Unit 5 · Topic 5.10 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 5.10: Introduction to Optimization Problems

Q: Do I always have to check the endpoints?

On a closed interval, yes. The extreme value can be at an endpoint even when the derivative there is not 0. On an interval with no endpoints, such as x > 0, there are none to check, so use the sign of the derivative instead.

Q: What if the derivative is never 0 on the interval?

If f' exists everywhere on the interval and is never 0, then f is always increasing or always decreasing there, so on a closed interval the maximum and minimum are at the two endpoints.

Q: Can I use a graphing calculator?

Only where the question allows one. There it can solve f'(x) = 0 numerically, but write the equation you solved, and still compare the candidates or check the sign of the derivative.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 5: 15 to 20% of the multiple-choice section

What you'll be able to do

Use the derivative to find the minimum or maximum value of a function on an interval, for a function you are given or one you build from a situation.

Key vocabulary

Term	What it means	The exam trap
Optimization problem	A problem that asks for the minimum or maximum value of a function on a given interval.	In a situation the function is usually not given. Build it from the description before you differentiate.
Minimum or maximum value	The smallest or largest output of a function over the interval, such as the least cost or the greatest area.	The value and the input where it occurs are two different numbers. Answer with the one the question names.
Given interval	The inputs a problem allows, either stated outright or set by the situation, such as lengths that cannot be negative.	Write the interval before solving, because it decides which inputs can be the answer.

Graph of h(x) = x cubed minus 12x, drawn solid on the interval from 0 to 3. The endpoints (0, 0) and (3, -9) and the critical point (2, -16) are marked as candidates. The other critical point, (-2, 16), has an open dot because x = -2 lies outside the interval.

Part 1 Think first How deep should the pen be?

Commit to a guess before the lesson. Guessing first, even wrongly, makes the answer easier to remember.

Predict first

A farmer has 120 meters of fence for a rectangular pen against the long wall of a barn. The wall forms one side, so the fence runs along only three sides. The two sides that meet the wall are each $x$ meters long. Which value of $x$ gives the pen the most area?

In short. An optimization problem asks for the minimum or maximum value of a function on an interval, and the derivative finds the inputs where that value can occur.

In a situation, first write the quantity as a function of one variable and state the interval of inputs the situation allows.
On a closed interval, compare the function at every critical point inside the interval and at both endpoints.
On an interval with no endpoints, a single critical point where the derivative changes sign gives the absolute extreme value.

Part 2 Learn it Three ideas, each with a check

5a. Optimizing means finding an extreme value on an interval

An optimization problem asks for the minimum or maximum value of a function on a given interval, such as the most area or the least cost. The derivative tells you where to look.

On a closed interval $[a, b]$ , a continuous function has an absolute maximum and an absolute minimum, and each one occurs at a critical point inside the interval or at an endpoint. That is the Candidates Test, and it turns optimization into three steps:

Find $f^{'} (x)$ and every input in $(a, b)$ where $f^{'} (x) = 0$ or $f^{'} (x)$ does not exist.
Evaluate $f$ at those inputs and at both endpoints.
The largest output is the maximum value, and the smallest is the minimum value.

Take $h (x) = x^{3} - 12 x$ on $[0, 3]$ , the graph at the top. $h^{'} (x) = 3 x^{2} - 12$ is 0 at $x = 2$ and $x = - 2$ , but $- 2$ is not in $[0, 3]$ , so it is not a candidate. The candidates give $h (0) = 0$ , $h (2) = - 16$ and $h (3) = - 9$ . The minimum value is $- 16$ , at $x = 2$ . The maximum value is 0, at the endpoint $x = 0$ , where $h^{'} (0) = - 12$ is not zero at all.

Each answer has two parts: the value, an output such as $- 16$ , and where it occurs, an input such as $x = 2$ . Read which one the question asks for.

Quick check · Calculate

No calculator. Let $g (x) = x^{3} - 3 x^{2}$ . What is the maximum value of $g$ on the interval $[- 1, 4]$ ?

Your answer

5b. Build one function of one variable

In a situation, nobody hands you the function. Problems that look different (a fence, a box) share one structure:

a quantity to optimize, such as an area or a cost, which often depends on two measurements;
a condition that ties those measurements together, such as a fixed length of fence or a fixed volume;
the interval of inputs the situation allows.

Use the condition to write one measurement in terms of the other and substitute. That leaves a function of one variable, which you can differentiate.

For the barn in Think first, let $x$ be each side that meets the wall and $y$ the side along it. The fence gives $2 x + y = 120$ , so $y = 120 - 2 x$ and

A (x) = x (120 - 2 x) .

No side can be negative, so $x \geq 0$ and $120 - 2 x \geq 0$ : the interval is $0 \leq x \leq 60$ .

Now the derivative does the rest. $A^{'} (x) = 120 - 4 x$ is 0 at $x = 30$ , and comparing $A (0) = 0$ , $A (30) = 1800$ and $A (60) = 0$ gives the maximum area, 1800 square meters, for a pen 30 meters deep and 60 meters long.

