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Solving
Optimization Problems

Learn to find maximum and minimum values of real-world quantities using derivatives and critical-point analysis.

AP Calculus AB

What Is Optimization?

Optimization is the process of finding the absolute maximum or minimum value of a function subject to given constraints. In calculus, we use derivatives to locate these extreme values efficiently.

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

Why Optimization Matters

Every real-world design decision involves trade-offs. Optimization lets us quantify the best choice. Here is a classic scenario: you have a fixed amount of fencing and want to enclose the largest area.

Width \( x \) Area \( A(x) \) Maximum \( x = 25 \) 0 50

Key Vocabulary

Objective Function
The quantity you want to maximize or minimize, expressed as a function.
Constraint
A condition that relates the variables and limits your choices.
Critical Point
A value where \( f'(x) = 0 \) or \( f'(x) \) is undefined.

The Extreme Value Theorem

This guarantees that optimization problems on closed domains always have a solution. The extreme values occur either at critical points in \( (a, b) \) or at the endpoints \( a \) and \( b \).

Abs. Max Abs. Min \( a \) \( b \)

The Closed Interval Method

To find the absolute extrema of a continuous function \( f \) on \( [a, b] \):

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

First vs. Second Derivative Test

First Derivative Test Second Derivative Test
When to use Always works at critical points When \( f''(c) \) is easy to compute
Requires Sign chart of \( f' \) \( f''(c) \neq 0 \)
Local max \( f' \) changes + to − \( f''(c) < 0 \)
Local min \( f' \) changes − to + \( f''(c) > 0 \)
Inconclusive Never (always gives answer) When \( f''(c) = 0 \)

The 5-Step Strategy

1
Draw a diagram and label all quantities with variables.
2
Write the objective function: the quantity to optimize.
3
Use the constraint to reduce to one variable.
4
Find the domain (realistic values for the remaining variable).
5
Differentiate, set \( f'(x) = 0 \), solve, and verify it is a max or min.

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 square sheet, cutting corners of side x gives V(x) = x(24 − 2x)². The maximum volume is 1,024 cm³ when x = 4 cm.

24 × 24 sheet-cut preset

For a 24 cm square sheet, cutting corners of side x gives V(x) = x(24 − 2x)². The maximum volume is 1,024 cm³ when x = 4 cm.For a 24 cm square sheet, cutting corners of side x gives V(x) = x(24 − 2x)². The maximum volume is 1,024 cm³ when x = 4 cm.

Example 1: Fencing Problem

A farmer has 200 m of fencing and wants to enclose a rectangular area along a river (no fencing needed on the river side). What dimensions maximize the area?

River (no fence) \( y \) \( y \) \( x \)

Example 1: Solution

1
Eliminate \( y \): From \( x + 2y = 200 \), we get \( y = \frac{200 - x}{2} \).
2
Substitute: \( A(x) = x \cdot \frac{200 - x}{2} = 100x - \frac{x^2}{2} \).
3
Domain: \( 0 < x < 200 \) (both sides must be positive).
4
Differentiate: \( A'(x) = 100 - x \). Set \( A'(x) = 0 \): \( x = 100 \).
5
Verify: \( A''(x) = -1 < 0 \), so \( x = 100 \) gives a maximum.

Example 2: Open Box Problem

A box with an open top is made from a 12 in. by 12 in. sheet of metal by cutting equal squares of side \( x \) from each corner and folding up the sides. Find \( x \) to maximize volume.

\( x \) \( x \) 12 in. \( 12 - 2x \) 12 in.

Example 2: Solution

1
Expand: \( V(x) = x(12-2x)^2 = x(144 - 48x + 4x^2) = 4x^3 - 48x^2 + 144x \).
2
Differentiate: \( V'(x) = 12x^2 - 96x + 144 = 12(x^2 - 8x + 12) \).
3
Solve: \( x^2 - 8x + 12 = 0 \Rightarrow (x-2)(x-6) = 0 \), so \( x = 2 \) or \( x = 6 \).
4
Check domain: \( x = 6 \) is an endpoint where \( V = 0 \). Only \( x = 2 \) is valid.
5
Verify: \( V''(2) = 24(2) - 96 = -48 < 0 \). Confirmed maximum.

Example 3: Minimum Distance

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

x y (3, 0) \( (x, x^2) \) \( d \)

Example 3: Solution

1
Objective: \( D(x) = (x-3)^2 + (x^2)^2 = x^4 + x^2 - 6x + 9 \).
2
Differentiate: \( D'(x) = 4x^3 + 2x - 6 \).
3
Solve \( D'(x) = 0 \): Factor: \( 2(2x^3 + x - 3) = 2(x-1)(2x^2+2x+3) = 0 \).
4
Analyze: \( 2x^2+2x+3 \) has discriminant \( 4 - 24 < 0 \), so the only real root is \( x = 1 \).
5
Verify: \( D''(1) = 12(1)^2 + 2 = 14 > 0 \). Confirmed minimum.

Common Mistakes

✕ Forgetting the Domain
A critical point outside the realistic domain is not a valid answer. Always determine \( x \)-bounds first.
✕ Not Verifying Max/Min
Solving \( f'(x) = 0 \) only finds candidates. Use the second derivative or endpoint check to confirm.
✕ Wrong Variable Elimination
Substituting incorrectly from the constraint leads to a wrong objective function. Double-check algebra.

Optimization Summary

Step Action Key Check
1. Setup Identify variables, draw diagram All quantities labeled?
2. Objective Write \( f(x) \) to optimize Correct formula?
3. Constraint Eliminate a variable One variable left?
4. Domain Find valid \( x \)-range Endpoints realistic?
5. Solve Set \( f'(x) = 0 \), solve In domain?
6. Verify Use \( f'' \) or endpoints Max or min confirmed?

Question 1 of 5

A farmer uses 120 ft of fencing to enclose a rectangular pen against a barn wall (no fencing on the barn side). Which function represents the area in terms of the width \( w \)?

Question 2 of 5

Given \( f(x) = 2x^3 - 9x^2 + 12x \) on \( [0, 4] \), what is the absolute maximum value?

Question 3 of 5

For the open-box problem, a square sheet has side length \( s \). After cutting corner squares of side \( x \), the volume is \( V(x) = x(s-2x)^2 \). What is the domain of \( x \)?

Question 4 of 5

If \( f'(c) = 0 \) and \( f''(c) > 0 \), what can you conclude about \( f(c) \)?

Question 5 of 5

The sum of two positive numbers is 50. Which value of \( x \) minimizes the expression \( S = x^2 + y^2 \)?