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Optimization Solution Methods

Finding and Verifying Extreme Values

You've set up the optimization problem: an objective function $f(x)$ on some domain. Now comes the payoff—finding where $f$ achieves its maximum or minimum.

The method you use depends on the domain. A closed interval $[a, b]$ allows the Closed Interval Method. An open or infinite domain requires the First Derivative Test for Absolute Extrema. Sometimes, the geometry of the problem makes the answer obvious.

Choosing the right verification method saves time and prevents errors.

Prerequisite Map

Quick Reference

Method Selection: A Decision Tree

graph TD
    Q1{"Is the domain a<br/>closed interval [a,b]?"}
    Q1 -->|Yes| M1["Use Closed<br/>Interval Method"]
    Q1 -->|No| Q2{"Only one<br/>critical point?"}

    Q2 -->|Yes| M2["Use First Derivative Test<br/>for Absolute Extrema"]
    Q2 -->|No| M3["Analyze behavior<br/>at all critical points<br/>and endpoints/limits"]

    style M1 fill:#e0f2fe
    style M2 fill:#fef3c7
    style M3 fill:#fce7f3

Method 1: Closed Interval Method

When to use: The domain is a closed interval $[a, b]$.

Procedure:

1. Find all critical numbers in $(a, b)$
2. Evaluate $f$ at each critical number
3. Evaluate $f$ at the endpoints $a$ and $b$
4. The largest value is the absolute maximum; the smallest is the absolute minimum

Why it works: The Extreme Value Theorem guarantees that a continuous function on a closed interval attains both a maximum and minimum. These must occur at critical points or endpoints.

Example: Fencing Problem

From the setup skill, we found: $$A(x) = 2400x - 2x^2, \quad 0 \le x \le 1200$$

Step 1: Find critical numbers. $$A'(x) = 2400 - 4x = 0 \Rightarrow x = 600$$

Step 2: Evaluate at critical number. $$A(600) = 2400(600) - 2(600)^2 = 1{,}440{,}000 - 720{,}000 = 720{,}000$$

Step 3: Evaluate at endpoints. $$A(0) = 0, \quad A(1200) = 2400(1200) - 2(1200)^2 = 0$$

Step 4: Compare values.

The absolute maximum area is $720{,}000$ ft² at $x = 600$ ft.

Method 2: First Derivative Test for Absolute Extrema

When to use: The domain is not closed (open interval, half-open, or unbounded) AND there is exactly one critical number.

The Test:

Suppose $c$ is the only critical number of a continuous function $f$ on an interval.

Why it works: If $f$ is increasing before $c$ and decreasing after, then $f(c)$ is the highest point—and with only one critical number, there's nowhere else for a higher value to occur.

Example: Minimum Surface Area

A cylindrical can must hold 1 liter (1000 cm³). The surface area is: $$S(r) = 2\pi r^2 + \frac{2000}{r}, \quad r > 0$$

Find the critical number: $$S'(r) = 4\pi r - \frac{2000}{r^2} = \frac{4\pi r^3 - 2000}{r^2}$$

Setting $S'(r) = 0$: $$4\pi r^3 = 2000 \Rightarrow r^3 = \frac{500}{\pi} \Rightarrow r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42 \text{ cm}$$

Verify it's a minimum:

• For $r < \sqrt[3]{500/\pi}$: $S'(r) < 0$ (decreasing)
• For $r > \sqrt[3]{500/\pi}$: $S'(r) > 0$ (increasing)

Pattern: negative → positive at critical point → absolute minimum.

Method 3: Domain Behavior Argument

When to use: When the physics or geometry makes the answer obvious, or when endpoint behavior forces an extremum.

The Argument: If $f(x) \to \infty$ as $x$ approaches both ends of the domain, then any local minimum must be an absolute minimum.

If $f(x) \to 0$ or $f(x) \to -\infty$ at both ends, then any local maximum must be an absolute maximum.

Example: Can Problem Revisited

For $S(r) = 2\pi r^2 + \frac{2000}{r}$ on $(0, \infty)$:

• As $r \to 0^+$: $S(r) \to \infty$ (the $2000/r$ term dominates)
• As $r \to \infty$: $S(r) \to \infty$ (the $2\pi r^2$ term dominates)

Since $S \to \infty$ at both ends and $S$ is continuous, there must be a minimum somewhere in between. With only one critical point, that's where the minimum occurs.

Method 4: Second Derivative Confirmation

When to use: As a quick check, especially when you want to confirm concavity at the critical point.

If $f'(c) = 0$ and:

• $f''(c) < 0$: local maximum at $c$
• $f''(c) > 0$: local minimum at $c$

Warning: This only confirms local extrema. You still need to argue that the local extremum is absolute (using Methods 2 or 3).

Example: Fencing Problem

$$A(x) = 2400x - 2x^2$$ $$A''(x) = -4 < 0 \text{ for all } x$$

Since $A'' < 0$ everywhere, the function is concave down. The single critical point at $x = 600$ must be an absolute maximum.

Comparison of Methods

Practice Problems

    A ──────────────────────────────
    │╲
  2 │  ╲
    │    ╲
    ├─────●───────────────────● B
    C     D                   (6 km from C)

CCI-Style Conceptual Questions

Mastery Checklist

• [ ] I can apply the Closed Interval Method on $[a, b]$
• [ ] I can state and apply the First Derivative Test for Absolute Extrema
• [ ] I can justify an absolute extremum using domain behavior arguments
• [ ] I can choose the appropriate method based on the domain type
• [ ] I can verify my answer makes physical sense in context

Mental Model

The Landscape Metaphor:

Imagine the graph of $f(x)$ as a hiking trail along a mountain ridge.

• Closed interval: You must start at the trailhead $a$ and end at $b$. The highest point is either at a peak (critical point) or at one of the trailheads.

• Open/infinite interval: The trail extends forever. If there's only one peak and the trail goes down to valleys on both sides, that peak is the highest point—guaranteed.

The domain tells you the boundaries of your hike. The derivative tells you where the peaks are.

Connections

Looking back:

• Optimization Problem Setup prepares the function to be analyzed
• Closed Interval Method (§3.1) and First Derivative Test (§3.3) provide the core techniques

Looking ahead:

• Business Optimization applies these methods to economics
• Related Rates (§3.9) uses similar translation skills but for rates instead of extrema

Real-world connections:

• Engineers verify designs by checking that critical points give minima for stress/cost
• The "only one critical point" condition often holds in physical problems due to natural constraints

Last updated: 2026-01-22