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Unit 1.3: Problem Formulation

Lesson 3 of 22 in the free Engineering Optimization notes on Siksha Sarovar, written by Rohit Jangra.

Unit 1.3: Optimal Problem Formulation

Formulating a problem is often more difficult than solving it. It involves translating a descriptive physical problem into a purely mathematical model.

Step-by-Step Procedure

  1. Identify Design Variables: What can you change? (e.g., Length, Width).
  2. Identify the Objective: What is the goal? (e.g., Minimize Cost). Express it in terms of variables.
  3. Identify Constraints: What are the rules? (e.g., Max Stress, Min Volume).
  4. Set Bounds: What are the limits? (e.g., Dimensions > 0).

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Detailed Example 1: Design of a Soft Drink Can

Problem: Design a closed cylindrical soft drink can to hold a volume of V = 330 ml, such that the material used (Total Surface Area) is minimized.

1. Design Variables:

  • Radius: r
  • Height: h
  • Vector: X = [r, h]ᵀ

2. Objective Function:

  • Minimizing material means minimizing Surface Area (A).
  • Top/Bottom Area: 2 * πr²
  • Side Area: 2πrh
  • Minimize f(r, h) = 2πr² + 2πrh

3. Constraints:

  • Volume must be exactly 330 ml.
  • Volume formula: V = πr²h
  • Eq Constraint: πr²h = 330 => πr²h - 330 = 0

4. Bounds:

  • Radius and height cannot be negative.
  • r > 0, h > 0

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Detailed Example 2: Rectangular Garden Fencing

Problem: You have 100 meters of fencing wire. You want to make a rectangular garden that has the maximum possible area.

1. Design Variables:

  • Length: x
  • Width: y

2. Objective Function:

  • Maximize Area A = x * y
  • (Or Minimize f(x, y) = -xy)

3. Constraints:

  • Total perimeter must be 100m.
  • Perimeter P = 2x + 2y
  • Eq Constraint: 2x + 2y = 100 => x + y - 50 = 0

4. Bounds:

  • x > 0, y > 0