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Solving
Optimization Problems

Master the art of finding maximum and minimum values using derivatives: from setting up constraints to identifying critical points and confirming extrema.

AP Calculus BC

What is Optimization?

Optimization is the process of finding the greatest or least value that a function can achieve, subject to certain conditions. In calculus, we use derivatives to locate these extreme values efficiently.

Maximize
Find the largest value: profit, area, volume, efficiency
Minimize
Find the smallest value: cost, distance, material, time

The Core Idea: Extrema and Derivatives

At a local maximum or minimum on a smooth curve, the tangent line is horizontal: \( f'(x) = 0 \). These are critical points. Endpoints of a closed interval can also be extrema.

x y local max local min \( f'=0 \) \( f'=0 \)

Key Vocabulary

Objective Function
The quantity you want to maximize or minimize, expressed as \( f(x) \).
Constraint
A condition that limits the variables, e.g. a fixed perimeter or budget.
Critical Point
A value \( x = c \) where \( f'(c) = 0 \) or \( f'(c) \) is undefined.

The Closed Interval Method

If \( f \) is continuous on a closed interval \([a, b]\), then the Extreme Value Theorem guarantees that \( f \) attains both an absolute maximum and an absolute minimum on \([a, b]\).

1
Find all critical points of \( f \) in \((a, b)\) by solving \( f'(x) = 0 \) and checking where \( f' \) is undefined.
2
Evaluate \( f \) at each critical point and at both endpoints \( x = a \) and \( x = b \).
3
Compare values: the largest is the absolute max; the smallest is the absolute min.

First vs Second Derivative Test

First Derivative Test Second Derivative Test
Requirement \( f'(c) = 0 \) \( f'(c) = 0 \)
Check Sign change of \( f' \) around \( c \) Sign of \( f''(c) \)
Local Max \( f' \) changes + to − \( f''(c) < 0 \)
Local Min \( f' \) changes − to + \( f''(c) > 0 \)
Inconclusive No sign change \( f''(c) = 0 \)

Open Domain vs Closed Interval

Closed Interval \([a,b]\)
Extreme Value Theorem applies. Check critical points and endpoints. Absolute extrema guaranteed.
Open / Unbounded Domain
No endpoints to check. Use the first or second derivative test. If only one critical point exists and it is a local extremum, it is the absolute extremum.

The 5-Step Strategy

1
Draw a diagram and label all quantities. Identify what to maximize or minimize.
2
Write the objective function in terms of two or more variables.
3
Use the constraint to eliminate variables until the objective is in one variable.
4
Differentiate, set \( f'(x) = 0 \), and solve for critical points.
5
Confirm the extremum (first/second derivative test or closed-interval check) and answer the question.

An open box taking shape on a designer's table

A cardboard net with four corner squares removed, partly folded walls, and a finished open-top box on a worktable.For an open-top square-base box with 1,200 cm² of material, x² + 4xh = 1,200. Volume is largest at base side 20 cm and height 10 cm, giving 4,000 cm³.

A different constraint: material budget

For an open-top square-base box with 1,200 cm² of material, x² + 4xh = 1,200. Volume is largest at base side 20 cm and height 10 cm, giving 4,000 cm³.For an open-top square-base box with 1,200 cm² of material, x² + 4xh = 1,200. Volume is largest at base side 20 cm and height 10 cm, giving 4,000 cm³.

Example 1: Maximum Area

A farmer has 200 m of fencing to enclose a rectangular field along a river (no fence needed on the river side). Find the dimensions that maximize the area.

River (no fence) x y y

Example 1: Verification

Let's confirm that \( y = 50 \) gives a maximum using both tests.

Second Derivative Test
\( A''(y) = -4 < 0 \) for all \( y \). Since \( A''(50) < 0 \), the critical point is a local maximum.
Closed Interval Check
Domain: \( 0 \leq y \leq 100 \). Values: \( A(0) = 0 \), \( A(50) = 5000 \), \( A(100) = 0 \). Largest is 5000.

Example 2: Minimum Surface Area

An open-top box with a square base must have a volume of 32 cm\(^3\). Find the dimensions that minimize the surface area.

x x h open top

Example 3: Closest Point on a Curve

Find the point on the parabola \( y = x^2 \) that is closest to the point \( (3, 0) \).

x y (3, 0) (1, 1) 0 1 2 3 1 4 9 \( y = x^2 \)

Why Minimize \( D^2 \) Instead of \( D \)?

The distance formula involves a square root: \( D = \sqrt{(x-3)^2 + x^4} \). Differentiating through the radical is messy. Since the square root is an increasing function, the value of \( x \) that minimizes \( D^2 \) also minimizes \( D \).

Minimize \( D \)
Requires chain rule through \( \sqrt{\phantom{x}} \): longer algebra, same critical points.
Minimize \( D^2 \)
Polynomial derivative: cleaner, faster, and produces the same answer.

Common Mistakes

✕ Forgetting Endpoints
On a closed interval, the absolute max or min may occur at an endpoint, not at a critical point.
✕ Wrong Variable Eliminated
Substituting the constraint incorrectly or solving for the wrong variable leads to an objective with two unknowns.
✕ Not Verifying the Type
Finding \( f'(c)=0 \) does not tell you whether it is a max or min. Always apply a derivative test or compare values.

Optimization Checklist

Step Action Watch For
1. Setup Diagram, label, identify objective Units, positive quantities
2. Model Write objective and constraint equations Correct formulas (area, volume, distance)
3. Reduce Eliminate variables using constraint Solve for the right variable
4. Differentiate Set \( f'(x) = 0 \), solve Don't forget quotient/chain rule
5. Confirm Derivative test or endpoint check Max vs min, answer the actual question

Question 1 of 5

A farmer uses 120 m of fencing to enclose a rectangular pen against a barn wall (3 sides fenced). If the side perpendicular to the barn has length \( x \), what value of \( x \) maximizes the area?

Question 2 of 5

To minimize the surface area of a closed cylindrical can with volume \( V \), the relationship between radius \( r \) and height \( h \) at the optimum is:

Question 3 of 5

Given \( f(x) = x^3 - 12x + 1 \) on \([-3, 5]\), what is the absolute maximum value?

Question 4 of 5

In a distance-minimization problem, why do we typically minimize \( D^2 \) instead of \( D \)?

Question 5 of 5

A continuous function on an open interval \((a, b)\) has exactly one critical point at \( x = c \), and \( f''(c) > 0 \). What can you conclude?