📐 Concept diagram

04-08 - Optimization

Phase: 4 | Subject: 04-08 Prerequisites: 04-07-applications-of-derivatives.md (first/second derivative tests) Next subject: 04-09-lhopitals-rule.md

Learning Objectives

By the end of this subject, you will be able to:

1. Find absolute extrema on closed intervals
2. Solve applied optimization problems
3. Apply Fermat's theorem
4. Apply Rolle's theorem
5. Apply the Mean Value Theorem

Core Content

Absolute Extrema on Closed Intervals

On a closed interval [a, b], a continuous function attains: - An absolute maximum (largest value) - An absolute minimum (smallest value)

These occur either at: 1. Critical points inside (a, b), OR 2. Endpoints a and b

Steps: 1. Find critical points in (a, b) by solving f'(x) = 0 2. Evaluate f at critical points and endpoints 3. Largest = absolute max, smallest = absolute min

Example: f(x) = x³ - 3x on [-2, 2]

f'(x) = 3x² - 3 = 0 → x = ±1 f(-2) = -8 + 6 = -2 f(-1) = -1 + 3 = 2 f(1) = 1 - 3 = -2 f(2) = 8 - 6 = 2

Absolute max: 2 at x = -1 and x = 2 Absolute min: -2 at x = -2 and x = 1

Fermat's Theorem

If f has a local extremum at c and f'(c) exists, then f'(c) = 0.

Converse is FALSE: f'(c) = 0 doesn't guarantee an extremum.

Example: f(x) = x³ at x = 0: f'(0) = 0, but no extremum (inflection).

Applied Optimization

General approach: 1. Draw a diagram, label variables 2. Write the objective function (what to maximise/minimise) 3. Write the constraint equation 4. Use constraint to eliminate one variable 5. Find derivative, set to zero, solve 6. Verify it's a maximum/minimum 7. Check endpoints if needed

Example: Maximise area of rectangle with perimeter 20.

1. Let sides be x and y. 2x + 2y = 20, so y = 10 - x
2. Area: A = x(10 - x) = 10x - x²
3. A' = 10 - 2x = 0 → x = 5
4. A'' = -2 < 0 → maximum
5. y = 5. Maximum area: 25 (a square!)

Rolle's Theorem

If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists c in (a, b) such that f'(c) = 0.

Geometric: There's a horizontal tangent somewhere between a and b.

Mean Value Theorem (MVT)

⚠️ THIS IS CRITICAL — MVT is the bridge between derivatives and function behaviour. It proves that if f' > 0, f is increasing; if f' < 0, f is decreasing. It underpins the Fundamental Theorem of Calculus and numerous convergence proofs.

If f is continuous on [a, b] and differentiable on (a, b), then there exists c in (a, b) such that:

$f'(c) = [f(b) - f(a)] / (b - a)
$

Interpretation: At some point, the instantaneous rate equals the average rate.

Example: f(x) = x² on [1, 3] Average rate: (9 - 1)/(3 - 1) = 8/2 = 4 f'(c) = 2c = 4, so c = 2. At x = 2, instantaneous rate = 4 = average rate.

Key Terms

• 04 08 Optimization
• Absolute Extrema on Closed Intervals
• Applied Optimization
• Correct: A)
• Correct: B)
• Correct: C)
• Example 1: Absolute extrema
• Example 2: Optimization
• Example 3: MVT application
• Fermat's Theorem
• Mean Value Theorem (MVT)
• Rolle's Theorem

Worked Examples

Example 1: Absolute extrema

f(x) = x⁴ - 4x² + 1 on [-2, 3]

f'(x) = 4x³ - 8x = 4x(x² - 2) = 0 → x = 0, ±√2 √2 ≈ 1.414 (inside [-2, 3]), -√2 ≈ -1.414 (inside)

Evaluate: f(-2) = 16 - 16 + 1 = 1 f(-√2) = 4 - 8 + 1 = -3 f(0) = 1 f(√2) = -3 f(3) = 81 - 36 + 1 = 46

Absolute max: 46 at x = 3 Absolute min: -3 at x = ±√2

Example 2: Optimization

Find dimensions of rectangle with maximum area inscribed in semicircle of radius r.

1. Rectangle width = 2x, height = y. Point (x, y) on circle: x² + y² = r²
2. y = √(r² - x²)
3. Area: A = 2x·√(r² - x²)
4. A' = 2√(r² - x²) + 2x · (-x)/√(r² - x²)
5. Set A' = 0: 2(r² - x²) - 2x² = 0 → 2r² = 4x² → x = r/√2
6. y = r/√2
7. Maximum area: 2(r/√2)(r/√2) = r²

The optimal rectangle is actually a square with side r/√2.

Example 3: MVT application

Verify the Mean Value Theorem for f(x) = x³ - x on [-1, 2].

