Lesson 6-2: Derivatives – Extrema and Optimization
Learning Goals
- Find critical points from and where derivative is undefined.
- Classify extrema via first/second derivative tests.
- Execute the closed-interval method to find absolute extrema.
- Model optimization problems with realistic constraints.
Assumptions
- Function differentiable where tests are applied.
- Constraints are explicit and define a valid domain.
- Interpret results in real units and feasibility.
First Derivative Test
If changes sign from positive to negative at a critical point, the point is a local maximum; from negative to positive, a local minimum. If no sign change, not an extremum.
Optimization Clinic Appendix
Clinic 1: Box Volume Maximization
- Objective: V = x(20-2x)(15-2x) from cut corners, domain 0 < x < 7.5.
- V'(x) = 12x² - 140x + 300, critical points: x = 3.33, x = 7.5.
- Test: V''(3.33) < 0 (max), check endpoints V(0) = V(7.5) = 0.
- Maximum: V(3.33) ≈ 592.6 cubic units, feasible for practical box.
Clinic 2: Fence Perimeter Problem
- Objective: A = xy with 2x + y = 240 constraint, domain x > 0.
- Substitute: A(x) = x(240-2x) = 240x - 2x², A'(x) = 240 - 4x.
- Critical point: x = 60, A''(x) = -4 < 0 confirms maximum.
- Optimal: 60×120 rectangle, area = 7200 sq units within budget.
Clinic 3: Can Surface Area
- Objective: S = 2πr² + 2πrh with V = πr²h = 355 constraint.
- Substitute h = 355/(πr²): S(r) = 2πr² + 710/r, S'(r) = 4πr - 710/r².
- Critical point: r³ = 710/(4π), r ≈ 3.84 cm, S''(r) > 0 (min).
- Minimum surface: r = 3.84 cm, h = 7.68 cm, real manufacturing size.
Clinic 4: Distance Minimization
- Objective: d = √[(x-2)² + (x²-1)²] from point (2,1) to parabola y = x².
- Minimize d²: f(x) = (x-2)² + (x²-1)², f'(x) = 2(x-2) + 4x(x²-1).
- Solve f'(x) = 0: 2x - 4 + 4x³ - 4x = 0 → 4x³ - 2x - 4 = 0.
- Numerical solution: x ≈ 1.22, minimum distance ≈ 0.86 units.
Clinic 5: Profit Maximization
- Revenue R(x) = 50x - 0.1x², Cost C(x) = 10x + 1000, domain x ≥ 0.
- Profit P(x) = R(x) - C(x) = 40x - 0.1x² - 1000, P'(x) = 40 - 0.2x.
- Critical point: x = 200 units, P''(x) = -0.2 < 0 confirms max.
- Maximum profit: P(200) = $3000 at production level = 200 units.
Clinic 6: Ladder Around Corner
- Objective: L = a/sin θ + b/cos θ where a,b are corridor widths.
- L'(θ) = -a cos θ/sin² θ + b sin θ/cos² θ, set equal to zero.
- Critical condition: (a/b) = (sin³ θ)/(cos³ θ) = tan³ θ.
- Minimum ladder length depends on specific corridor dimensions.
Clinic 7: Inventory Cost Model
- Cost C(x) = (D/x)×S + (x/2)×H where D=demand, S=setup, H=holding.
- C'(x) = -DS/x² + H/2, critical point: x = √(2DS/H).
- EOQ formula: x* = √(2DS/H), C''(x*) > 0 confirms minimum.
- Optimal order quantity minimizes total inventory costs.
Clinic 8: Pipeline Route
- Cost = land portion + underwater portion with different unit costs.
- Total C(x) = c₁√(a² + x²) + c₂√(b² + (d-x)²) where x is split point.
- C'(x) = c₁x/√(a² + x²) - c₂(d-x)/√(b² + (d-x)²) = 0.
- Solve numerically: optimal split minimizes construction cost.
Clinic 9: Window Design
- Perimeter P = 2x + y + πx/2, Area A = xy + πx²/8, P fixed.
- Express y = P - 2x - πx/2, substitute into A(x).
- A'(x) = P - 4x - πx = 0, critical point x = P/(4+π).
- Maximum area window has optimal rectangle-semicircle ratio.
Clinic 10: Reflector Design
- Light path: source to reflector point to target, minimize total distance.
- Use Fermat's principle: angle of incidence = angle of reflection.
- Set up distance function and differentiate to find reflection point.
- Optimal design achieves minimum travel time for light ray.
Second Derivative Test
Concavity and Extrema
If and exists near c: implies a local maximum; implies a local minimum; is inconclusive.
Inflection Points
Points where concavity changes sign are inflection points. Often found by solving or where is undefined and confirming sign change.
