Unit 5.10 – Introduction to Optimization Problems

AP® Calculus AB & BC | Using Derivatives to Maximize and Minimize Real-World Quantities

Why This Matters: Optimization problems are among the most practical applications of calculus! They ask you to find the maximum or minimum value of a quantity under given constraints—like maximizing profit, minimizing cost, finding the largest area, or the shortest distance. These problems combine all your derivative skills: finding critical points, testing for extrema, and understanding function behavior. Optimization is used everywhere: engineering design, business decisions, physics, economics, and medicine. Mastering this topic means you can solve real-world problems that matter!

🎯 What is Optimization?

DEFINITION

Optimization is the process of finding the maximum or minimum value of a quantity (objective function) subject to certain conditions (constraints).

The Optimization Problem Structure:

Objective: The quantity you want to maximize or minimize (e.g., area, volume, profit, cost, distance)
Constraint: The conditions or limitations that must be satisfied (e.g., fixed perimeter, budget limit, material constraint)
Variables: The quantities that can change
Goal: Find the values of variables that optimize the objective while satisfying constraints

Classic Examples:

Maximum area: What dimensions give the largest rectangular garden with 100 feet of fencing?
Minimum cost: What dimensions minimize the cost to build a cylindrical can of fixed volume?
Maximum profit: How many items should be produced to maximize revenue minus cost?
Shortest distance: What point on a curve is closest to a given point?
Maximum volume: What size square should be cut from corners to make a box with largest volume?

📋 The Standard Optimization Procedure

The Systematic 7-Step Process:

Understand the problem:
- Read carefully—what are you maximizing or minimizing?
- What are the constraints?
- Draw a diagram if possible (highly recommended!)
Identify variables:
- Choose variables for all changing quantities
- Label your diagram
Write the primary equation:
- Express the quantity to optimize as a function
- This is your objective function
Write the constraint equation(s):
- Express relationships between variables
- Use given conditions, formulas, or geometry
Reduce to one variable:
- Use constraints to eliminate all but one variable
- Substitute into primary equation
- Get function of single variable: \(f(x)\)
Find the optimal value:
- Find \(f'(x)\)
- Find critical points: solve \(f'(x) = 0\)
- Check endpoints if domain is restricted
- Use First or Second Derivative Test to verify max/min
Answer the question:
- Find all requested values (not just \(x\)!)
- Include units
- Verify answer makes physical sense

🔑 Key Equations:

Primary Equation (Objective Function)

The function you want to optimize. Examples:

Area: \(A = lw\) (rectangle), \(A = \pi r^2\) (circle)
Volume: \(V = lwh\) (box), \(V = \pi r^2 h\) (cylinder)
Profit: \(P = R - C\) (Revenue - Cost)
Distance: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
Surface Area: \(S = 2lw + 2lh + 2wh\) (box)

Constraint Equation

The relationship that must be satisfied. Examples:

Fixed perimeter: \(2l + 2w = 100\)
Fixed volume: \(\pi r^2 h = 1000\)
Budget constraint: \(C_1 x + C_2 y = B\)
Point on curve: \(y = f(x)\)

📖 Comprehensive Worked Examples

Example 1: Maximum Area with Fixed Perimeter

Problem: A farmer has 200 feet of fencing to enclose a rectangular garden. What dimensions will maximize the area?

Solution:

Step 1: Understand & Diagram

Want to: MAXIMIZE area

Constraint: Perimeter = 200 feet

Diagram:
┌─────────────┐
│                   │ w
│                   │
└─────────────┘
        l

Step 2: Variables

Let \(l\) = length, \(w\) = width

Step 3: Primary Equation (Objective)

\[ A = lw \]

(Want to maximize area)

Step 4: Constraint Equation

\[ 2l + 2w = 200 \]

Simplify:

\[ l + w = 100 \]

\[ l = 100 - w \]

Step 5: Reduce to One Variable

Substitute \(l = 100 - w\) into \(A = lw\):

\[ A(w) = (100 - w)w = 100w - w^2 \]

Domain: \(0 < w < 100\) (physical constraint)

Step 6: Find Optimal Value

Find derivative:

\[ A'(w) = 100 - 2w \]

Set equal to zero:

\[ 100 - 2w = 0 \]

\[ w = 50 \text{ feet} \]

Verify it's a maximum using Second Derivative Test:

\[ A''(w) = -2 < 0 \]

Since \(A''(50) < 0\), this is a MAXIMUM ✓

Step 7: Answer the Question

When \(w = 50\):

\[ l = 100 - 50 = 50 \text{ feet} \]

Maximum area:

\[ A = 50 \times 50 = 2500 \text{ square feet} \]

Answer: Dimensions of 50 ft × 50 ft (a square!) maximize the area at 2,500 square feet.

Example 2: Minimum Surface Area with Fixed Volume

Problem: A cylindrical can must hold 1000 cm³. Find the radius and height that minimize the amount of material (surface area).

