Stacking efficiency? Material Surface Area Direct material cost reduction.
III. The "Real World" Defense
Trade-off Analysis: Form vs. Function
If you chose dimensions slightly different from the calculus-ideal, why? Consider grip, retail shelf space, or consumer psychology.
The Economic Impact
Explain how much money/material this design saves compared to a "guessed" or "standard" design.
Packaging Optimization Sequence / Lesson 05 / Presentation Aid
Design Defense Rubric Presentation Rubric
Optimized Design Defense
Manufacturing Optimization Unit
Final Assessment
Criteria Exceeds (4) Meets (3) Developing (2-1) Calculus Rigor Derivative is perfect; Critical point solving is clear and includes 1st Derivative Test verification. Derivative is correct; Critical points found correctly but missing formal verification. Errors in differentiation or algebra prevent finding the correct optimal point. Modeling Precision Dimensions are calculated exactly; units are included; constraints (domain) are fully addressed. Dimensions are mostly correct; units included; physical constraints mentioned. Dimensions are unrealistic or missing; units are omitted. Applied Rationale Deep analysis of trade-offs (hand grip, shelf space, material cost vs. labor). Identifies basic trade-offs between math-ideal and real-world use. Fails to connect math to real-world manufacturing or consumer needs. Communication Professional pitch; clear visuals; handle complex "CFO" questions with confidence. Clear presentation; mostly readable visuals; answers some follow-up questions. Presentation is disorganized; visuals are cluttered; cannot justify numbers.
STAKEHOLDER FEEDBACK (TEACHER NOTES)
Strengths
Opportunities for Revision
Final Score: _______ / 16
Packaging Optimization Sequence / Lesson 05 / Assessment
