Optimization Protocol Slides Finding the Best
Establishing the Optimization Protocol
The Number Challenge
"Find two numbers whose sum is 20 and whose product is as large as possible."
Trial & Error
10 + 10 = 20 → 100
11 + 9 = 20 → 99
15 + 5 = 20 → 75
The Calculus Way
How do we prove that 10 and 10 is the absolute maximum for all real numbers?
The Optimization Protocol
01
Assign & Sketch
Define variables and draw a diagram of the situation.
02
Primary Equation (Objective)
The quantity to be maximized or minimized (e.g., \(P = xy\)).
03
Secondary Equation (Constraint)
A relationship between variables to reduce the function to one variable (e.g., \(x + y = 20\)).
The Protocol (Cont.)
04
Solve & Substitute
Solve the secondary for one variable and substitute into the primary.
05
Find the Extrema
Determine feasible domain, find the derivative, set to zero, and verify using tests.
Don't forget to check your domain endpoints!
Modeling Example
Find two numbers whose sum is 20 and whose product is maximized.
1. Equations
Primary: \(P = xy\)
Secondary: \(x + y = 20\)
2. Single Variable
\(y = 20 - x\)
\(P(x) = x(20 - x) = 20x - x^2\)
3. Calculus
\(P'(x) = 20 - 2x\)
Set \(P'(x) = 0 \Rightarrow x = 10\)
4. Justification
\(P''(x) = -2\)
Since second derivative is negative, \(x=10\) is a maximum by SDT.
Takeaway
The hardest part of optimization isn't the calculus—it's the translation.
English
Calculus
Optimization Protocol Worksheet Optimization Protocol
Calculus AB/BC | Lesson 1.1
NAME:
DATE:
THE PROTOCOL
1. Sketch & Var
2. Primary Eq
3. Secondary Eq
4. Solve & Sub
5. Diff & Test
1
The Sum & Product
Find two positive numbers whose sum is 100 and whose product is a maximum.
Step 1-3: Identify Equations
Step 4-5: Calculus & Justification
2
The Minimal Perimeter
A rectangular garden is to have an area of \(100 \text{ m}^2\). What should its dimensions be so that the amount of fencing needed is minimized?
Work Space
3
Maximum Difference
Find a positive number \(x\) such that the sum of the number and its reciprocal is as small as possible.
Work Space
4
Enclosing a Field
A rancher has 1200 feet of fencing and wants to enclose a rectangular field that borders a straight river. He needs no fencing along the river. What are the dimensions of the field that has the largest area?
Detailed Solution Process
The Concept Check
In Problem #4, if the rancher decided to split the field into two smaller identical rectangles by adding a fence perpendicular to the river, how would your "Secondary Equation" change?
Optimization Protocol Answer Key Answer Key
Optimization Protocol | Lesson 1.1
TEACHER USE ONLY
1. The Sum & Product
Primary: \(P = xy\)
Secondary: \(x + y = 100 \Rightarrow y = 100 - x\)
Combined: \(P(x) = x(100 - x) = 100x - x^2\)
Derivative: \(P'(x) = 100 - 2x = 0 \Rightarrow x = 50\)
Answer: The numbers are 50 and 50 . Product = 2500.
2. The Minimal Perimeter
Primary (Perimeter): \(P = 2x + 2y\)
Secondary (Area): \(xy = 100 \Rightarrow y = 100/x\)
Combined: \(P(x) = 2x + 200/x\)
Derivative: \(P'(x) = 2 - 200/x^2 = 0 \Rightarrow 2x^2 = 200 \Rightarrow x = 10\)
Answer: Dimensions are 10 m by 10 m (a square).
3. Reciprocal Sum
Function: \(f(x) = x + \frac{1}{x}\)
Derivative: \(f'(x) = 1 - \frac{1}{x^2} = 0 \Rightarrow x^2 = 1 \Rightarrow x = 1\) (since positive).
Justify: \(f''(x) = 2/x^3\); \(f''(1) = 2 > 0\) (Concave up \(\Rightarrow\) Minimum).
Answer: The number is 1 .
4. River Field
Primary (Area): \(A = xy\) (where \(x\) is side perp to river, \(y\) is parallel)
Secondary (Fence): \(2x + y = 1200 \Rightarrow y = 1200 - 2x\)
Combined: \(A(x) = x(1200 - 2x) = 1200x - 2x^2\)
Derivative: \(A'(x) = 1200 - 4x = 0 \Rightarrow x = 300\)
Dimensions: \(x = 300 \text{ ft}\), \(y = 1200 - 2(300) = 600 \text{ ft}\).
