Optimization with Derivatives: Word Problems

Overview

• Topic: Applications of Differential Calculus — Maxima and Minima (word problems).
• Focus: Step-by-step method for solving optimization problems using derivatives.
• Examples covered: (1) Maximize x^2 y given x + y = 300, (2) Maximize volume of an open-top box from 3"×8" cardboard, (3) Maximize area of rectangle from 16 ft wire.

Steps For Solving Maxima/Minima Word Problems

• Draw a diagram if necessary to relate variables visually.
• Write an equation representing the quantity to maximize or minimize.
• Use relationships between variables to express the quantity as a single-variable function.
• Differentiate the function and set the derivative to zero to find critical points.
• Use the second derivative test to classify critical points as maxima or minima.
• Check endpoints or constraints if variables are restricted (e.g., non-negative).

Example 1: Maximize x^2 y Subject To x + y = 300

• Problem: Find non-negative x, y with x + y = 300 that maximize f = x^2 y.
• Constraint: y = 300 − x.
• Single-variable function: f(x) = x^2(300 − x) = 300x^2 − x^3.
• First derivative: f'(x) = 600x − 3x^2.
• Solve f'(x) = 0: 3x(200 − x) = 0 → x = 0 or x = 200.
• Second derivative: f''(x) = 600 − 6x.
• Classify:
  • f''(0) = 600 > 0 → x = 0 is a local minimum.
  • f''(200) = 600 − 1200 = −600 < 0 → x = 200 is a local maximum.
• Corresponding y: y = 300 − 200 = 100.
• Answer: Numbers are x = 200 and y = 100 (maximum of x^2 y).

Example 2: Maximize Volume Of Open-Top Box From 3"×8" Cardboard

• Problem: Cut equal squares of side x from each corner of a 3"×8" rectangle, fold sides to form open-top box. Find x giving maximum volume.
• Dimensions after cutting:
  • Length = 8 − 2x.
  • Width = 3 − 2x.
  • Height = x.
• Volume function: V(x) = x(8 − 2x)(3 − 2x).
• Expanded: V(x) = 4x^3 − 22x^2 + 24x.
• First derivative: V'(x) = 12x^2 − 44x + 24.
• Simplify by dividing by 4: 3x^2 − 11x + 6 = 0.
• Factor/solve: (3x − 2)(x − 3) = 0 → x = 2/3 or x = 3.
• Validity: x must satisfy 0 < x < min(3/2, 8/2)=1.5, so x = 3 is invalid (too large).
• Second derivative: V''(x) = 24x − 44.
• Evaluate at x = 2/3: V''(2/3) = 24(2/3) − 44 = 16 − 44 = −28 < 0 → local maximum.
• Answer: Cut squares of side x = 2/3 inch.

Example 3: Maximize Area Of Rectangle Formed From 16 ft Wire

• Problem: A 16-foot wire forms a rectangle. Find dimensions maximizing area.
• Let width = x, length = y.
• Perimeter constraint: 2(x + y) = 16 → x + y = 8 → y = 8 − x.
• Area function: A(x) = x(8 − x) = 8x − x^2.
• First derivative: A'(x) = 8 − 2x.
• Solve A'(x) = 0 → 8 − 2x = 0 → x = 4.
• Second derivative: A''(x) = −2 < 0 → x = 4 gives a maximum.
• Corresponding y: y = 8 − 4 = 4.
• Answer: Rectangle is 4 ft by 4 ft (a square) for maximal area.

Key Terms And Definitions

• Critical Point: x where f'(x) = 0 or f' undefined.
• Second Derivative Test:
  • If f''(c) > 0 → local minimum at c.
  • If f''(c) < 0 → local maximum at c.
• Constraint Equation: Relationship between variables used to reduce function to one variable.
• Valid/Feasible Solution: Critical point must satisfy original problem constraints (non-negativity, dimensions).

Summary Table: Examples And Results

Action Items / Next Steps

• Practice: Solve additional optimization problems using the six-step approach.
• Check Constraints: Always verify critical points are within feasible domain.
• Visualize: Draw diagrams to understand variable relationships before algebraic work.
• Verify: Use endpoints or physical constraints when domain is limited.