Optimization Problems & 5-Step Solution Method

Optimization Problems Overview

Optimization seeks the absolute maximum or minimum of a quantity.
- Absolute (global) extrema = highest/lowest values a function can attain.
- In real‐world models we seldom study a function “in a vacuum.”
Typical goals:
- Maximize profits, speed, enclosed area, efficiency, revenue.
- Minimize cost, material usage, waste, loss, decay rate.
Clues in problem wording: “best,” “least,” “most,” “fastest,” “slowest,” “highest,” “lowest,” “maximum,” “minimum.” The presence of superlatives → likely an optimization setup.
Optimization problems are usually constrained: limited resources, fixed budgets, physical bounds, etc.
Core calculus connection: extrema are found by locating critical points of a function (derivative =0 or undefined) and checking endpoints when the domain is closed.

Identify Quantities & Restrictions
- List every relevant measurable quantity.
- Pin down explicit constraints (length of wire, budget cap, physical laws, …).
- Decide which single quantity you must optimize (area, volume, cost,…).
Choose Independent Variables
- Select variables allowed to vary freely.
- Typical independent variables: lengths, widths, production levels, time, temperature.
- Non-controllable figures (e.g., unit price of materials) are parameters rather than variables.
Write Relationships
- Use geometry, physics, chemistry, economics, or problem-supplied formulas.
- Express every quantity (including constraints) in terms of the chosen variables.
Form a Single-Variable Function & Domain
- Substitute constraints to eliminate extra variables, ending with one variable.
- Determine an appropriate domain (usually closed interval):
  - Physical realism: lengths \ge0, masses \ge0.
  - Constraint substitution may bound the variable.
Apply Calculus to Find Extrema
- Compute derivative; set f'(x)=0 (critical numbers) or note where f'(x) undefined.
- Evaluate objective function at:
1. Each critical number within the domain.
2. Each domain endpoint.
- Pick the largest (or smallest) resulting value, according to the problem.

Couple owns 100 m of wire fencing.
Fence will form three sides of a rectangle; the fourth side is an existing building (no fence needed there).
Objective: maximize enclosed area.

Quantities:
- Fence side along the building → no material.
- Remaining three sides → two equal widths x and one length y.
- Area to maximize: A.
Constraint: available fencing = 100\text{ m}.

Solve the constraint for one variable, e.g.
y = 100 - 2x
Substitute into area formula:
A(x) = x(100 - 2x) = 100x - 2x^2
Domain: physical lengths x, y \ge 0.
- x \ge 0.
- y=100-2x \ge 0 \implies 0 \le x \le 50.
- Therefore x \in [0,50].

Derivative:
A'(x) = 100 - 4x
Critical number:
A'(x)=0 \Rightarrow 100-4x=0 \Rightarrow x=25
Evaluate A(x) at critical number and endpoints:
- x=0:\; A=100(0)-2(0)^2=0
- x=25:\; A=100(25)-2(25)^2 = 2500 - 1250 = 1250\text{ m}^2
- x=50:\; A=100(50)-2(50)^2 = 5000 - 5000 = 0
Absolute maximum = 1250\text{ m}^2 at x=25\text{ m},\; y=100-2(25)=50\text{ m}.

Endpoint areas are 0 because choosing x=0 (no width) or x=50 (no length) degenerates the garden.
The optimal shape uses half the fence for the long side (50 m) and splits the remaining half equally (25 m each) for the two widths.
This matches intuitive patterns: for rectangles with a given perimeter, a more “square-ish” shape maximizes area. The building side plays the role of the fourth 50 m.

Many optimization problems reduce to turning a multi-variable situation into a single-variable calculus problem through substitution.
Constraints provide necessary linkage; without them the quantity often has no finite maximum or minimum (e.g.
unlimited fence → unbounded area).
Always interpret endpoints physically. Zero area at an endpoint often reveals a degenerate configuration.
The geometric idea that, for fixed perimeter, a square maximizes area leads here to a rectangle with the side ratio y : x = 2 : 1 because only three sides consume perimeter.
Method adapts easily to:
- Minimizing cost with different material prices (introduce separate cost coefficients in the perimeter equation).
- Volume problems (boxes without tops, etc.).
- Time/velocity trade-offs in physics.

Resource limitations (only 100 m of fence) are realistic constraints; ignoring them could waste material or money.
Optimization helps households/businesses use materials efficiently, reducing environmental impact of excess production.

Constraint equation: 100 = 2x + y linear in two variables.
Objective function after substitution: quadratic A(x)=100x-2x^2.
Derivative zero gives critical point x=25.
Closed interval [0,50] ensures Extreme Value Theorem applies → guaranteed absolute max/min.

Memorize the 5-step framework; practice mapping word descriptions to variables & equations.
Clearly distinguish between independent variables, parameters, constraints, and objective function.
Look for physical bounds to craft a closed domain; do not rely solely on algebraic domain.
Derivative test + endpoint checks = reliable method for 1-variable absolute extrema.
Interpret final answers contextually (units, feasibility, reasonableness).