Unit 5Finding Maximum/Minimum Values

Concept 4.201Built for May 2027 AP Exam

Finding Maximum/Minimum Values

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

This page is about the answer rather than the method. Finding a maximum value means producing a number, and finding a maximum location means producing an input, and the two are graded separately. When a model carries a parameter, the maximum usually comes out as an expression in that parameter, and the final step is to set that expression equal to a supplied number and solve. That two-stage structure is what the released BC questions actually test. The most expensive habit here is stopping at the critical number, because the number is the answer and the input is only the address.

On the AP Exam

What you’ll see: A model with a constant in it, a stated maximum value, and a request for the constant, or a bare model and a request for the largest value it can take.
What you’ll do: Optimize the model symbolically, express the extreme value in terms of the constant, then solve the resulting equation for that constant.
Memory hook: Optimize first, solve second. The maximum is an expression until the last line turns it into a number.
Use it on the exam: For MCQs: The distractors give the location when the value was asked for, so the noun in the question is the whole test. For FRQs: On 2021 Q3(b) the largest cross-sectional radius is stated as $1.2$ and the task is to recover $c$ , which is exactly the two-stage structure described here.

Concept Review

Value and Location

Figure 19 — A model with a parameter, its single peak, and the horizontal line the peak must touch. The peak height is an expression in the parameter, and the line supplies the number that fixes it.

The location of an extreme is an input, and the value is an output. A stem that asks "at what time" wants the input, and a stem that asks "what is the greatest number" wants the output. Answering the other one is the most common way to lose a point that was already earned by the calculus.

A curve with a single peak marked by a dot, a dashed horizontal line through the peak labelled with the largest value 1.2 and a dashed vertical line down to the x-axis labelled x equals root two, with cards showing the peak height 2c and the resulting equation

Extremes With a Parameter

Situation	What the derivative gives	What the last step is
$y=c\,x\sqrt{4-x^{2}}$	the peak sits at $x=\sqrt{2}$ for every $c>0$	set the peak height $2c$ equal to the stated value
$A=k\,x(10-x)$	the peak sits at $x=5$ for every $k>0$	the peak value is $25k$
$C(t)=a+be^{-t}$	no interior critical number if $b\neq0$	the extreme is at an endpoint or a limit

The pattern is that the location often does not depend on the parameter while the value always does. Recognising that saves time, because the location can be found once and reused.

When There Is No Closed Interval

Without a stated closed interval, the candidate list is the critical numbers only, and the type of each one decides whether it is a maximum. A critical number whose second derivative is negative is a maximum on the whole domain when the function has no other critical numbers, which is the usual situation in these parameter problems.

Verifying Rather Than Assuming

The derivative vanishing is not enough, and the check is cheap. A sign change from positive to negative, or a negative second derivative, settles the type. When the model is a product with a square root, the second derivative is unpleasant and the sign change is the better tool.

Reporting the Value

A finished answer for this topic reads as a sentence with the value first and the location second: "the largest radius is $1.2$ inches, occurring at $x=\sqrt{2}$ ". That sentence answers both readings of the question at once, and it costs one line.

Common Misconception Check

Myth 1: "The critical number is the maximum value." Truth: It is the input where the maximum occurs, and the value is the function evaluated there.
Myth 2: "The parameter changes the location." Truth: In most of these models it scales the value only, so the location is independent of it.
Myth 3: "A maximum must be verified at the endpoints." Truth: Only when the domain is closed, and these parameter models usually have an open domain.

Topic-Specific Study Method

Two Stages, Two Lines

Borrowed idea: Two stages. Split the work into an optimization stage and a solving stage. The first stage produces the extreme as an expression. The second turns that expression into a number. Keeping them on separate lines makes the structure visible. So what does the grader look for? A critical number with a justification, an expression for the extreme, and one equation that produces the constant.

Stage	What you write	What it earns
optimize	" $y'=0$ at $x=\sqrt{2}$ "	the location
evaluate	" $y(\sqrt{2})=2c$ "	the extreme as an expression
constrain	" $2c=1.2$ "	the equation the stem implies
solve	" $c=0.6$ "	the answer

① Name which one the stem wants. Underline "value" or "at what time" before starting, because the two paths diverge at the last line.

② Carry the parameter symbolically. Substituting a number early throws away the expression the last step needs.

③ Write the extreme as its own line. " $y(\sqrt{2})=2c$ " on its own line is the step a reader can credit.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find the maximum value of $f(x)=x^{2}-4x+1$ on $[0,5]$ .

Standard Execution:

Locate. $f'(x)=2x-4=0$ gives $x=2$ , which is inside the interval.

