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Introduction to
Optimization Problems

Learn how to use derivatives to find maximum and minimum values of functions under real-world constraints.

AP Calculus BC

What Is Optimization?

Optimization is the process of finding the best possible value of a function: the greatest profit, the shortest distance, the least material, or the largest area. In calculus, we translate "best" into finding the absolute maximum or minimum of a function on a given domain.

Why Optimization Matters

Engineering
Minimize material cost while maintaining structural strength.
Business
Maximize revenue or profit given production constraints.
Science
Find the trajectory, dosage, or design that produces the best outcome.

The Optimization Landscape

x f(x) local max local min absolute max endpoint endpoint f'(x) = 0

The Closed Interval Method

To find the absolute maximum and minimum of a continuous function \( f \) on a closed interval \([a, b]\):

1
Find all critical points of \( f \) in \( (a, b) \): solve \( f'(x) = 0 \) and locate where \( f'(x) \) is undefined.
2
Evaluate \( f \) at each critical point and at the endpoints \( x = a \) and \( x = b \).
3
Compare all values. The largest is the absolute maximum; the smallest is the absolute minimum.

Open-Domain Optimization

When the domain is not a closed interval (e.g., \( x > 0 \)), you cannot simply check endpoints. Instead, use the First or Second Derivative Test to classify each critical point.

First Derivative Test
Check the sign of \( f'(x) \) on each side of the critical point. A sign change from \( + \) to \( - \) means a local max.
Second Derivative Test
If \( f'(c) = 0 \) and \( f''(c) < 0 \), then \( c \) is a local max. If \( f''(c) > 0 \), it is a local min.

Closed vs. Open Domain

Closed Interval \([a,b]\) Open / Unbounded Domain
Guaranteed extrema? Yes (Extreme Value Theorem) Not guaranteed
Check endpoints? Always No endpoints to check
Primary method Evaluate & compare 1st or 2nd Derivative Test
When to use Physical constraints give a finite interval Domain is naturally open or infinite

The 5-Step Strategy

1
Draw a diagram and label all variables.
2
Write the objective function: the quantity to maximize or minimize.
3
Use the constraint to eliminate variables so the objective is in one variable.
4
Differentiate, set \( f'(x) = 0 \), and solve for critical points.
5
Verify using the 1st/2nd Derivative Test or endpoint comparison, and answer the question asked.

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 a 24 cm × 16 cm sheet, V(x) = x(24 − 2x)(16 − 2x), with 0 < x < 8. The maximizing cut is x = (20 − 4√7)/3 cm, about 3.14 cm.

24 × 16 sheet-cut preset

For a 24 cm × 16 cm sheet, V(x) = x(24 − 2x)(16 − 2x), with 0 < x < 8. The maximizing cut is x = (20 − 4√7)/3 cm, about 3.14 cm.For a 24 cm × 16 cm sheet, V(x) = x(24 − 2x)(16 − 2x), with 0 < x < 8. The maximizing cut is x = (20 − 4√7)/3 cm, about 3.14 cm.

Example 1: Fencing a Garden

A farmer has 120 meters of fencing and wants to enclose a rectangular garden against a barn wall (so only three sides need fencing). What dimensions maximize the area?

BARN WALL y y x no fence needed

Example 1: Solution

1
Eliminate a variable. From \( x + 2y = 120 \), we get \( x = 120 - 2y \).
2
Substitute. \( A(y) = (120 - 2y) \cdot y = 120y - 2y^2 \).
3
Differentiate. \( A'(y) = 120 - 4y \). Set \( A'(y) = 0 \): \( y = 30 \).
4
Verify. \( A''(y) = -4 < 0 \), so \( y = 30 \) is a maximum.

Example 2: Open-Top Box

An open-top box with a square base must have a volume of 32 cubic inches. Find the dimensions that minimize the surface area (material used).

x x h open top

Example 2: Solution

1
Eliminate \( h \). From \( x^2 h = 32 \): \( h = \frac{32}{x^2} \).
2
Substitute. \( S(x) = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x} \), for \( x > 0 \).
3
Differentiate. \( S'(x) = 2x - \frac{128}{x^2} \). Set to 0: \( 2x = \frac{128}{x^2} \Rightarrow x^3 = 64 \Rightarrow x = 4 \).
4
Verify. \( S''(x) = 2 + \frac{256}{x^3} \). At \( x = 4 \): \( S''(4) = 2 + 4 = 6 > 0 \). Confirmed minimum.

Example 3: Closest Point on a Curve

Find the point on the parabola \( y = x^2 \) that is closest to the point \( (0, 1) \).

x y (0, 1) closest d 1 1

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

A powerful shortcut: since the square root function is increasing, the value of \( x \) that minimizes \( D^2 \) also minimizes \( D \). This avoids differentiating a square root.

Minimizing \( D \)
\( D = \sqrt{(x - a)^2 + (y - b)^2} \). Chain rule produces messy fractions.
Minimizing \( D^2 \)
\( D^2 = (x - a)^2 + (y - b)^2 \). A polynomial: clean and easy to differentiate.

Common Mistakes

✕ Forgetting Endpoints
On a closed interval, the absolute extremum may occur at an endpoint, not at a critical point.
✕ Wrong Variable Eliminated
Using the constraint incorrectly or substituting into the constraint instead of the objective function.
✕ Skipping Verification
Finding \( f'(x) = 0 \) does not guarantee a max or min. You must confirm with the 2nd Derivative Test or sign analysis.

Optimization Checklist

Step Action Key Question
1 Diagram & variables What are the unknowns?
2 Objective function What am I optimizing?
3 Constraint → one variable What equation links variables?
4 Differentiate & solve Where is \( f'(x) = 0 \)?
5 Verify & conclude Is it a max or min? Did I answer the question?

Question 1 of 5

A farmer has 200 meters of fencing and wants to enclose a rectangular field against a river (no fence needed on that side). What is the maximum area the farmer can enclose?

Question 2 of 5

When finding the absolute extrema of \( f(x) \) on a closed interval \([a, b]\), which of the following is required?

Question 3 of 5

For an open-top box with a square base and fixed volume \( V \), the surface area is \( S = x^2 + \frac{4V}{x} \). Which value of \( x \) minimizes \( S \)?

Question 4 of 5

Why is it valid to minimize \( D^2 \) instead of \( D \) when finding the shortest distance to a curve?

Question 5 of 5

A continuous function \( f \) on the open interval \( (0, \infty) \) has exactly one critical point at \( x = 3 \), where \( f''(3) > 0 \). What can we conclude?