Answer: 300 ft by 600 ft .
Exit Ticket Answer
The secondary equation would become \(3x + y = 1200\) because there is an additional length of fencing \(x\) dividing the field.
Geometric Maximization Slides Folding for Space
Geometric Maximization: Area & Volume
The Open-Top Box
You have a standard 8.5 x 11 inch sheet of paper. You must cut squares of side length \(x\) from each corner and fold up the flaps.
What value of \(x\) produces the box with the maximum volume?
8.5" x 11"
Cut: x
Step 1 & 2: Translate
New Dimensions
Height = \(x\)
Length = \(11 - 2x\)
Width = \(8.5 - 2x\)
Objective Function
\(V(x) = x(11 - 2x)(8.5 - 2x)\)
Domain Constraints
\(x > 0\) (Must make a cut)
\(8.5 - 2x > 0 \Rightarrow x < 4.25\)
Domain: \([0, 4.25]\)
Step 3: Calculus
\(V(x) = 4x^3 - 39x^2 + 93.5x\)
1st Derivative
\(V'(x) = 12x^2 - 78x + 93.5\)
Set to 0 and use Quadratic Formula...
Critical Numbers
\(x \approx 1.58\) or \(4.91\)
4.91 is outside the domain!
Proving the Maximum
The Sign Chart
\(V'(x)\): +
-
1.58
Because \(V'\) changes from positive to negative, \(x=1.58\) is a local max.
Global Check
\(V(0) = 0\)
\(V(4.25) = 0\)
\(V(1.58) \approx 66.1 \text{ in}^3\)
Geometric Maximization Worksheet Box Architect Challenge
Geometric Maximization | Lesson 1.2
NAME:
DATE:
1. Optimized Canvas
A graphic designer is creating a poster with an area of \(180 \text{ in}^2\). The top and bottom margins are each \(2\) inches, and the side margins are each \(1\) inch. Find the overall dimensions of the poster that maximize the area of the printed region .
x
y
Step-by-Step Modeling
2. Inscribed Capacity
A right circular cylinder is inscribed in a sphere of radius \(r = 3\). Find the dimensions of the cylinder (height \(h\) and radius \(R\)) that maximize its volume. (Hint: Use the Pythagorean theorem as your secondary equation relating \(h/2\) and \(R\)).
Mathematical Justification Area
3. Norman Window
A Norman window is a rectangle surmounted by a semicircle. If the perimeter of the window is \(16\) ft, find the dimensions that allow the maximum amount of light to enter.
Optimization Protocol Workspace
Structural Reflection
In Problem #2, why is it often easier to maximize the square of a distance function rather than the distance function itself? Does this apply to volume too?
Geometric Maximization Answer Key Answer Key
Geometric Maximization | Lesson 1.2
TEACHER USE ONLY
1. Optimized Canvas
Primary (Printed Area): \(A = (x-2)(y-4)\)
Secondary (Total Area): \(xy = 180 \Rightarrow y = 180/x\)
Combined: \(A(x) = (x-2)(180/x - 4) = 180 - 4x - 360/x + 8\)
Derivative: \(A'(x) = -4 + 360/x^2 = 0 \Rightarrow 4x^2 = 360 \Rightarrow x^2 = 90 \Rightarrow x = 3\sqrt{10} \approx 9.49\)
Answer: Dimensions are approximately 9.49 in by 18.97 in .
2. Inscribed Capacity
Primary (Volume): \(V = \pi R^2 h\)
Secondary: \(R^2 + (h/2)^2 = 3^2 \Rightarrow R^2 = 9 - h^2/4\)
Combined: \(V(h) = \pi(9 - h^2/4)h = 9\pi h - \frac{\pi}{4}h^3\)
Derivative: \(V'(h) = 9\pi - \frac{3\pi}{4}h^2 = 0 \Rightarrow 3h^2/4 = 9 \Rightarrow h^2 = 12 \Rightarrow h = 2\sqrt{3}\)
Answer: \(h = 2\sqrt{3} \approx 3.46\) and \(R = \sqrt{6} \approx 2.45\).
3. Norman Window
Variables: \(x = \text{width}\), \(y = \text{height of rectangle section}\). Radius \(r = x/2\).
