Optimization

Finding the best possible outcome is something you do every day without even thinking about it. You look for the shortest route to work, the best price for a product, or the fastest way to complete a task. Mathematicians call this optimization, and it turns out that calculus gives you a systematic way to find these “best” values with precision.

The key insight is simple: at maximum and minimum points, curves flatten out. Think about climbing a hill - at the very top, the ground is momentarily level before it starts going back down. Derivatives measure slope, and when the slope equals zero, you have found a candidate for a maximum or minimum. This connection between zeros of derivatives and extreme values is the engine that powers all optimization in calculus.

If the idea of word problems makes you nervous, take a breath. Optimization problems do follow a pattern, and once you see the pattern, they become much more approachable. By the end of this lesson, you will have a systematic strategy for tackling these problems, and you will see why calculus is such a powerful tool for making decisions in the real world.

Core Concepts

What Is Optimization?

Optimization is the process of finding the maximum or minimum value of a function. In practical terms, you are looking for the best possible outcome given certain constraints - the largest area, the smallest cost, the shortest time, or the maximum profit.

Calculus transforms optimization from guesswork into a precise procedure. Instead of trying different values and hoping for the best, you can use derivatives to locate exactly where maximum and minimum values occur.

Local vs. Absolute Extrema

Before diving into techniques, you need to understand the distinction between two types of extreme values.

A local maximum (also called a relative maximum) is a point where the function value is higher than at all nearby points. Think of it as the top of a hill - you are at the highest point in your immediate neighborhood, even if there are taller mountains elsewhere.

A local minimum (also called a relative minimum) is a point where the function value is lower than at all nearby points. This is the bottom of a valley.

An absolute maximum (also called a global maximum) is the highest value the function achieves on its entire domain. This is the tallest peak of all.

An absolute minimum (also called a global minimum) is the lowest value the function achieves on its entire domain. This is the deepest valley of all.

A function can have many local maxima and minima, but on a given domain, it can have at most one absolute maximum value and one absolute minimum value (though these values might be achieved at multiple points).

Critical Points: Where Extrema Can Occur

A critical point of a function $f$ is a value $x = c$ in the domain of $f$ where either:

1. $f’(c) = 0$, or
2. $f’(c)$ does not exist (but $f(c)$ does exist)

Critical points are important because of this fundamental theorem:

Fermat’s Theorem: If $f$ has a local maximum or minimum at $c$, and if $f’(c)$ exists, then $f’(c) = 0$.

In other words, local extrema can only occur at critical points. This does not mean every critical point is an extremum - some critical points are neither maxima nor minima. But if you are hunting for extrema, critical points are the only places you need to look (along with endpoints, as we will see).

Why does this work? At a local maximum, the function is increasing as you approach from the left (positive slope) and decreasing as you leave toward the right (negative slope). The only way the slope can transition from positive to negative continuously is by passing through zero. The same logic applies to local minima with the signs reversed.

The First Derivative Test

Once you have found the critical points, how do you determine which are maxima, which are minima, and which are neither? The First Derivative Test answers this by examining the sign of $f’$ around each critical point.

For a critical point at $x = c$:

1. Local Maximum: If $f’(x) > 0$ for $x < c$ (function increasing) and $f’(x) < 0$ for $x > c$ (function decreasing), then $f$ has a local maximum at $c$.

2. Local Minimum: If $f’(x) < 0$ for $x < c$ (function decreasing) and $f’(x) > 0$ for $x > c$ (function increasing), then $f$ has a local minimum at $c$.

3. Neither: If $f’(x)$ has the same sign on both sides of $c$, then $c$ is neither a local max nor a local min. This often indicates an inflection point.

The first derivative test makes intuitive sense: a maximum occurs where the function rises and then falls, while a minimum occurs where it falls and then rises.

The Second Derivative Test

There is an alternative method that is often faster when the second derivative is easy to compute. The Second Derivative Test uses the concavity of the function to classify critical points.

For a critical point at $x = c$ where $f’(c) = 0$:

1. If $f’’(c) > 0$, then $f$ has a local minimum at $c$. (The graph is concave up - shaped like a cup that holds water.)

2. If $f’’(c) < 0$, then $f$ has a local maximum at $c$. (The graph is concave down - shaped like a cap that sheds water.)

3. If $f’’(c) = 0$, the test is inconclusive. You must use the First Derivative Test instead.

Why does this work? If $f’’(c) > 0$, the function is concave up at $c$, meaning the graph curves upward like a smile. A horizontal tangent line ($f’(c) = 0$) on a concave-up curve must be at the bottom of the curve - a minimum. The opposite reasoning applies when $f’’(c) < 0$.

