MathIsimple

Lesson 6-2: Derivatives – Extrema and Optimization

Learning Goals

  • Find critical points from f(x)=0f'(x)=0 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 f(x)f'(x) 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 f(c)=0f'(c)=0 and ff'' exists near c: f(c)<0f''(c)<0 implies a local maximum; f(c)>0f''(c)>0 implies a local minimum; f(c)=0f''(c)=0 is inconclusive.

Inflection Points

Points where concavity changes sign are inflection points. Often found by solving f(x)=0f''(x)=0 or whereff'' is undefined and confirming sign change.

Closed-Interval Absolute Extrema

  1. Find critical points in the open interval where f(x)=0f'(x)=0 or undefined.
  2. Evaluate ff at critical points and endpoints.
  3. Compare values: largest is absolute maximum, smallest is absolute minimum.

Worked Example: Polynomial

For f(x)=x33x29x+5f(x)=x^3-3x^2-9x+5, compute extrema.

f(x)=3x26x9=3(x3)(x+1)f'(x)=3x^2-6x-9=3(x-3)(x+1)

Critical points: x=-1, x=3.

f(x)=6x6f''(x)=6x-6

f(1)=12<0f''(-1)=-12<0 → local maximum at x=-1; f(3)=12>0f''(3)=12>0 → 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 S=x(202x)S=x(20-2x) for0<x<100<x<10. Critical x=5x=5 → maximum area 50 m².

Box with Square Base (No Lid)

Surface area 108 m², base ximesxx imes x, height h. Constraint x2+4xh=108x^2+4xh=108h=(108x2)/(4x)h=(108-x^2)/(4x); volume V=x2h=27xfrac14x3V=x^2h=27x- frac{1}{4}x^3 maximized at x=6, h=3, V=108.

Guided Practice

Polynomial Mix

  1. For f(x)=x^4-4x^2+1, find intervals of increase/decrease.
  2. Classify critical points and identify inflection points.
  3. Sketch a qualitative graph with extrema and concavity.

Rational Function

  1. For g(x)=(x+1)/(x^2+1), locate extrema and concavity.
  2. Determine asymptotic behavior and discuss global extrema.
  3. Explain why no vertical asymptotes occur.

Exponential-Log

  1. For h(x)=xexh(x)=xe^{-x}, find absolute max on [0,)[0,\infty).
  2. For p(x)=xlnxp(x)=x\ln x (x>0x>0), locate critical points and nature.
  3. Compare growth/decay impact on extrema.

Trigonometric

  1. For q(x)=sin x - 1/2 sin 2x on [0,2π], find extrema.
  2. Discuss concavity intervals using q''(x).
  3. 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.

  1. Express h in terms of r using volume constraint.
  2. Differentiate surface area S(r) and find the minimizer.
  3. 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.

  1. Model time as function of landing point x; set domain.
  2. Compute derivative and solve for optimal x.
  3. 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

  1. Find f'(x) and solve for candidates
  2. Use sign chart to classify local extrema
  3. Check inflection via f''(x)
  4. Summarize with a qualitative sketch

Case B: Rational with Hole

  1. Determine domain and removable discontinuity
  2. Locate critical points excluding the hole
  3. Classify using first/second derivative tests
  4. State limits near the hole and behavior at infinity

Case C: Exponential-Log Mix

  1. Differentiate and solve f'(x)=0 numerically
  2. Discuss existence/uniqueness of the maximizer
  3. Check concavity to confirm classification
  4. Report extremum value with units

Practice Bank

Bank A: First Derivative Test

  1. Build sign chart around candidates
  2. Identify increase/decrease intervals
  3. State local extrema with coordinates

Bank B: Second Derivative Test

  1. Compute f'' at each candidate
  2. Classify min/max/undetermined
  3. Handle f''=0 via monotonicity

Bank C: Closed Interval

  1. Evaluate endpoints
  2. Compare with interior candidates
  3. Report absolute extrema with units

Bank D: Modeling

  1. Form objective+constraints from context
  2. Reduce to single variable
  3. 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.