Solution:

Step 1-2: Setup

Want to: MINIMIZE surface area

Constraint: Volume = 1000 cm³

Variables: \(r\) = radius, \(h\) = height

Step 3: Primary Equation

Surface area of cylinder (top + bottom + side):

\[ S = 2\pi r^2 + 2\pi rh \]

Step 4: Constraint

Volume of cylinder:

\[ \pi r^2 h = 1000 \]

Solve for \(h\):

\[ h = \frac{1000}{\pi r^2} \]

Step 5: Substitute

\[ S(r) = 2\pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2} \]

\[ S(r) = 2\pi r^2 + \frac{2000}{r} \]

Domain: \(r > 0\)

Step 6: Optimize

\[ S'(r) = 4\pi r - \frac{2000}{r^2} \]

Set equal to zero:

\[ 4\pi r - \frac{2000}{r^2} = 0 \]

\[ 4\pi r = \frac{2000}{r^2} \]

\[ 4\pi r^3 = 2000 \]

\[ r^3 = \frac{2000}{4\pi} = \frac{500}{\pi} \]

\[ r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42 \text{ cm} \]

Verify minimum with Second Derivative Test:

\[ S''(r) = 4\pi + \frac{4000}{r^3} \]

Since \(r > 0\), we have \(S''(r) > 0\) → MINIMUM ✓

Step 7: Complete Answer

\[ r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42 \text{ cm} \]

\[ h = \frac{1000}{\pi r^2} = \frac{1000}{\pi \cdot \frac{500}{\pi}} = \frac{1000\pi}{500} = 2\sqrt[3]{\frac{500}{\pi}} \approx 10.84 \text{ cm} \]

Notice: \(h = 2r\) (height equals diameter!)

Answer: Radius \(\approx\) 5.42 cm, Height \(\approx\) 10.84 cm minimize surface area.

Example 3: Maximum Product with Fixed Sum

Problem: Find two positive numbers whose sum is 60 and whose product is as large as possible.

Solution:

Step 1-2: Setup

Let the numbers be \(x\) and \(y\)

Want to: MAXIMIZE \(xy\)

Constraint: \(x + y = 60\)

Step 3-5: Set Up Function

From constraint: \(y = 60 - x\)

Product to maximize:

\[ P(x) = x(60 - x) = 60x - x^2 \]

Domain: \(0 < x < 60\)

Step 6: Optimize

\[ P'(x) = 60 - 2x \]

\[ 60 - 2x = 0 \quad \Rightarrow \quad x = 30 \]

\[ P''(x) = -2 < 0 \quad \Rightarrow \quad \text{Maximum} \]

Step 7: Answer

When \(x = 30\): \(y = 60 - 30 = 30\)

Maximum product: \(30 \times 30 = 900\)

Answer: The numbers are both 30, giving maximum product of 900.

📚 Common Optimization Problem Types

Classic Categories You'll Encounter:

Common Optimization Scenarios
Problem Type	Typical Objective	Common Constraint	Key Formula
Fencing/Perimeter	Maximize area	Fixed perimeter	\(A = lw\), \(P = 2l + 2w\)
Box/Container	Maximize volume OR minimize surface area	Fixed volume OR fixed material	\(V = lwh\), \(S = 2lw + 2lh + 2wh\)
Revenue/Profit	Maximize profit or revenue	Price-demand relationship	\(P = R - C\), \(R = px\)
Distance	Minimize distance	Point on curve	\(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
Inscribed/Circumscribed	Maximize/minimize area or volume	Geometric relationship	Varies by shape
Time/Rate	Minimize time or cost	Speed or rate constraint	\(t = \frac{d}{v}\)

💡 Tips, Tricks & Strategies

✅ Essential Optimization Tips:

ALWAYS draw a diagram: Label everything—variables, constants, relationships
Identify what to optimize: Read carefully—max or min? Area, volume, cost?
Write equations clearly: Separate primary equation from constraint
Reduce to ONE variable: You can only optimize with one variable
Find the domain: Physical constraints often limit variable ranges
Check endpoints: If domain is closed interval, evaluate at endpoints too
Verify max/min: Use First or Second Derivative Test
Answer what's asked: Don't just find \(x\)—find all requested quantities
Include units: Always state units in final answer
Common sense check: Does your answer make physical sense?

🔥 Problem-Solving Strategies:

Strategy 1: Simplify Distance Problems

Instead of minimizing \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\), minimize \(d^2\) (eliminates square root!). Same critical points, easier derivative.

Strategy 2: Use Geometry

Look for similar triangles, Pythagorean theorem, circle properties, etc. to establish constraint equations.

Strategy 3: Check Special Cases

Often optimal shapes are symmetric (squares, cubes) or have special proportions (height = diameter for cylinder).

