Candle Problem: Step Three — Set Up
Volume links
Once
SA becomes a function of
Candle Problem: Step Four — Explore
SA = 2πr² + 2πrh — substitute constraint for
| SA (cm²) | Fits box? | ||
|---|---|---|---|
| 3 | 8.84 | 222 | ✓ |
| 4 | 4.97 | 226 | ✓ |
| 5 | 3.18 | 257 | ✓ |
| 6 | 2.21 | 309 | ✗ |
Minimum SA near
Candle Problem: Step Five — Interpret
Winner:
Verify:
- Diameter 6 cm < 12 cm ✓
- Height 8.84 cm < 12 cm ✓
- Volume
cm³ ✓
Math gives a solution. Interpret checks it's a good one.
Check-In: Setting Up a Design Problem
A storage shed must:
- Hold at least 10 m³
- Fit through a gate that is 3 m wide
- Use minimum material (minimize SA)
Identify:
a) Two constraints (as inequalities)
b) The objective (as a phrase or formula)
Think before advancing.
Answer: Constraints and Objective for Shed
Constraints (Step 2):
| Requirement | Math |
|---|---|
| Volume at least 10 m³ | |
| Fits through 3 m gate |
Objective (Step 3):
Minimize surface area: SA = 2ℓw + 2ℓh + 2wh
Many valid constraint interpretations exist —
depth, height limits depend on what the design specifies.
From Any Design to the Best Design
The modeling cycle finds valid designs.
Optimization finds the best design within constraints.
Key question: among all cylinders with volume 400 cm³,
which dimensions give the minimum surface area?
→ This is the classic can optimization problem
Volume Constraint Reduces Two Variables to One
For a cylinder with fixed volume V = 400 cm³:
- Choosing
determines entirely - Two-variable problem collapses to one variable
This is why optimization is possible:
once you pick
Setting Up the Can Optimization Problem
Volume
Building the Can Optimization Table
| SA (cm²) | Cost (cents) | ||
|---|---|---|---|
| 2 | 31.83 | 425 | 127 |
| 3 | 14.15 | 323 | 97 |
| 4 | 7.96 | 301 | 90 |
| 5 | 5.09 | 317 | 95 |
| 6 | 3.54 | 360 | 108 |
Minimum surface area near
The SA vs r Curve Shows the Minimum
Minimum SA occurs at
Optimal Cylinder Has Height Equal to Diameter
At the minimum,
| Quantity | Value |
|---|---|
| Optimal radius | |
| Optimal height | |
| Height/diameter ratio |
This holds for any fixed volume — the optimal ratio is a geometric property, not specific to 400 cm³.
Optimization Check-In: Working at r = 5
A can must hold 1000 cm³ of paint.
At
a) What is
b) What is SA?
Set up and compute before advancing.
Answer: Can Dimensions at r = 5 cm
Given:
Box Optimization: Setting Up the Equations
Open-top square-base box,
Box Optimization: Table Finds the Minimum
| SA (cm²) | ||
|---|---|---|
| 5 | 40.0 | 825 |
| 10 | 10.0 | 500 |
| 12 | 6.9 | 478 |
| 15 | 4.4 | 492 |
Minimum near
Why Real Cans Differ from Optimal Cans
The mathematical optimum gives
- Shelf: tall cans fit more units per row
- Label: more height = more branding space
- Stacking: certain proportions are more stable
The model doesn't include every constraint — always interpret.
Key Takeaways from Deck One
✓ Modeling cycle: Define → Constrain → Set Up → Explore → Interpret
✓ Volume constraint → one variable; optimal cylinder
Watch out:
- Constraint (fixed) vs. objective (minimize) — always label both
- Always interpret: does the solution make physical sense?
Preview: Constraint Design and Grid Systems
Deck 2 builds on today's modeling cycle:
- Chunk 3: Constraint-based design — shipping boxes, water tanks, packaging efficiency
- Chunk 4: Geometry in 2D design — typographic grids, golden ratio, page proportions
Design shapes not only containers but every page and screen.
Click to begin the narrated lesson
Apply geometry to design problems