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Introduction to
Optimization Problems

Learn how to use derivatives to find maximum and minimum values of real-world quantities: from maximizing area to minimizing cost.

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 identify these extreme values efficiently.

Why Does Optimization Matter?

Engineering
Minimize material cost while maximizing structural strength
Business
Maximize profit or revenue given production constraints
Science
Find the shortest path, greatest volume, or least energy configuration

Key Vocabulary

1
Objective function: the function you want to maximize or minimize, e.g. \( A(x) \) for area
2
Constraint: a condition that limits the variables, e.g. a fixed perimeter \( 2x + 2y = 100 \)
3
Critical point: a value \( c \) where \( f'(c) = 0 \) or \( f'(c) \) is undefined
4
Feasible domain: the interval of \( x \)-values that make physical sense for the problem

The Closed Interval Method

1
Find all critical points of \( f \) in \( (a, b) \) by solving \( f'(x) = 0 \)
2
Evaluate \( f \) at each critical point and at the endpoints \( x = a \) and \( x = b \)
3
The largest value is the absolute max; the smallest is the absolute min

Visualizing Extreme Values

x f(x) a b Local Min Local Max Abs Min Abs Max (endpoint)

First vs. Second Derivative Test

First Derivative Test Second Derivative Test
What you check Sign change of \( f'(x) \) around \( c \) Sign of \( f''(c) \)
Local max \( f' \) changes + to − \( f''(c) < 0 \)
Local min \( f' \) changes − to + \( f''(c) > 0 \)
When it fails Never fails (always conclusive) Inconclusive if \( f''(c) = 0 \)
Best for General use, any critical point Quick check when \( f'' \) is easy to compute

How to Solve an Optimization Problem

1
Draw a diagram and assign variables to all unknown quantities
2
Write the objective function (the quantity to maximize or minimize) in terms of your variables
3
Use the constraint to eliminate one variable, reducing the objective to a function of one variable
4
Determine the feasible domain: the interval of physically meaningful values
5
Take the derivative, set \( f'(x) = 0 \), and solve for critical points
6
Evaluate and compare \( f \) at critical points and endpoints to find the optimal value

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.Suppose an open box is made from a 12 cm square sheet. Cutting corners of side x gives V(x) = x(12 − 2x)² for 0 < x < 6; the maximum is 128 cm³ at x = 2 cm.

Corner cuts determine the box dimensions

A 12 cm square net has four equal corner cuts and dashed fold lines, with the resulting 8 × 8 × 2 cm box and volume 128 cm³.Cutting 2 cm squares from each corner of a 12 cm square sheet leaves an 8 × 8 cm base and 2 cm walls. The resulting volume 128 cm³ is the maximum of V(x) = x(12 − 2 x)² on 0 < x < 6.Open full diagram

12 × 12 sheet-cut preset

Suppose an open box is made from a 12 cm square sheet. Cutting corners of side x gives V(x) = x(12 − 2x)² for 0 < x < 6; the maximum is 128 cm³ at x = 2 cm.Suppose an open box is made from a 12 cm square sheet. Cutting corners of side x gives V(x) = x(12 − 2x)² for 0 < x < 6; the maximum is 128 cm³ at x = 2 cm.

Designing a three-sided enclosure beside a river

Selected v2 has one landscape, three fence runs and an unfenced river edge; generated graph and duplicate inset were removed.With the river side unfenced, 120 m of fencing gives x + 2y = 120. The largest rectangular area is 1,800 m², with x = 60 m and y = 30 m.

Designing a three-sided enclosure beside a river

With the river side unfenced, 120 m of fencing gives x + 2y = 120. The largest rectangular area is 1,800 m², with x = 60 m and y = 30 m.With the river side unfenced, 120 m of fencing gives x + 2y = 120. The largest rectangular area is 1,800 m², with x = 60 m and y = 30 m.

Example 1: Maximum Area Fence

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

River (no fence) x y y \( A = xy \)

Example 1: Solution

1
From the constraint: \( y = \frac{200 - x}{2} \), so \( A(x) = x \cdot \frac{200 - x}{2} = 100x - \frac{x^2}{2} \)
2
Feasible domain: \( 0 \leq x \leq 200 \) (both dimensions must be non-negative)
3
Differentiate: \( A'(x) = 100 - x \). Set \( A'(x) = 0 \): \( x = 100 \)
4
Check: \( A(0) = 0 \), \( A(100) = 5000 \), \( A(200) = 0 \). Maximum area is \(5000 \text{ m}^2\)

Example 1: The Area Function

x A(x) 0 50 100 150 200 0 2500 5000 Max: (100, 5000)

Example 2: Minimum Material Box

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

open top x x h \( V = x^2 h = 32 \) \( S = x^2 + 4xh \)

