Unit 5.11 – Solving Optimization Problems

AP® Calculus AB & BC | Advanced Techniques and Complex Applications

Why This Matters: Now we dive deep into solving optimization problems! This topic goes beyond the basics—you'll tackle complex geometric problems, business applications, distance minimization, and situations requiring advanced problem-solving strategies. You'll learn to handle implicit constraints, multiple variables, and real-world scenarios that require careful mathematical modeling. These problems test your complete understanding of derivatives, critical thinking, and ability to translate words into mathematics. Master these techniques and you'll excel on AP® exams and real-world applications!

🎯 Advanced Problem-Solving Framework

The Enhanced Optimization Strategy

Comprehensive Approach:

Read and Re-read: Understand what's being optimized and under what constraints
Visualize: Draw detailed diagram with labels
Identify:
- Quantity to optimize (objective function)
- Given constraints (equations relating variables)
- Domain restrictions (physical limits)
Express mathematically:
- Primary equation: $Q = f(x, y, ...)$
- Constraint equation(s): $g(x, y, ...) = k$
Eliminate variables: Solve constraint for one variable, substitute
Create single-variable function: $Q(x)$ or $Q(y)$
Find critical points:
- Compute $Q'(x)$
- Solve $Q'(x) = 0$
- Check where $Q'(x)$ is undefined
Test critical points AND endpoints:
- Use First or Second Derivative Test
- Check boundary values if domain is closed
Verify and interpret:
- Does answer make physical sense?
- Calculate ALL requested quantities
- Include proper units

📖 Advanced Worked Examples

Example 1: Box with Open Top from Square Material

Problem: A box with a square base and open top is to be made from a square piece of cardboard by cutting equal squares from each corner and folding up the sides. If the cardboard is 12 inches on each side, what size squares should be cut to maximize the volume?

Solution:

Step 1: Understand & Visualize

Start with 12" × 12" cardboard. Cut $x \times x$ squares from corners.

Before cutting:
┌───────────────┐
│ x │         │ x │
├───┼─────────┼───┤
│   │         │   │
│   │ 12 - 2x │   │
│   │         │   │
├───┼─────────┼───┤
│ x │         │ x │
└───────────────┘
After folding: box with height x

Step 2: Variables and Constraints

Let $x$ = side length of cut squares (also = height of box)
After cutting, base = $(12 - 2x) \times (12 - 2x)$
Height = $x$
Domain: $0 < x < 6$ (can't cut more than half)

Step 3: Primary Equation

Volume of box:

\[ V(x) = (\text{length})(\text{width})(\text{height}) = (12-2x)(12-2x)(x) \]

\[ V(x) = (12-2x)^2 \cdot x \]

\[ V(x) = (144 - 48x + 4x^2) \cdot x = 144x - 48x^2 + 4x^3 \]

Step 4: Find Critical Points

\[ V'(x) = 144 - 96x + 12x^2 = 12(12 - 8x + x^2) = 12(x^2 - 8x + 12) \]

Factor:

\[ V'(x) = 12(x - 2)(x - 6) \]

Critical points: $x = 2, 6$

Since domain is $0 < x < 6$, only $x = 2$ is in the interior.

Step 5: Verify Maximum

Use Second Derivative Test:

\[ V''(x) = -96 + 24x \]

\[ V''(2) = -96 + 48 = -48 < 0 \]

Since $V''(2) < 0$, we have a MAXIMUM at $x = 2$ ✓

Check endpoints:

$V(0) = 0$ (no box)
$V(2) = 144(2) - 48(4) + 4(8) = 288 - 192 + 32 = 128$ cubic inches
$V(6) = 0$ (no box)

Step 6: Complete Answer

Cut squares of side length: 2 inches

Resulting box dimensions: $8" \times 8" \times 2"$

Maximum volume: 128 cubic inches

Answer: Cut 2-inch squares from each corner to maximize volume at 128 in³.

Example 2: Minimum Distance from Point to Curve

Problem: Find the point on the parabola $y = x^2$ that is closest to the point $(0, 1)$.

Solution:

Step 1: Setup

Want to minimize distance from $(x, x^2)$ on parabola to $(0, 1)$.