1. f is a polynomial, so continuous on [-1, 2] and differentiable on (-1, 2). ✓
2. f(-1) = -1 + 1 = 0. f(2) = 8 - 2 = 6.
3. Average rate: [f(2) - f(-1)]/(2 - (-1)) = 6/3 = 2.
4. f'(x) = 3x² - 1. Set 3c² - 1 = 2 → 3c² = 3 → c² = 1 → c = 1 (c = -1 is at the left endpoint, not in (-1, 2)).
5. Check: f'(1) = 2, which equals the average rate. MVT confirmed.

Quiz

Q1: What does the concept of Absolute Extrema on Closed Intervals primarily refer to in this subject?

A) The definition and application of Absolute Extrema on Closed Intervals B) A visual representation of Absolute Extrema on Closed Intervals C) A historical anecdote about Absolute Extrema on Closed Intervals D) A computational error related to Absolute Extrema on Closed Intervals

Correct: A)

• If you chose A: Absolute Extrema on Closed Intervals is defined as: the definition and application of absolute extrema on closed intervals. The other options describe different aspects that are not the primary focus. Correct!
• If you chose B: This is incorrect. Absolute Extrema on Closed Intervals is defined as: the definition and application of absolute extrema on closed intervals. The other options describe different aspects that are not the primary focus.
• If you chose C: This is incorrect. Absolute Extrema on Closed Intervals is defined as: the definition and application of absolute extrema on closed intervals. The other options describe different aspects that are not the primary focus.
• If you chose D: This is incorrect. Absolute Extrema on Closed Intervals is defined as: the definition and application of absolute extrema on closed intervals. The other options describe different aspects that are not the primary focus.

Q2: What is the primary purpose of Applied Optimization?

A) It replaces all other methods in this domain B) It is primarily a historical notation system C) It is used only in advanced research contexts D) It is used to applied optimization in mathematical analysis

Correct: D)

• If you chose A: This is incorrect. Applied Optimization serves the purpose described in the correct answer. The other options misrepresent its role.
• If you chose B: This is incorrect. Applied Optimization serves the purpose described in the correct answer. The other options misrepresent its role.
• If you chose C: This is incorrect. Applied Optimization serves the purpose described in the correct answer. The other options misrepresent its role.
• If you chose D: Applied Optimization serves the purpose described in the correct answer. The other options misrepresent its role. Correct!

Q3: Which statement about Mean Value Theorem (MVT) is TRUE?

A) Mean Value Theorem (MVT) is mentioned only as a historical footnote B) Mean Value Theorem (MVT) is an advanced topic beyond this subject's scope C) Mean Value Theorem (MVT) is a fundamental concept covered in this subject D) Mean Value Theorem (MVT) is not related to this subject

Correct: C)

• If you chose A: This is incorrect. Mean Value Theorem (MVT) is a fundamental concept covered in this subject. This subject covers Mean Value Theorem (MVT) as part of its core content.
• If you chose B: This is incorrect. Mean Value Theorem (MVT) is a fundamental concept covered in this subject. This subject covers Mean Value Theorem (MVT) as part of its core content.
• If you chose C: Mean Value Theorem (MVT) is a fundamental concept covered in this subject. This subject covers Mean Value Theorem (MVT) as part of its core content. Correct!
• If you chose D: This is incorrect. Mean Value Theorem (MVT) is a fundamental concept covered in this subject. This subject covers Mean Value Theorem (MVT) as part of its core content.

Q4: Based on the worked examples in this subject, what is the correct result?

A) 2. Min B) The inverse of the correct answer C) An unrelated numerical value D) A different result from a common mistake

Correct: A)

• If you chose A: The worked examples show that the result is 2. Min. The other options represent common errors. Correct!
• If you chose B: This is incorrect. The worked examples show that the result is 2. Min. The other options represent common errors.
• If you chose C: This is incorrect. The worked examples show that the result is 2. Min. The other options represent common errors.
• If you chose D: This is incorrect. The worked examples show that the result is 2. Min. The other options represent common errors.

Q5: How are Mean Value Theorem (MVT) and Fermat'S Theorem related?

A) Mean Value Theorem (MVT) is the inverse of Fermat'S Theorem B) Mean Value Theorem (MVT) and Fermat'S Theorem are closely related concepts C) Mean Value Theorem (MVT) is a special case of Fermat'S Theorem D) Mean Value Theorem (MVT) and Fermat'S Theorem are completely unrelated topics

Correct: B)

• If you chose A: This is incorrect. Both Mean Value Theorem (MVT) and Fermat'S Theorem are covered in this subject as interconnected topics.
• If you chose B: Both Mean Value Theorem (MVT) and Fermat'S Theorem are covered in this subject as interconnected topics. Correct!
• If you chose C: This is incorrect. Both Mean Value Theorem (MVT) and Fermat'S Theorem are covered in this subject as interconnected topics.
• If you chose D: This is incorrect. Both Mean Value Theorem (MVT) and Fermat'S Theorem are covered in this subject as interconnected topics.