Example 2: Solution

1
Eliminate \( h \): from \( x^2 h = 32 \), we get \( h = \frac{32}{x^2} \)
2
Substitute: \( S(x) = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x} \), domain \( x > 0 \)
3
Differentiate: \( S'(x) = 2x - \frac{128}{x^2} \). Set \( S'(x) = 0 \): \( 2x = \frac{128}{x^2} \Rightarrow x^3 = 64 \Rightarrow x = 4 \)
4
Verify: \( S''(x) = 2 + \frac{256}{x^3} > 0 \) for all \( x > 0 \), so \( x = 4 \) is a minimum

Example 3: Closest Point on a Curve

Find the point on \( y = \sqrt{x} \) that is closest to the point \( (3, 0) \).

x y 0 1 2 3 4 5 1 2 (3, 0) Closest point d

Common Mistakes to Avoid

✕ Forgetting the Domain
A critical point outside the feasible domain is meaningless. Always identify valid \( x \)-values first.
✕ Not Checking Endpoints
On a closed interval, the absolute max or min may occur at an endpoint, not at a critical point.
✕ Optimizing Two Variables
Forgetting to use the constraint to eliminate one variable before differentiating. Always reduce to one variable.

Summary of Optimization Steps

Step Action Key Question
1. Setup Draw diagram, label variables What am I optimizing?
2. Model Write objective and constraint What equation relates my variables?
3. Reduce Use constraint to get one variable Which variable should I eliminate?
4. Domain Find feasible interval What values make physical sense?
5. Calculus Differentiate, set equal to zero Where is \( f'(x) = 0 \)?
6. Verify Compare critical points and endpoints Is this a max or min?

Worked practice: Section 9, Practice Problem 4 — Applied

Work through this original lesson problem, then reveal the solution and illustrations.

A manufacturer produces cylindrical cans that must hold exactly 500 cm³ of liquid. The material for the top and bottom costs $0.04 per cm², and the material for the curved side costs $0.02 per cm². (a) Express the total cost C as a function of the radius r. (b) Find the radius r that minimizes the total cost. (c) Find the corresponding height h. (d) Justify that your answer gives a minimum.

Show answer and explanation

Rubric (5 points): Point 1 — Setting up the cost function: The top and bottom each have area πr², costing 0.04(2πr²) = 0.08πr². The lateral surface has area 2πrh, costing 0.02(2πrh) = 0.04πrh. Total cost: C = 0.08πr² + 0.04πrh. Point 2 — Eliminating h using the volume constraint: From V = πr²h = 500, we get h = 500/(πr²). Substituting: C(r) = 0.08πr² + 0.04πr · 500/(πr²) = 0.08πr² + 20/r, for r > 0. Point 3 — Finding the critical point: C′(r) = 0.16πr − 20/r². Setting C′(r) = 0: 0.16πr = 20/r², so r³ = 20/(0.16π) = 125/π, giving r = (125/π)^(1/3) ≈ 3.414 cm. Point 4 — Finding the corresponding height: h = 500/(πr²) = 500/(π(125/π)^(2/3)). This simplifies to h ≈ 13.66 cm. Point 5 — Justification that this is a minimum: C″(r) = 0.16π + 40/r³. Since both terms are positive for all r > 0, C″(r) > 0 for all r in the domain. In particular, C″ at the critical point is positive, so the Second Derivative Test confirms a local minimum. Since this is the only critical point on (0, ∞) and C → ∞ as r → 0⁺ and as r → ∞, it is also the absolute minimum.

Cost C(r)=0.08πr²+20/r for a closed cylinder of volume 500 cm³. The minimum occurs at r about 3.414 cm, with h=4r.
At fixed 500 cm³ volume and the stated unequal material costs, the least-cost closed can has h=4 r.
A cylinder separated into two circular ends priced at 0.04 per cm² and a side rectangle priced at 0.02 per cm². The rectangle has dimensions 2πr by h.
A cylinder separated into two circular ends priced at 0.04 per cm² and a side rectangle priced at 0.02 per cm². The rectangle has dimensions 2πr by h.

Question 1 of 5

A farmer encloses a rectangular area with 120 m of fencing on three sides (the fourth side is a wall). If \( x \) is the side parallel to the wall, which expression gives the area as a function of one variable?

Question 2 of 5

If \( f(x) = 3x^2 - 12x + 7 \) on the interval \( [0, 5] \), what is the absolute minimum value of \( f \)?

Question 3 of 5

An open-top cylindrical can must hold \( 500\pi \text{ cm}^3\). Which equation correctly relates the surface area to the radius \( r \)?

Question 4 of 5

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

Question 5 of 5

A function \( f \) is continuous on \( [1, 6] \) with critical points at \( x = 2 \) and \( x = 4 \). Given \( f(1) = 3 \), \( f(2) = 7 \), \( f(4) = 1 \), \( f(6) = 5 \), what is the absolute maximum?