Strategy 4: Closed Interval Method

If domain is \([a, b]\), evaluate function at:

All critical points in \((a, b)\)
Both endpoints \(a\) and \(b\)
Compare all values

📐 Essential Formulas Reference

Quick Formula Reference:

Geometric Formulas for Optimization
Shape	Area/Volume	Perimeter/Surface Area
Rectangle	\(A = lw\)	\(P = 2l + 2w\)
Circle	\(A = \pi r^2\)	\(C = 2\pi r\)
Triangle	\(A = \frac{1}{2}bh\)	\(P = a + b + c\)
Box	\(V = lwh\)	\(S = 2lw + 2lh + 2wh\)
Cylinder	\(V = \pi r^2 h\)	\(S = 2\pi r^2 + 2\pi rh\)
Sphere	\(V = \frac{4}{3}\pi r^3\)	\(S = 4\pi r^2\)
Cone	\(V = \frac{1}{3}\pi r^2 h\)	\(S = \pi r^2 + \pi r\ell\) (\(\ell\) = slant height)

❌ Common Mistakes to Avoid

Mistake 1: Not drawing a diagram (makes it much harder to see relationships!)
Mistake 2: Trying to optimize with two variables (must reduce to one!)
Mistake 3: Forgetting to check endpoints when domain is restricted
Mistake 4: Not verifying that critical point is actually a max/min
Mistake 5: Mixing up primary equation and constraint equation
Mistake 6: Ignoring physical constraints (e.g., negative dimensions impossible)
Mistake 7: Finding \(x\) but not answering what's actually asked
Mistake 8: Arithmetic errors when substituting constraint into primary equation
Mistake 9: Not simplifying before taking derivative (makes it harder!)
Mistake 10: Forgetting units in final answer
Mistake 11: Using wrong formula (e.g., cylinder vs. cone)
Mistake 12: Not checking if answer makes sense (negative area? too large?)

📝 Practice Problems

Set A: Basic Optimization

Find two positive numbers whose sum is 20 and whose product is maximum.
A rectangle has perimeter 40 cm. What dimensions maximize the area?
Find the point on the line \(y = 2x + 1\) closest to the origin.

Answers:

Both numbers are 10 (product = 100)
10 cm × 10 cm (square; area = 100 cm²)
\(\left(-\frac{2}{5}, \frac{1}{5}\right)\)

Set B: Applied Problems

A box with square base and open top must have volume 32,000 cm³. Find dimensions that minimize surface area.
A farmer wants to fence a rectangular area along a river (one side needs no fence). With 800 m of fence, what dimensions maximize area?

Answers:

Base: 40 cm × 40 cm, Height: 20 cm
400 m parallel to river, 200 m perpendicular (area = 80,000 m²)

Set C: Challenging

A rectangle is inscribed in a semicircle of radius 10. What dimensions maximize its area?
A poster must have 50 cm² of print area with 4 cm margins on top/bottom and 2 cm margins on sides. What dimensions minimize total poster area?

Answers:

Width: \(10\sqrt{2}\) cm, Height: \(5\sqrt{2}\) cm
Print dimensions: 5 cm × 10 cm (total poster: 9 cm × 18 cm)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Clear setup: Define variables, write equations clearly
Constraint elimination: Show substitution work to get one variable
Derivative shown: Write \(f'(x) = \ldots\) explicitly
Critical points found: Solve \(f'(x) = 0\) with work shown
Justification: Verify max/min using derivative test
Endpoint checking: If applicable, check boundaries
Complete answer: Find ALL requested values with units
Logical flow: Clear progression from setup to answer

💯 Earning Full Credit:

1-2 points: Correct setup (primary and constraint equations)
1 point: Reducing to single variable correctly
1 point: Finding derivative and critical point(s)
1 point: Justifying that critical point gives max/min
1 point: Complete, correct final answer with units
Key: Show ALL work—partial credit available for correct process!

⚡ Quick Reference Card

Optimization Quick Reference
Step	Action	Key Question
1. Understand	Read & draw diagram	What am I optimizing?
2. Variables	Define all variables	What can change?
3. Primary Eq	Write objective function	Formula for what I'm optimizing?
4. Constraint	Write relationship	What's the limitation?
5. One Variable	Substitute constraint	Can I express as \(f(x)\)?
6. Optimize	Find \(f'(x) = 0\), verify max/min	Where is derivative zero?
7. Answer	Find all values, include units	Did I answer the question?

Master Optimization! This powerful application of derivatives lets you solve real-world problems by finding maximum or minimum values under constraints. The systematic 7-step process is key: understand the problem (draw diagram!), identify variables, write the primary equation (what to optimize), write the constraint equation (relationship between variables), use constraint to reduce to one variable, find critical points by setting \(f'(x) = 0\), and verify max/min using derivative tests. Always check endpoints if domain is restricted! Common problem types include maximizing area with fixed perimeter, minimizing surface area with fixed volume, and distance problems. Remember: draw diagrams, label everything, simplify before differentiating, verify your answer makes sense, and include units. Optimization problems synthesize everything from Unit 5 and appear frequently on AP® exams! 🎯✨