Quick check · Multiple choice

No calculator. A wire 24 centimeters long is bent into a rectangle of width $x$ centimeters. Which function and interval give the area for every width that leaves no side negative, and for no other width, so you can find the largest area?

5c. When the interval has no endpoints

Sometimes the situation allows every positive input. A rectangle of area 36 square meters can have any width $x > 0$ , with length $\frac{36}{x}$ , so its perimeter is

P (x) = 2 x + \frac{72}{x}, x > 0.

There are no endpoints to compare, so use the sign of the derivative. $P^{'} (x) = 2 - \frac{72}{x^{2}}$ is 0 at $x = 6$ (and at $x = - 6$ , which is not a width). $P^{'} (x) < 0$ for $0 < x < 6$ and $P^{'} (x) > 0$ for $x > 6$ , so $P$ decreases all the way to $x = 6$ and increases after it. No width gives a smaller perimeter, so the minimum perimeter is $P (6) = 24$ meters.

This argument works because $x = 6$ is the only critical point on the interval. A single critical point where the derivative changes sign gives the absolute extreme value on that interval. A critical point on its own is only a place to look.

Whichever argument you use, finish by answering the question that was asked: a value with its units (24 meters), the input where it happens (a width of 6 meters), or both.

Quick check · Decide and justify

No calculator. A workshop's cost per chair is $C (x) = 4 x + \frac{100}{x}$ dollars when it builds $x$ chairs a day, for $x > 0$ . The only critical point is $x = 5$ , where $C^{'} (5) = 0$ . A classmate says $C$ has its least value on $x > 0$ at $x = 5$ . Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Comparing only the critical points and forgetting the endpoints of a closed interval.	The largest or smallest value can sit at an endpoint, where the derivative need not be 0.	List both endpoints with the critical points, and evaluate the function at every one.
Keeping a critical point that lies outside the interval, such as a negative length.	Its output is not a value the situation allows, so it cannot be the answer.	Write the interval before you solve, and keep only the critical points inside it.
Giving the input where the extreme value occurs when the question asks for the value, or the other way round.	Where the maximum occurs is an input, such as a width of 30 meters. The maximum value is an output, such as 1800 square meters.	Before you answer, reread what the question asks for, and give that quantity with its units.

Part 3 Watch it done The largest rectangle under a parabola

Watch each step, then do the last one yourself.

A rectangle has its base on the $x$ -axis and its two upper corners on the parabola $y = 12 - x^{2}$ . Find the largest possible area of the rectangle.

Step 1 · Name a variable and write the area
Put the upper right corner at $(x, 12 - x^{2})$ with $x \geq 0$ . The parabola is symmetric about the $y$ -axis, so the base runs from $- x$ to $x$ : the width is $2 x$ and the height is $12 - x^{2}$ . The area is $A (x) = 2 x (12 - x^{2}) = 24 x - 2 x^{3}$ .
Step 2 · State the interval
The corner must be on the parabola at or above the $x$ -axis, so $x \geq 0$ and $12 - x^{2} \geq 0$ . The interval is $0 \leq x \leq \sqrt{12}$ . $A$ is a polynomial, so it is continuous on this closed interval, and its largest value is at a critical point or an endpoint.
Step 3 · Find the critical points
$A^{'} (x) = 24 - 6 x^{2} = 0$ when $x^{2} = 4$ , so $x = 2$ or $x = - 2$ . Only $x = 2$ is in $[0, \sqrt{12}]$ .
Why this step?
Why is $x = - 2$ left out, when $A^{'} (- 2) = 0$ too?
Here $x$ is the coordinate of the right-hand corner, so the interval is $0 \leq x \leq \sqrt{12}$ . An input outside the interval describes no rectangle in this setup, so its output is not a candidate.
Your step · Calculate

No calculator. Finish it. For $A (x) = 24 x - 2 x^{3}$ on $0 \leq x \leq \sqrt{12}$ , the candidates are $x = 0$ , $x = 2$ and $x = \sqrt{12}$ . What is the largest possible area of the rectangle?

Your answer

Part 4 Try it Two new situations

Try it · Calculate

No calculator. A rectangular garden with an area of 200 square meters is fenced on all four sides. One side of the garden faces the street, and fence on that side costs 12 dollars per meter. Fence on the other three sides costs 4 dollars per meter. What is the least possible cost of the fence, in dollars?

Your answer

Try it · Match

No calculator. Each situation asks for an extreme value of one quantity. Tap a situation, then the function of $x$ alone that gives that quantity.

Part 5 Explain it Write it the way the exam asks

Free response is handwritten on the exam, so write your answer on paper first. Nothing you type here leaves this page. Then compare with the model answer and score yourself on each point.

Free response is 50% of the exam score. This one is shaped like Question 4, from free-response Part B (calculator not permitted).

Free-response practice · Question 4 style

No calculator. An open-top box with a square base is to hold 4000 cubic centimeters. The base is $x$ centimeters on each side, and the box is $h$ centimeters tall. The material covers the base and the four sides.

(a) Write an expression for $h$ in terms of $x$ . Verify that the area of material used is $S (x) = x^{2} + \frac{16000}{x}$ square centimeters.