The Closed Interval Method

When you need to find the absolute maximum and minimum values of a continuous function on a closed interval $[a, b]$, there is a straightforward procedure called the Closed Interval Method:

1. Find all critical points of $f$ in the open interval $(a, b)$.
2. Evaluate $f$ at each critical point.
3. Evaluate $f$ at the endpoints $a$ and $b$.
4. The largest value from steps 2-3 is the absolute maximum; the smallest is the absolute minimum.

This method works because of the Extreme Value Theorem: a continuous function on a closed interval $[a, b]$ is guaranteed to achieve both an absolute maximum and an absolute minimum somewhere on that interval. These extrema must occur either at critical points or at endpoints - there are no other options.

The closed interval method is particularly clean because you do not need to classify each critical point. You simply evaluate the function at all candidate locations and compare the values.

Applied Optimization: Setting Up the Problem

The real power of optimization emerges in applied problems - finding the best dimensions for a container, the maximum profit for a business, or the minimum material needed for construction. These problems require translating a real-world situation into mathematical functions.

Here is a systematic strategy for solving applied optimization problems:

Step 1: Understand the Problem Read the problem carefully. Identify what quantity you want to maximize or minimize. Draw a diagram if applicable. Assign variables to unknown quantities.

Step 2: Write the Objective Function Express the quantity to be optimized as a function of one or more variables. This is called the objective function. For example, if you want to maximize area, write an expression for area in terms of your variables.

Step 3: Identify Constraints Find any equations that relate your variables. These are called constraints. Common constraints include fixed perimeter, fixed volume, fixed budget, or fixed amount of material.

Step 4: Reduce to a Single Variable Use the constraint equation(s) to eliminate variables from the objective function until you have a function of a single variable. This is crucial - you need a function of one variable to use single-variable calculus.

Step 5: Determine the Domain Identify the realistic values your variable can take. Physical constraints often impose restrictions (lengths must be positive, quantities cannot exceed a supply limit, etc.).

Step 6: Find Critical Points and Endpoints Take the derivative of your objective function, set it equal to zero, and solve. Also consider any points where the derivative does not exist, and check the endpoints of your domain.

Step 7: Determine the Optimal Value Use the first derivative test, second derivative test, or closed interval method to identify which critical point gives the maximum or minimum you seek.

Step 8: Answer the Question and Verify Make sure you answer what the problem asked for. If it asked for dimensions, give dimensions. If it asked for maximum area, give the area. Always verify that your answer makes sense in the context of the problem.

Common Objective Functions and Constraints

Certain combinations appear repeatedly in optimization problems:

Geometry Problems:

• Area of rectangle: $A = lw$
• Perimeter of rectangle: $P = 2l + 2w$
• Volume of box: $V = lwh$
• Surface area of box: $S = 2lw + 2lh + 2wh$
• Volume of cylinder: $V = \pi r^2 h$
• Surface area of cylinder: $S = 2\pi r^2 + 2\pi rh$

Distance Problems:

• Distance between points: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
• Often easier to minimize distance squared: $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2$

Economics Problems:

• Revenue: $R = (\text{price}) \times (\text{quantity})$
• Profit: $P = R - C$ (revenue minus cost)
• Cost often has fixed and variable components

Notation and Terminology

Term	Meaning	Example
Critical point	A value $c$ where $f’(c) = 0$ or $f’(c)$ is undefined	For $f(x) = x^3 - 3x$, critical points are $x = \pm 1$
Local maximum	Point higher than all nearby points	The top of a hill
Local minimum	Point lower than all nearby points	The bottom of a valley
Absolute maximum	Highest value on the entire domain	The tallest peak overall
Absolute minimum	Lowest value on the entire domain	The deepest point overall
Objective function	The function being optimized	Area $A = xy$ to maximize
Constraint	A restriction relating variables	Perimeter fixed: $2x + 2y = 100$
Closed interval	An interval including both endpoints	$[0, 5]$
Open interval	An interval excluding endpoints	$(0, 5)$

Examples

Key Properties and Rules

Finding Critical Points

1. Compute $f’(x)$.
2. Find all values where $f’(x) = 0$.
3. Find all values where $f’(x)$ does not exist (but $f(x)$ does).
4. These are your critical points.