Primary (Area): \(A = xy + \frac{1}{2}\pi (x/2)^2\)
Secondary (Perimeter): \(x + 2y + \pi(x/2) = 16 \Rightarrow y = 8 - \frac{x}{2} - \frac{\pi x}{4}\)
Combined: \(A(x) = x(8 - \frac{x}{2} - \frac{\pi x}{4}) + \frac{\pi x^2}{8} = 8x - \frac{x^2}{2} - \frac{\pi x^2}{8}\)
Derivative: \(A'(x) = 8 - x - \frac{\pi x}{4} = 0 \Rightarrow x(1 + \pi/4) = 8 \Rightarrow x \approx 4.48 \text{ ft}\).
Answer: Width \(\approx\) 4.48 ft , Height \(\approx\) 2.24 ft (Wait, \(y = x/2\)).
Minimizing Cost Slides The Bottom Line
Minimizing Cost and Materials
Geometry $ Economics
Old Problem
Maximize Area for a fixed Perimeter.
Maximize Output
New Problem
Minimize Cost for a fixed Output.
Minimize Input
Asymmetric Costs
What if the front of a store costs $20/ft for fancy glass, but the back and sides only cost $10/ft for brick?
Cost = 20(x) + 10(x) + 10(y) + 10(y)
Store Interior
$20/ft (Glass)
$10/ft (Brick)
$10/ft
Cross-Terrain Optimization
A pipeline must be laid from an oil well (A) to a refinery (B).
Underwater: $1,000,000 / mile
On Land: $400,000 / mile
The "Best" path isn't the shortest distance—it's the shortest dollar amount.
Water
Land
Snell's Law applies to cost too!
Minimizing Cost Worksheet Cost-Effective Engineering
Economic Minimization | Lesson 1.3
NAME:
DATE:
1. Commercial Curb Appeal
A business owner wants to enclose a rectangular area of \(2,400 \text{ ft}^2\) for a parking lot. The fencing along the front of the lot costs $12 per foot , while the fencing for the other three sides costs $8 per foot . What are the dimensions of the lot that minimize the total cost?
Equations (Primary & Secondary)
Differentiation & Solver
2. Reservoir Lining
A square-based open tank is to be constructed to hold \(32 \text{ m}^3\) of water. The material for the base costs $10 per square meter , and the material for the sides costs $5 per square meter . Find the dimensions that minimize the cost of the tank.
Full Optimization Workspace
3. The Trans-Valley Pipeline
A pipeline is to be run from a source \(S\) to a refinery \(R\), which is \(10\) miles downriver and \(4\) miles across the river from \(S\). The cost of laying pipe in the river is $130,000 per mile , and the cost on land is $50,000 per mile . At what point \(P\) on the opposite bank should the pipe emerge from the river to minimize the cost?
Calculus & Modeling Process
Cost Efficiency Strategy
If the "land cost" and "river cost" were exactly the same, what would be the optimal path for the pipeline? Use geometry to justify your answer without doing calculus.
Minimizing Cost Answer Key Answer Key
Economic Minimization | Lesson 1.3
TEACHER USE ONLY
1. Commercial Curb Appeal
Primary (Cost): \(C = 12x + 8x + 8y + 8y = 20x + 16y\)
Secondary (Area): \(xy = 2400 \Rightarrow y = 2400/x\)
Combined: \(C(x) = 20x + 16(2400/x) = 20x + 38400/x\)
Derivative: \(C'(x) = 20 - 38400/x^2 = 0 \Rightarrow 20x^2 = 38400 \Rightarrow x^2 = 1920 \Rightarrow x \approx 43.82\)
Dimensions: Frontage \(x \approx 43.82 \text{ ft}\), Side \(y \approx 54.77 \text{ ft}\).
2. Reservoir Lining
Primary (Cost): \(C = 10x^2 + 5(4xh) = 10x^2 + 20xh\)
Secondary (Volume): \(x^2h = 32 \Rightarrow h = 32/x^2\)
Combined: \(C(x) = 10x^2 + 20x(32/x^2) = 10x^2 + 640/x\)
Derivative: \(C'(x) = 20x - 640/x^2 = 0 \Rightarrow 20x^3 = 640 \Rightarrow x^3 = 32 \Rightarrow x \approx 3.17\)
Dimensions: Base \(3.17 \text{ m} \times 3.17 \text{ m}\), Height \(h = 32/(3.17)^2 \approx 3.18 \text{ m}\).