(b) Find the value of $x$ , for $x > 0$ , at which $S$ has its minimum value. Justify your answer.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) The volume is $x^{2} h = 4000$ , so $h = \frac{4000}{x^{2}}$ . The base has area $x^{2}$ and each of the four sides has area $x h$ , so $S = x^{2} + 4 x h = x^{2} + 4 x \cdot \frac{4000}{x^{2}} = x^{2} + \frac{16000}{x}$ .

(b) $S^{'} (x) = 2 x - \frac{16000}{x^{2}}$ . Setting $S^{'} (x) = 0$ gives $2 x^{3} = 16000$ , so $x^{3} = 8000$ and $x = 20$ . $S^{'} (x) < 0$ for $0 < x < 20$ and $S^{'} (x) > 0$ for $x > 20$ . Since $x = 20$ is the only critical point for $x > 0$ and $S^{'}$ changes from negative to positive there, $S$ has its minimum value at $x = 20$ .

(c) $S (20) = 400 + \frac{16000}{20} = 400 + 800 = 1200$ square centimeters, and the height is $h = \frac{4000}{20^{2}} = 10$ centimeters.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Writes $h = \frac{4000}{x^{2}}$ and substitutes it into $S = x^{2} + 4 x h$ to get $x^{2} + \frac{16000}{x}$ .
Finds $S^{'} (x) = 2 x - \frac{16000}{x^{2}}$ and solves $S^{'} (x) = 0$ to get $x = 20$ .
Justifies the minimum: $x = 20$ is the only critical point for $x > 0$ , and $S^{'}$ changes from negative to positive there.
Gives the minimum area as 1200 square centimeters and the height as 10 centimeters, with units.

Part 6 Check yourself A medicine in the blood

The two case questions use the same function. The table lists only some of its values.

After a dose of a medicine, its concentration in a patient's blood $t$ hours later is $c (t) = \frac{30 t}{t^{2} + 9}$ milligrams per liter, for $0 \leq t \leq 8$ . The table gives some values, rounded to three decimal places.

Concentration of the medicine after the dose
$t$ (hours)	$c (t)$ (milligrams per liter)
0	0
1	3
2	4.615
4	4.8
6	4
8	3.288

Case question 1 · Calculate

No calculator. What is the maximum concentration of the medicine on $0 \leq t \leq 8$ , in milligrams per liter?

Your answer

Case question 2 · Multiple choice

No calculator. Let $t = k$ be the input in $0 < t < 8$ where $c^{'} (t) = 0$ . Which argument justifies that the maximum concentration on $0 \leq t \leq 8$ occurs at $t = k$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds the largest area of a rectangle with a perimeter of 36 meters. Tap the first line with a mistake.

Check yourself · Sort it

No calculator. A function $V$ is differentiable for all real numbers, and $V^{'} (x) = 0$ only at $x = - 2$ , $x = 2$ and $x = 9$ . You want the maximum value of $V$ on $0 \leq x \leq 6$ . Tap an input, then say whether you must evaluate $V$ there.

A candidate to test

Not a candidate

Part 7 Keep it Bring back earlier ideas

Optimization stands on the first derivative test from this unit and the chain rule from Unit 3. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. The derivative of a function $f$ is $f^{'} (x) = (x - 1) (x - 4)^{2}$ . Which statement about $f$ at $x = 4$ is true?

Bring it back · Calculate

No calculator. Let $f (x) = (x^{2} + 1)^{3}$ . Find $f^{'} (1)$ .

Your answer

In three lines

An optimization problem asks for the minimum or maximum value of a function on an interval.
Build the function in one variable from the situation, and state the interval its inputs can take.
Compare the values at every candidate, or use the sign of the derivative when there are no endpoints.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a free-response part asks for a minimum or maximum, write the derivative, set it equal to 0, and list every candidate, including the endpoints of a closed interval, before you name the answer with its units. On an interval with no endpoints, justify the answer with the sign of the derivative on each side of the only critical point.

Questions students ask

Do I always have to check the endpoints?

On a closed interval, yes. The extreme value can be at an endpoint even when the derivative there is not 0. On an interval with no endpoints, such as $x > 0$ , there are none to check, so use the sign of the derivative instead.

What if the derivative is never 0 on the interval?

If $f^{'}$ exists everywhere on the interval and is never 0, then $f$ is always increasing or always decreasing there, so on a closed interval the maximum and minimum are at the two endpoints.

How do I know whether I found a maximum or a minimum?

On a closed interval, compare the values at the candidates: the largest is the maximum and the smallest is the minimum. With no endpoints, the sign of the derivative tells you: falling then rising is a minimum, rising then falling is a maximum.

Why does the interval matter so much?

It decides which critical points count and whether there are endpoints to test. A critical point outside the interval, such as a negative length, is not a possible answer.

Can I use a graphing calculator?

Only where the question allows one. There it can solve $f^{'} (x) = 0$ numerically, but write the equation you solved, and still compare the candidates or check the sign of the derivative.

Course alignment, for teachers

AP Calculus AB topic 5.10, Unit 5: Analytical Applications of Differentiation.

Skill 2.A: Identify common underlying structures in problems involving different contextual situations.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.