Q6: What is a common pitfall when working with Rolle'S Theorem?

A) A common mistake is confusing Rolle'S Theorem with a similar concept B) The main error with Rolle'S Theorem is using it when it is not needed C) Rolle'S Theorem is always computed the same way in all contexts D) Rolle'S Theorem has no common misconceptions

Correct: A)

• If you chose A: Students often confuse Rolle'S Theorem with similar-sounding or related concepts. Pay attention to the precise definitions. Correct!
• If you chose B: This is incorrect. Students often confuse Rolle'S Theorem with similar-sounding or related concepts. Pay attention to the precise definitions.
• If you chose C: This is incorrect. Students often confuse Rolle'S Theorem with similar-sounding or related concepts. Pay attention to the precise definitions.
• If you chose D: This is incorrect. Students often confuse Rolle'S Theorem with similar-sounding or related concepts. Pay attention to the precise definitions.

Q7: When should you apply Example 1: Absolute Extrema?

A) Use Example 1: Absolute Extrema only in pure mathematics contexts B) Avoid Example 1: Absolute Extrema unless explicitly instructed C) Apply Example 1: Absolute Extrema to solve problems in this subject's domain D) Example 1: Absolute Extrema is not practically useful

Correct: C)

• If you chose A: This is incorrect. Example 1: Absolute Extrema is a practical tool used throughout this subject to solve relevant problems.
• If you chose B: This is incorrect. Example 1: Absolute Extrema is a practical tool used throughout this subject to solve relevant problems.
• If you chose C: Example 1: Absolute Extrema is a practical tool used throughout this subject to solve relevant problems. Correct!
• If you chose D: This is incorrect. Example 1: Absolute Extrema is a practical tool used throughout this subject to solve relevant problems.

Practice Problems

1. Absolute extrema of f(x) = x² on [-1, 2] Answer: f'(x) = 2x = 0 → x = 0. f(-1) = 1, f(0) = 0, f(2) = 4. Max = 4 at x = 2. Min = 0 at x = 0.

2. Maximise area of rectangle with perimeter 40. Answer: y = 20 - x. A = x(20 - x). A' = 20 - 2x = 0 → x = 10. Square with sides 10×10, area 100.

3. MVT for f(x) = x³ on [0, 2] Answer: f'(c) = (8 - 0)/2 = 4. 3c² = 4, c = 2/√3 ≈ 1.155.

4. Find dimensions of rectangle with maximum area, given it must fit inside a right triangle with legs 3 and 4 (one corner at the right angle). Answer: Place triangle with legs along axes: y = 4 - (4/3)x. Rectangle width=x, height=y. Area A = x(4 - 4x/3) = 4x - 4x²/3. A' = 4 - 8x/3 = 0 → x = 3/2. y = 4 - 2 = 2. Dimensions: 3/2 × 2, area = 3.

5. Apply Rolle's theorem to f(x) = x² - 4x on [0, 4]. Find c. Answer: f(0) = 0, f(4) = 0. f'(x) = 2x - 4 = 0 → c = 2. Tangent is horizontal at x = 2.

Summary

Key takeaways:

• Absolute extrema on [a, b]: check critical points AND endpoints
• Fermat: local extremum → f'(c) = 0 (if f' exists)
• Optimization: write objective, eliminate variables, differentiate, solve
• Rolle's Theorem: f(a) = f(b) → horizontal tangent somewhere
• MVT: average rate = instantaneous rate at some point

Pitfalls

• Forgetting endpoints when finding absolute extrema. On a closed interval [a,b], the absolute max/min can occur at endpoints, not just at critical points. Checking only where f' = 0 can miss the actual extreme values.
• Assuming f'(c) = 0 implies an extremum. Fermat's theorem says extremum ⇒ f'(c) = 0, but the converse is false. f(x) = x³ at x = 0 has f'(0) = 0 but no extremum — it's an inflection point.
• Skipping the verification step in optimization. After finding a critical point, always confirm it's a max (or min, as needed) using the second derivative test or first derivative test. A critical point might be the opposite of what you want.
• Confusing Rolle's Theorem with MVT. Rolle's requires f(a) = f(b) and guarantees a horizontal tangent (f'(c) = 0). MVT works for any f(a) and f(b) and guarantees f'(c) equals the average rate.
• Writing the objective function without using the constraint. In applied problems, always use the constraint to reduce the objective to one variable before differentiating. Trying to optimise with two variables still in the expression won't work.

Next Steps

Next up: 04-09-lhopitals-rule.md