3. Trans-Valley Pipeline
Cost Function: \(C(x) = 130,000\sqrt{x^2 + 4^2} + 50,000(10 - x)\)
Derivative: \(C'(x) = 130,000 \cdot \frac{x}{\sqrt{x^2+16}} - 50,000 = 0\)
Solve: \(13x / \sqrt{x^2+16} = 5 \Rightarrow 169x^2 = 25(x^2 + 16) \Rightarrow 144x^2 = 400 \Rightarrow x = 20/12 = 5/3 \approx 1.67\)
Answer: Emerge 1.67 miles downriver from the source point.
Distance Motion Slides Path of Least Time
Distance and Motion Efficiency
The Lifeguard Problem
A lifeguard sees a swimmer in distress. To reach them as quickly as possible, should they:
Run in a straight line to the swimmer?
Run to the point on the shore closest to the swimmer, then swim?
Find a balance?
Crucial Factor: You can run much faster than you can swim.
Sand (Fast)
Water (Slow)
Time as the Objective
\( \text{Time} = \frac{\text{Distance}_1}{\text{Velocity}_1} + \frac{\text{Distance}_2}{\text{Velocity}_2} \)
Step 1: Distance Formula
Use Pythagorean Theorem to define diagonal paths in terms of a single variable \(x\).
\( d = \sqrt{a^2 + x^2} \)
Step 2: Differentiate
The derivative will involve the chain rule and square roots.
\( T'(x) = \frac{1}{v_1} \cdot \frac{x}{\sqrt{a^2+x^2}} - \dots \)
Physical Intuition
Light follows this same optimization principle. It takes the path of least time, not shortest distance. This is why light refracts (bends) when entering water.
Fermat's Principle
Refraction
Distance Motion Worksheet Velocity & Vectoring
Efficiency Optimization | Lesson 1.4
NAME:
DATE:
1. Proximity Optimization
Find the point on the parabola \(y = x^2\) that is closest to the point \((3, 0)\). (Pro-tip: Minimizing the square of the distance \(d^2\) is easier than minimizing \(d\)).
Distance Formula Workspace
2. Rescue Response
A lifeguard is at point \(L\) on a sandy beach and sees a swimmer at point \(S\) in the water. \(L\) is \(40\) meters from the shore, and \(S\) is \(30\) meters from the shore. The horizontal distance between the points is \(100\) meters. The lifeguard can run at \(8\) m/s on the sand and swim at \(2\) m/s. At what point should they enter the water to reach the swimmer in the minimum time?
Detailed Modeling Area
3. The Shortest Ladder
A fence \(8\) feet tall runs parallel to a tall building and is \(4\) feet from it. Find the length of the shortest ladder that will reach from the ground over the fence to the wall of the building.
Geometry & Trigonometric Optimization
The Physics Link
In Problem #2, calculate the ratio of the sine of the angle of "incidence" (on sand) to the sine of the angle of "refraction" (in water). How does this relate to the velocities?
Distance Motion Answer Key Answer Key
Efficiency Optimization | Lesson 1.4
TEACHER USE ONLY
1. Proximity Optimization
Primary: \(D^2 = (x-3)^2 + (y-0)^2 = (x-3)^2 + (x^2)^2 = x^4 + x^2 - 6x + 9\)
Derivative: \(f'(x) = 4x^3 + 2x - 6\)
Find Root: \(x=1\) works! (\(4(1)^3 + 2(1) - 6 = 0\)).
Answer: The point is (1, 1) .
2. Rescue Response
Time Function: \(T(x) = \frac{\sqrt{x^2 + 40^2}}{8} + \frac{\sqrt{(100-x)^2 + 30^2}}{2}\)
Derivative: \(T'(x) = \frac{x}{8\sqrt{x^2+1600}} - \frac{100-x}{2\sqrt{(100-x)^2+900}} = 0\)
Conceptual Solve: This yields \(\sin \theta_1 / v_1 = \sin \theta_2 / v_2\). Using numerical methods or simplification: \(x \approx 97.4 \text{ m}\).
Answer: Enter water 97.4 meters from the starting horizontal position (very close to the swimmer's perpendicular point to take advantage of running speed).
3. Shortest Ladder
Variables: \(L^2 = (x+4)^2 + y^2\). By similar triangles: \(y/8 = (x+4)/x \Rightarrow y = 8(x+4)/x\).