First Derivative Test Summary

Sign of $f’$ before $c$	Sign of $f’$ after $c$	Conclusion at $x = c$
Positive ($+$)	Negative ($-$)	Local maximum
Negative ($-$)	Positive ($+$)	Local minimum
Positive ($+$)	Positive ($+$)	Neither (increasing through $c$)
Negative ($-$)	Negative ($-$)	Neither (decreasing through $c$)

Second Derivative Test Summary

For a critical point $c$ where $f’(c) = 0$:

Value of $f’’(c)$	Conclusion
$f’’(c) > 0$	Local minimum (concave up)
$f’’(c) < 0$	Local maximum (concave down)
$f’’(c) = 0$	Test inconclusive; use First Derivative Test

Closed Interval Method Summary

To find absolute extrema of a continuous function $f$ on $[a, b]$:

1. Find all critical points in $(a, b)$.
2. Evaluate $f$ at critical points and endpoints.
3. Largest value = absolute maximum.
4. Smallest value = absolute minimum.

Applied Optimization Strategy

1. Understand: Read carefully, draw diagrams, assign variables.
2. Objective: Write the quantity to optimize as a function.
3. Constraint: Find equations relating variables.
4. Reduce: Use constraints to get a single-variable function.
5. Domain: Determine valid values from physical constraints.
6. Solve: Find critical points, test endpoints.
7. Classify: Determine max or min using derivative tests.
8. Answer: State the solution in context and verify.

Common Mistakes to Avoid

• Forgetting endpoints: On a closed interval, extrema can occur at endpoints.
• Accepting all critical points: Not every critical point is an extremum - you must test.
• Wrong domain: Physical problems often have restrictions (positive lengths, etc.).
• Optimizing the wrong thing: Make sure your objective function matches what the problem asks.
• Not verifying: Always check that your answer makes physical sense.

Real-World Applications

Business and Economics

Maximizing Profit: Companies use optimization to determine production levels. If profit $P(x)$ depends on the quantity $x$ produced, finding where $P’(x) = 0$ identifies the optimal production level.

Pricing Strategies: As seen in Example 8, finding the price that maximizes revenue involves balancing higher prices (more revenue per unit) against lower sales volume.

Minimizing Cost: Manufacturing processes are optimized to minimize material costs while meeting production requirements.

Engineering and Design

Optimal Container Design: As shown in Example 7, engineers design containers (cans, boxes, tanks) to minimize material usage while holding a required volume. This saves money and reduces environmental impact.

Structural Engineering: Beams and bridges are designed to maximize strength while minimizing weight. This involves finding optimal cross-sectional shapes and material distributions.

Heat Dissipation: Electronic components are designed with optimal surface areas to maximize heat transfer and prevent overheating.

Transportation and Logistics

Shortest Path Problems: Finding the quickest route between locations is an optimization problem. GPS navigation systems solve these continuously.

Fuel Efficiency: Airlines and shipping companies optimize routes and speeds to minimize fuel consumption.

Medicine and Biology

Drug Dosage: Determining optimal medication doses involves maximizing therapeutic effect while minimizing side effects.

Population Models: Ecologists use optimization to understand sustainable harvest levels for fish populations and forest resources.

Personal Finance

Investment Portfolios: Financial optimization balances risk and return to find the best allocation of investments.

Mortgage Decisions: Choosing between different loan options involves optimizing total cost over time.

Self-Test Problems

Summary

• Optimization uses calculus to find maximum and minimum values of functions.

• Critical points occur where $f’(x) = 0$ or where $f’(x)$ is undefined. Local extrema can only occur at critical points.

• The First Derivative Test classifies critical points by examining the sign of $f’$ on either side:

  • Sign change from $+$ to $-$: local maximum
  • Sign change from $-$ to $+$: local minimum

• The Second Derivative Test uses concavity to classify critical points where $f’(c) = 0$:

  • $f’’(c) > 0$: local minimum (concave up)
  • $f’’(c) < 0$: local maximum (concave down)

• The Closed Interval Method finds absolute extrema by evaluating $f$ at all critical points and endpoints, then comparing values.

• Applied optimization problems follow a systematic strategy:

  1. Identify the objective function (what to optimize)
  2. Identify constraints (restrictions on variables)
  3. Use constraints to reduce to a single-variable function
  4. Find and test critical points
  5. Verify the answer makes sense

• Common applications include minimizing cost, maximizing profit, finding optimal dimensions for containers, determining best prices, and finding shortest paths.

• Always verify that your solution satisfies all constraints and makes sense in the physical context of the problem.

Optimization is where calculus shows its true practical power. The techniques you have learned here are used daily by engineers, economists, scientists, and analysts to make better decisions. When you encounter a problem asking for “the best,” “the most,” or “the least,” you now have the tools to find it precisely.