Closed-Interval Absolute Extrema
- Find critical points in the open interval where or undefined.
- Evaluate at critical points and endpoints.
- Compare values: largest is absolute maximum, smallest is absolute minimum.
Worked Example: Polynomial
For , compute extrema.
Critical points: x=-1, x=3.
→ local maximum at x=-1; → local minimum at x=3.
Values: f(-1)=10, f(3)=-22.
Optimization Modeling
Fence Problem (No Roof)
Perimeter 20 m, three sides of a rectangle (one side along a wall). Maximize area for. Critical → maximum area 50 m².
Box with Square Base (No Lid)
Surface area 108 m², base , height h. Constraint →; volume maximized at x=6, h=3, V=108.
Guided Practice
Polynomial Mix
- For f(x)=x^4-4x^2+1, find intervals of increase/decrease.
- Classify critical points and identify inflection points.
- Sketch a qualitative graph with extrema and concavity.
Rational Function
- For g(x)=(x+1)/(x^2+1), locate extrema and concavity.
- Determine asymptotic behavior and discuss global extrema.
- Explain why no vertical asymptotes occur.
Exponential-Log
- For , find absolute max on .
- For (), locate critical points and nature.
- Compare growth/decay impact on extrema.
Trigonometric
- For q(x)=sin x - 1/2 sin 2x on [0,2π], find extrema.
- Discuss concavity intervals using q''(x).
- Interpret geometric meaning on the unit circle.
Mixed Optimization Cases
Packaging Design
Minimize surface area of a closed cylinder of fixed volume V. Derive r and h for minimal material and discuss manufacturability.
- Express h in terms of r using volume constraint.
- Differentiate surface area S(r) and find the minimizer.
- Conclude r:h ratio and practical implications.
Logistics Path
Minimize travel time across road (speed v1) and field (speed v2) between two points separated by a river bank.
- Model time as function of landing point x; set domain.
- Compute derivative and solve for optimal x.
- Discuss sensitivity when v2→v1 and v2≪v1.
Proceed to rates of change and advanced optimization in Lesson 6-3.
Theorems & Proof Sketches
Fermat’s Theorem
If f has a local extremum at c and is differentiable at c, then f'(c)=0.
Sketch: Compare one-sided difference quotients; signs must be ≥0 and ≤0.
Second Derivative Test
If f'(c)=0 and f''(c)≠0, the sign of f''(c) determines local min/max.
Sketch: Use Taylor expansion to second order near c.
Extended Examples & Case Studies
Case A: Quartic with Two Wells
- Find f'(x) and solve for candidates
- Use sign chart to classify local extrema
- Check inflection via f''(x)
- Summarize with a qualitative sketch
Case B: Rational with Hole
- Determine domain and removable discontinuity
- Locate critical points excluding the hole
- Classify using first/second derivative tests
- State limits near the hole and behavior at infinity
Case C: Exponential-Log Mix
- Differentiate and solve f'(x)=0 numerically
- Discuss existence/uniqueness of the maximizer
- Check concavity to confirm classification
- Report extremum value with units
Practice Bank
Bank A: First Derivative Test
- Build sign chart around candidates
- Identify increase/decrease intervals
- State local extrema with coordinates
Bank B: Second Derivative Test
- Compute f'' at each candidate
- Classify min/max/undetermined
- Handle f''=0 via monotonicity
Bank C: Closed Interval
- Evaluate endpoints
- Compare with interior candidates
- Report absolute extrema with units
Bank D: Modeling
- Form objective+constraints from context
- Reduce to single variable
- Solve and justify feasibility
FAQ (Extended)
Q: When do first- and second-derivative tests disagree?
When f''(c)=0. Use sign changes of f' or higher-order/monotonicity analysis.
Q: Absolute vs local extrema?
Absolute extrema are best on the whole domain; locals are best in a neighborhood. On closed intervals, compare endpoints and critical points.
Additional Projects
Project A: Cost with Volume Constraint
- Form cost function from material prices
- Use volume constraint to eliminate a variable
- Minimize and interpret sensitivity
Project B: Travel Time Minimization
- Model time with different media speeds
- Differentiate and solve for optimal split
- Discuss limiting cases v2→v1 and v2≪v1
Project C: Revenue Management
- Price–demand model R(x)=x·p(x)
- Find optimal quantity and price
- Check concavity and units
Problem Set
Proof & Theory
- State and explain Fermat’s Theorem on stationary points.
- Relate Rolle’s Theorem and MVT to optimization logic.
- Discuss when second derivative test fails and alternatives.
Applications
- Revenue management with price elasticity assumptions.
- Material minimization under strength constraints.
- Travel-time minimization with piecewise speeds.