Step 2: Distance Formula

\[ d = \sqrt{(x - 0)^2 + (x^2 - 1)^2} = \sqrt{x^2 + (x^2 - 1)^2} \]

Trick: Minimize $d^2$ instead (same critical points, easier calculus!)

\[ D(x) = d^2 = x^2 + (x^2 - 1)^2 = x^2 + x^4 - 2x^2 + 1 = x^4 - x^2 + 1 \]

Step 3: Find Critical Points

\[ D'(x) = 4x^3 - 2x = 2x(2x^2 - 1) \]

\[ 2x(2x^2 - 1) = 0 \]

Solutions:

$x = 0$
$2x^2 - 1 = 0 \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm\frac{1}{\sqrt{2}} = \pm\frac{\sqrt{2}}{2}$

Step 4: Test Critical Points

Using Second Derivative Test:

\[ D''(x) = 12x^2 - 2 \]

$x$	$D''(x)$	Conclusion	$D(x)$
$0$	$-2 < 0$	Local max	$1$
$\pm\frac{\sqrt{2}}{2}$	$12(\frac{1}{2}) - 2 = 4 > 0$	Local min	$\frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}$

Step 5: Identify Closest Points

Minimum distance occurs at $x = \pm\frac{\sqrt{2}}{2}$

For $x = \frac{\sqrt{2}}{2}$: $y = x^2 = \frac{1}{2}$

For $x = -\frac{\sqrt{2}}{2}$: $y = x^2 = \frac{1}{2}$

Minimum distance: $d = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

Answer: Points $\left(\pm\frac{\sqrt{2}}{2}, \frac{1}{2}\right)$ are closest to $(0, 1)$, with minimum distance $\frac{\sqrt{3}}{2}$.

Example 3: Norman Window (Semicircle + Rectangle)

Problem: A Norman window consists of a rectangle surmounted by a semicircle. If the perimeter of the window is 30 feet, find the dimensions that admit the most light (maximize area).

Solution:

Step 1: Diagram & Variables

Let $r$ = radius of semicircle = half the width of rectangle

Let $h$ = height of rectangular portion

Width of rectangle = $2r$

Step 2: Area (to maximize)

\[ A = \text{Rectangle} + \text{Semicircle} = 2rh + \frac{1}{2}\pi r^2 \]

Step 3: Perimeter Constraint

Perimeter consists of:

Bottom: $2r$
Two sides: $2h$
Semicircular top: $\frac{1}{2}(2\pi r) = \pi r$

\[ 2r + 2h + \pi r = 30 \]

\[ h = \frac{30 - 2r - \pi r}{2} = \frac{30 - r(2 + \pi)}{2} \]

Step 4: Substitute to Get Single Variable

\[ A(r) = 2r \cdot \frac{30 - r(2 + \pi)}{2} + \frac{\pi r^2}{2} \]

\[ A(r) = r[30 - r(2 + \pi)] + \frac{\pi r^2}{2} \]

\[ A(r) = 30r - 2r^2 - \pi r^2 + \frac{\pi r^2}{2} \]

\[ A(r) = 30r - 2r^2 - \frac{\pi r^2}{2} = 30r - r^2\left(2 + \frac{\pi}{2}\right) \]

Step 5: Optimize

\[ A'(r) = 30 - 2r\left(2 + \frac{\pi}{2}\right) = 30 - r(4 + \pi) \]

\[ 30 - r(4 + \pi) = 0 \]

\[ r = \frac{30}{4 + \pi} \approx 4.2 \text{ feet} \]

Verify maximum:

\[ A''(r) = -(4 + \pi) < 0 \quad \Rightarrow \quad \text{Maximum} ✓ \]

Step 6: Find All Dimensions

\[ r = \frac{30}{4 + \pi} \]

\[ h = \frac{30 - r(2 + \pi)}{2} = \frac{30 - \frac{30(2+\pi)}{4+\pi}}{2} = \frac{30}{4+\pi} = r \]

Interesting result: $h = r$!

Answer: Radius $r = \frac{30}{4+\pi} \approx 4.2$ ft, Height $h = \frac{30}{4+\pi} \approx 4.2$ ft.

Example 4: Business/Economics - Profit Maximization

Problem: A company can sell $x$ items per week at a price of $p = 200 - 0.5x$ dollars each. The cost to produce $x$ items is $C(x) = 50x + 10000$ dollars. How many items should be produced to maximize profit?

Solution:

Step 1: Profit Function

Profit = Revenue - Cost

\[ P(x) = R(x) - C(x) \]

Revenue = (price)(quantity):

\[ R(x) = px = (200 - 0.5x)x = 200x - 0.5x^2 \]

Therefore:

\[ P(x) = (200x - 0.5x^2) - (50x + 10000) = 200x - 0.5x^2 - 50x - 10000 \]

\[ P(x) = 150x - 0.5x^2 - 10000 \]

Step 2: Find Critical Points

\[ P'(x) = 150 - x \]

\[ 150 - x = 0 \quad \Rightarrow \quad x = 150 \]

Step 3: Verify Maximum

\[ P''(x) = -1 < 0 \quad \Rightarrow \quad \text{Maximum} ✓ \]

Step 4: Calculate Maximum Profit

\[ P(150) = 150(150) - 0.5(150)^2 - 10000 \]

\[ P(150) = 22500 - 11250 - 10000 = \$1250 \]

At this production level:

Price: $p = 200 - 0.5(150) = \$125$
Revenue: $R = 150 \times 125 = \$18,750$
Cost: $C = 50(150) + 10000 = \$17,500$

Answer: Produce 150 items per week, sell at $125 each for maximum profit of $1,250.