Function: \(L^2 = (x+4)^2 + \frac{64(x+4)^2}{x^2} = (x+4)^2 (1 + 64/x^2)\).
Derivative: Setting \(dL/dx = 0\) leads to \(x = 4\sqrt[3]{4}\) or using \(\theta\): \(L = 8\csc\theta + 4\sec\theta\). \(dL/d\theta = -8\csc\theta\cot\theta + 4\sec\theta\tan\theta = 0\).
Answer: \(\tan^3 \theta = 2 \Rightarrow \theta \approx 51.5^\circ\). Length \(L \approx 16.65 \text{ ft}\).
Packaging Design Slides Packaging Profits
Industrial Packaging Design Capstone
The Engineering Brief
"Our company manufactures 100 million beverage cans per year. If we can reduce the aluminum used per can by just 1%, we save over $1.5 million annually. Design a can that holds exactly 355mL with minimum surface area."
355
Volume (cm³)
Surface Area
Minimize
$1.5M
Potential Savings
Cylindrical Constraints
Primary: Surface Area
\( SA = 2\pi r^2 + 2\pi rh \)
(Top/Bottom + Side)
Secondary: Volume
\( V = \pi r^2 h = 355 \)
h
r
The Second Derivative Test
In engineering, we must prove our result is the absolute minimum.
Concavity Check
If \( f'(c) = 0 \) and \( f''(c) > 0 \), the function is concave up at \( c \), which guarantees a relative minimum .
Real-World Logic
Why aren't actual soda cans these "optimal" dimensions?
• Grip ergonomics (Width)
• Vending machine standards
• Metal thickness at seams
Packaging Proposal Project Engineering Division
DIMENSIONAL PROPOSAL
PROJECT: CAN-MIN-SA
Lead Engineer
Date of Analysis
1. Objective Statement
The objective of this analysis is to determine the optimal radius (\(r\)) and height (\(h\)) for a cylindrical aluminum container with a fixed volume of \(355 \text{ cm}^3\) such that the total surface area is minimized.
2. Mathematical Modeling
Primary/Secondary Equations
Derivative & Critical Numbers
3. Proof of Minimization
Apply the Second Derivative Test or a First Derivative Sign Chart to justify that your critical number yields an absolute minimum surface area.
4. Final Specification
Optimal Radius
Optimal Height
Total Surface Area
Engineering Recommendation
Compare your calculated dimensions to a standard soda can (Radius: ~3.3 cm, Height: ~12.3 cm). Why might your "optimal" dimensions differ from what is found in stores? Provide one structural or economic reason.
Packaging Design Answer Key Teacher Guide & Key
Packaging Design Capstone | Lesson 1.5
KEY & RUBRIC
Calculus Solution: The Optimal Can
1. Setup:
\(SA = 2\pi r^2 + 2\pi rh\)
\(V = \pi r^2 h = 355 \Rightarrow h = 355/(\pi r^2)\)
2. Substitute:
\(SA(r) = 2\pi r^2 + 2\pi r(355/(\pi r^2))\)
\(SA(r) = 2\pi r^2 + 710/r\)
3. Differentiate:
\(SA'(r) = 4\pi r - 710/r^2 = 0\)
\(4\pi r^3 = 710 \Rightarrow r^3 = 710/(4\pi) \approx 56.5\)
\(r = \sqrt[3]{56.5} \approx 3.84 \text{ cm}\)
4. Height:
\(h = 355/(\pi \cdot 3.84^2) \approx 7.67 \text{ cm}\)
Note: The optimal height is exactly twice the radius (\(h = 2r\)).
Assessment Rubric
Criterion Target Performance Points Equation Modeling Correctly identifies Primary (SA) and Secondary (V) equations. Substitutes correctly. /10 Calculus Execution Correct derivative, power rule application, and solving for \(r\) using cube root. /10 Justification Uses 2nd Derivative Test (\(SA'' = 4\pi + 1420/r^3\)) to prove minimum. /10 Reflection Insightful comparison between math-ideal and real-world can shapes. /10
Why the Difference?
Standard cans are taller and narrower than the "mathematical optimal." This is primarily due to: 1. Ergonomics: Humans find a 6-7cm diameter easier to grip than an 8cm one. 2. Seaming: Most metal is used in the top/bottom seams. The "optimal" math assumes uniform thickness everywhere. 3. Cooling: A taller can has more surface area (non-optimal) which actually helps the beverage cool faster in a fridge!
