Introduction to Optimization — AP® Calculus AB

Part 1 · Hook

Why this matters

A farmer with a fixed roll of fence wants the biggest pen; a theater wants the ticket price that brings in the most money. Both are asking for the absolute maximum of a function. The calculus is the tools you already have; the new skill is turning the words into that function.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. On a closed interval, where can the absolute maximum of a continuous function occur?

At an endpoint or a critical point
Only where $f' = 0$
Only at an endpoint
Where $f'' = 0$

Show the answer

The Candidates Test checks both.

Correct: At an endpoint or a critical point:
Only where $f' = 0$ :
Only at an endpoint:
Where $f'' = 0$ :

2. $f'(c) = 0$ and $f''(c) < 0$ mean

a relative maximum at $c$
a relative minimum at $c$
a point of inflection at $c$
nothing

Show the answer

The Second Derivative Test.

Correct: a relative maximum at $c$ :
a relative minimum at $c$ :
a point of inflection at $c$ :
nothing:

Part 4 · See it

See it first

Left: a rectangular pen against a river, split by a middle fence; the three fences perpendicular to the river have length x, the far side has length y, and 3x + y = 600 feet. Right: the graph of A(x) = x(600 - 3x) for 0 ≤ x ≤ 200, zero at both ends, highest at (100, 30000). — The fence fixes $3x + y = 600$ ; the area becomes a function of $x$ alone, $A(x) = x(600 - 3x)$ on $[0, 200]$ , with its highest point at $x = 100$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A problem asks for the largest or smallest of somethingname that quantity and write the objective function for it
The objective function has two variablesuse the given fact (a fixed perimeter, a fixed product) to write one variable in terms of the other
The variable must make physical sensefind the domain: lengths and amounts are not negative
Find the critical points and compare (or use one-critical-point reasoning)the absolute extremum, justified, answered in context with units

Part 6 · Key ideas

Key ideas

Optimization: finding the absolute maximum or minimum of a quantity in context.
The objective function is the quantity to optimize, written in one variable with a domain.
Justify the absolute extremum: Candidates Test on a closed interval, or one critical point that is a relative extremum.
Answer the question asked: the dimensions or the price, with units, not just $x$ .

Part 7 · Misconception

A common mistake

The wrong idea: Find where the derivative is zero, and that is the answer.

What actually happens: A critical point is only a candidate. It could be a minimum when you want a maximum (the shipping cost above), or an endpoint could win. Justify that it is the absolute extremum, and answer in context.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

A two-part pen against a river

A farmer has 600 feet of fence to build a rectangular pen against a straight river, split in two by a middle fence parallel to the ends. No fence is needed along the river. Each end and the divider has length

x

feet; the side opposite the river has length

y

feet.

A rectangular pen against a straight river, divided in two by a middle fence parallel to the ends. The two ends and the divider each have length x; the side opposite the river has length y. No fence along the river. Total fence: 600 feet. — LevlPrep original diagram.

1. Which function gives the area of the pen in terms of $x$ alone?

$A(x) = x(600 - 2x)$
$A(x) = x(600 - 3x)$
$A(x) = x\left(300 - \frac{3x}{2}\right)$
$A(x) = 3x + y$

Show the answer

Fence: two ends and a divider ( $3x$ ) plus one long side ( $y$ ): $3x + y = 600$ , so $y = 600 - 3x$ . Area $A = xy = x(600 - 3x)$ .

$A(x) = x(600 - 2x)$ : That forgets the middle fence: there are three pieces of length $x$ .
Correct: $A(x) = x(600 - 3x)$ : The fence is $3x + y = 600$ , so $y = 600 - 3x$ and $A = xy$ .
$A(x) = x\left(300 - \frac{3x}{2}\right)$ : That fences the river side too ( $3x + 2y = 600$ ).
$A(x) = 3x + y$ : $3x + y$ is the length of fence, and it still has two variables.

2. What is the domain of $A(x)$ that makes sense for the pen?

$0 \le x \le 200$
$0 \le x \le 600$
$0 \le x \le 300$
$x \ge 0$

Show the answer

Both $x \ge 0$ and $y = 600 - 3x \ge 0$ , so $0 \le x \le 200$ . (The ends give zero area, so the maximum is inside.)

Correct: $0 \le x \le 200$ : $y = 600 - 3x \ge 0$ needs $x \le 200$ ; lengths cannot be negative.
$0 \le x \le 600$ : At $x = 600$ , $y = 600 - 1800 < 0$ .
$0 \le x \le 300$ : At $x = 300$ , $y = -300$ .
$x \ge 0$ : For $x > 200$ , $y$ would be negative.

3. Find the value of $x$ that gives the largest area. Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$A'(x) = 600 - 6x = 0$ gives $x = 100$ feet. $A''(x) = -6 < 0$ , and $A(0) = A(200) = 0$ are smaller, so $x = 100$ gives the largest area.

Answer: 100 feet

4. Find the largest possible area of the pen. Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$A(100)$ $= 100(600 - 300)$ $= 30{,}000$ square feet, with $y = 300$ feet.

Answer: 30000 square feet

5. Which argument shows that $x = 100$ gives the absolute maximum area on $[0, 200]$ ?

$A'(100) = 0$ , so $A(100)$ is the largest area
$A$ is continuous on $[0, 200]$ , and $A(0) = 0$ , $A(100) = 30{,}000$ , $A(200) = 0$ : the largest candidate is $A(100)$
$A(100) > 0$ , so it is the maximum
$x = 100$ is the midpoint of $[0, 200]$

Show the answer

The candidates are $x = 0$ , $x = 200$ (endpoints) and $x = 100$ (the critical point). $A(100) = 30{,}000$ is the largest, so it is the absolute maximum.

$A'(100) = 0$ , so $A(100)$ is the largest area: A critical point alone could be a minimum or neither.
Correct: $A$ is continuous on $[0, 200]$ , and $A(0) = 0$ , $A(100) = 30{,}000$ , $A(200) = 0$ : the largest candidate is $A(100)$ : A Candidates Test: endpoints and the one critical point.
$A(100) > 0$ , so it is the maximum: Many areas are positive; being positive does not make one largest.
$x = 100$ is the midpoint of $[0, 200]$ : The midpoint has no special meaning; it happens to work here.

6. A student finds that a cost function $C(x)$ on $(0, \infty)$ has one critical point, $x = 4$ , and that $C''(x) > 0$ for all $x > 0$ . What can she conclude?

$C(4)$ is a relative minimum, but a lower cost may exist
$C(4)$ is the absolute minimum cost
$C(4)$ is the absolute maximum cost
Nothing, because the interval has no endpoints

Show the answer

$C''(4) > 0$ makes $x = 4$ a relative minimum, and it is the only critical point on the interval, so $C(4)$ is the absolute minimum.

$C(4)$ is a relative minimum, but a lower cost may exist: With one critical point and concave up everywhere, nothing can be lower.
Correct: $C(4)$ is the absolute minimum cost: One critical point that is a relative minimum on an interval gives the absolute minimum there.
$C(4)$ is the absolute maximum cost: Concave up at a flat point is a minimum.
Nothing, because the interval has no endpoints: The one-critical-point argument works on open intervals.

Part 9 · Summary

Summary

Optimization finds the absolute maximum or minimum of a quantity in context. Write the objective function, reduce it to one variable using the given facts, find its domain, find the critical points, and justify the absolute extremum with candidates or one-critical-point reasoning. Answer in context, with units.

Part 10 · Up next

What comes next

Part 11 · Connections