Example 5: Minimizing Material Cost (Multiple Materials)

Problem: A rectangular storage container with open top has volume 10 m³. The base costs $10 per m² and the sides cost $6 per m². Find dimensions that minimize the cost.

Solution:

Step 1: Variables

Let $x$ = length, $y$ = width, $h$ = height (all in meters)

Step 2: Cost Function

\[ C = (\text{base cost}) + (\text{side cost}) \]

\[ C = 10xy + 6(2xh + 2yh) = 10xy + 12xh + 12yh \]

Step 3: Volume Constraint

\[ xyh = 10 \quad \Rightarrow \quad h = \frac{10}{xy} \]

Step 4: Substitute

\[ C(x,y) = 10xy + 12x \cdot \frac{10}{xy} + 12y \cdot \frac{10}{xy} \]

\[ C(x,y) = 10xy + \frac{120}{y} + \frac{120}{x} \]

For square base ($x = y$), let $x = y$:

\[ C(x) = 10x^2 + \frac{120}{x} + \frac{120}{x} = 10x^2 + \frac{240}{x} \]

Step 5: Optimize

\[ C'(x) = 20x - \frac{240}{x^2} \]

\[ 20x - \frac{240}{x^2} = 0 \]

\[ 20x = \frac{240}{x^2} \]

\[ 20x^3 = 240 \]

\[ x^3 = 12 \quad \Rightarrow \quad x = \sqrt[3]{12} \approx 2.29 \text{ m} \]

Verify minimum:

\[ C''(x) = 20 + \frac{480}{x^3} > 0 \quad \Rightarrow \quad \text{Minimum} ✓ \]

Step 6: Find All Dimensions

\[ x = y = \sqrt[3]{12} \approx 2.29 \text{ m} \]

\[ h = \frac{10}{x^2} = \frac{10}{12^{2/3}} = \frac{10}{\sqrt[3]{144}} \approx 1.91 \text{ m} \]

Minimum cost:

\[ C = 10(12^{2/3}) + \frac{240}{\sqrt[3]{12}} \approx \$157.43 \]

Answer: Base $2.29 \text{ m} \times 2.29 \text{ m}$, Height $1.91 \text{ m}$ minimizes cost at approximately $157.43.

💡 Advanced Problem-Solving Strategies

🔥 Expert Techniques:

Strategy 1: Minimize Square Instead of Square Root

For distance problems: $d = \sqrt{f(x)}$, minimize $d^2 = f(x)$ instead.

Same critical points, simpler derivative!

Strategy 2: Use Symmetry

If problem has symmetry, assume optimal solution is symmetric.

Rectangle with fixed perimeter → square maximizes area
Cylinder with fixed volume → $h = 2r$ minimizes surface area

Strategy 3: Implicit Constraints

Some constraints come from geometry or physics:

Lengths must be positive
Similar triangles create ratios
Pythagorean theorem for right triangles

Strategy 4: Check Endpoints!

If domain is $[a, b]$, always evaluate at endpoints AND critical points.

Use Closed Interval Method.

Strategy 5: Simplify Before Differentiating

Expand, factor, or algebraically simplify the function before taking derivative.

Makes calculus much easier!

📚 Common Optimization Scenarios

Classic Problem Types:

Optimization Problem Library
Scenario	Setup	Key Insight
Fencing with Wall	3 sides need fence, 1 side is wall	Optimal: width = 2 × depth
Cylinder (volume fixed)	Minimize surface area	Optimal: $h = 2r$ (diameter)
Inscribed Rectangle	In circle/ellipse	Use geometry/trig to relate variables
Poster with Margins	Minimize total area	Add margins to print dimensions
Light Intensity	$I \propto \frac{k}{d^2}$	Minimize/maximize intensity
Travel Time	Different speeds in regions	$T = \frac{d_1}{v_1} + \frac{d_2}{v_2}$
Cone in Sphere	Maximize volume	Use Pythagorean theorem

🎯 Essential Tips & Tricks

✅ Master Checklist:

Draw first, calculate second: Always sketch the situation
Label everything: Variables, constants, relationships
Identify the objective clearly: What exactly are you optimizing?
Write constraints explicitly: Don't skip this step
Reduce to single variable: Essential for optimization
Simplify before differentiating: Algebra now saves calculus later
Domain matters: Physical constraints limit variable ranges
Check endpoints: Maximum/minimum might be at boundary
Verify max/min: Use First or Second Derivative Test
Answer the question: Find ALL requested values, not just $x$
Units matter: Always include units in final answer
Sanity check: Does answer make physical sense?

❌ Common Mistakes to Avoid

Mistake 1: Not drawing a diagram (biggest source of errors!)
Mistake 2: Confusing what to optimize with the constraint
Mistake 3: Failing to eliminate all but one variable
Mistake 4: Forgetting domain restrictions from physical reality
Mistake 5: Not checking endpoints when domain is closed interval
Mistake 6: Assuming critical point is maximum without verification
Mistake 7: Algebraic errors when substituting constraint
Mistake 8: Taking derivative of wrong function (optimize objective, not constraint!)
Mistake 9: Finding $x$ but not calculating final answer
Mistake 10: Using wrong formula (sphere vs. cylinder, etc.)
Mistake 11: Ignoring units or mixing units
Mistake 12: Not checking if answer makes sense (negative length?)

📝 Advanced Practice Problems

Set A: Geometric Optimization

A rectangle is inscribed in a circle of radius 5. Find dimensions that maximize its area.
A cylindrical can is to hold 1000 cm³ and have a plastic lid (cost = 2×base cost). Minimize total cost.
Find dimensions of rectangle with largest area that fits inside triangle with base 10 and height 8.

Answers:

$5\sqrt{2} \times 5\sqrt{2}$ (square with side $5\sqrt{2}$)
$r = \sqrt[3]{\frac{1000}{3\pi}}$, $h = 2r$
Width = 5, Height = 4 (half the triangle dimensions)

Set B: Applied Problems

Light pole 20 ft high, person 6 ft tall walking away at 5 ft/s. At what rate is tip of shadow moving?
Farmer has 2000 ft fence. Wants rectangular field divided by parallel fence. Maximize total area.

Answers:

Related rates, not optimization: $\frac{50}{7}$ ft/s
Dimensions: 500 ft × 333.33 ft; Area = 166,667 ft²

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Clear problem setup: Define variables with labels
Explicit equations: Primary and constraint equations clearly written
Substitution shown: Work for eliminating variables
Single-variable function: Clearly state $Q(x)$
Derivative calculated: Show $Q'(x)$ explicitly
Critical points found: Solve $Q'(x) = 0$ with work
Justification provided: First/Second Derivative Test or endpoint check
Complete answer: All requested quantities with units
Logical flow: Clear progression from setup to conclusion

💯 Maximizing Your Score:

Show ALL work: Partial credit available for correct process
Label clearly: "Let $x$ = ..." makes grading easier
Write equations: Don't just solve—show the equation first
Justify conclusions: State which test you used
Answer in context: Relate back to the problem statement
Check reasonableness: Briefly note if answer makes sense

⚡ Ultimate Quick Reference

Optimization Quick Reference Guide
Problem Element	What to Do	Why It Matters
Diagram	Always draw; label everything	Visualizes relationships
Objective	Identify quantity to optimize	This becomes $Q(x)$
Constraint	Write equation relating variables	Allows variable elimination
Single Variable	Substitute to get $Q(x)$	Can't optimize 2 variables
Domain	Identify physical limits	May restrict critical points
Derivative	Find $Q'(x)$; solve $Q'(x)=0$	Locates critical points
Verification	Test; check endpoints	Confirms max/min
Answer	All values with units	Answers the question!

Master Advanced Optimization! Solving complex optimization problems requires systematic thinking and strong calculus skills. Always start with a clear diagram and identify what you're optimizing (objective function) versus the constraints (relationships between variables). The key is reducing to a single-variable function by substituting constraints, then using derivatives to find critical points. Remember advanced strategies: minimize $d^2$ instead of $d$ for distance problems, use symmetry when possible, check endpoints for closed intervals, and always verify using derivative tests. Common scenarios include cutting corners for boxes, inscribed shapes, Norman windows, business profit, and minimum cost with different material prices. Watch for physical domain restrictions—lengths must be positive! Always complete the solution by finding ALL requested values with proper units. Show every step clearly on AP® exams for maximum partial credit. Practice these challenging problems until the pattern recognition becomes automatic—that's when you've truly mastered optimization! 🎯✨

\(x\)	\(D''(x)\)	Conclusion	\(D(x)\)
\(0\)	\(-2 < 0\)	Local max	\(1\)
\(\pm\frac{\sqrt{2}}{2}\)	\(12(\frac{1}{2}) - 2 = 4 > 0\)	Local min	\(\frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}\)