Evaluate. $f(0)=1$ , $f(2)=-3$ and $f(5)=6$ .

Compare. The smallest value is $-3$ and the largest is $6$ , which occurs at the right endpoint.

Report. The maximum value on $[0,5]$ is $6$ at $x=5$ , and the minimum value is $-3$ at $x=2$ .

Exam Variation (Context / Parameter Shift): A curve is given by $y=c\,x\sqrt{4-x^{2}}$ for $0\le x\le2$ with $c>0$ , and the largest value of $y$ is $1.2$ . Find $c$ and state where the maximum occurs.

Solution: Differentiating gives $y'=\frac{c(4-2x^{2})}{\sqrt{4-x^{2}}}$ , which vanishes at $x=\sqrt{2}$ independently of $c$ . The maximum value is $y(\sqrt{2})=c\sqrt{2}\cdot\sqrt{2}=2c$ , so $2c=1.2$ and $c=0.6$ . The maximum occurs at $x=\sqrt{2}$ , and the value there is $1.2$ .

Common Traps & Failure Signals:

The location reported (Signal: " $\sqrt{2}$ " as the answer to "what is the largest value"): the input was given instead of the output.
The parameter dropped (Signal: a numeric extreme with no reference to $c$ ): the expression was evaluated too early.
The type unchecked (Signal: a minimum offered as the maximum): no sign change or second derivative was examined.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Stages Merged: You optimized and solved in one motion, so the extreme was never written as an expression and the equation that fixes the parameter was never formed.
Location Reported: You treated the critical number as the answer, which is the address rather than the value the stem asked for.
Type Assumed: You accepted a zero as a maximum.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; One released BC set states a maximum value and asks for the parameter that produces it, and the optimization itself is a single derivative with an interior critical number
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; Counted where the scored answer is an extreme value of a supplied model rather than an extreme on a stated closed interval, which is counted in 5.7
FRQ Years Appeared: 2021 (Q3b)

Practice Examples

MCQ 1

The Value, Not the Address

What is the minimum value of $f(x)=x^{2}-4x+1$ on $[0,5]$ ?

MCQ 2

Locking a Parameter

The maximum value of $y=k\sin x$ on $(0,\pi)$ is $3$ . What is $k$ ?

MCQ 3

Where the Peak Sits

For $y=c\,x\sqrt{4-x^{2}}$ with $c>0$ on $[0,2]$ , at which $x$ does the maximum occur?

FRQ 1 • Optimize, Then Solve

Let $y=c\,x\sqrt{4-x^{2}}$ for $0\le x\le2$ , where $c$ is a positive constant.

(a) Find $\frac{dy}{dx}$ and show that the only critical number in $(0,2)$ is $x=\sqrt{2}$ .

(b) Show that $x=\sqrt{2}$ gives the maximum value of $y$ on $[0,2]$ .

(c) Show that the maximum value of $y$ is $2c$ .

(d) The largest value of $y$ is known to be $1.2$ . Find $c$ and state where the maximum occurs.

Cheatsheet

Maximum and Minimum Values

	What it wants	What to write
"what is the maximum value"	an output	" $y(\sqrt{2})=1.2$ "
"at what time is it greatest"	an input	" $t=3$ "
"the largest value is $1.2$ "	an equation to solve	" $2c=1.2$ "
"justify that it is a maximum"	a sign or second-derivative check	" $y'$ changes from positive to negative"

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 5.20 Implicit Differentiation for Analysis

Unit 5Finding Maximum/Minimum Values

Concept 4.201Built for May 2027 AP Exam

Finding Maximum/Minimum Values

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A model with a constant in it, a stated maximum value, and a request for the constant, or a bare model and a request for the largest value it can take.
What you’ll do: Optimize the model symbolically, express the extreme value in terms of the constant, then solve the resulting equation for that constant.
Memory hook: Optimize first, solve second. The maximum is an expression until the last line turns it into a number.
Use it on the exam: For MCQs: The distractors give the location when the value was asked for, so the noun in the question is the whole test. For FRQs: On 2021 Q3(b) the largest cross-sectional radius is stated as $1.2$ and the task is to recover $c$ , which is exactly the two-stage structure described here.

Concept Review

Value and Location

Extremes With a Parameter

Situation	What the derivative gives	What the last step is
$y=c\,x\sqrt{4-x^{2}}$	the peak sits at $x=\sqrt{2}$ for every $c>0$	set the peak height $2c$ equal to the stated value
$A=k\,x(10-x)$	the peak sits at $x=5$ for every $k>0$	the peak value is $25k$
$C(t)=a+be^{-t}$	no interior critical number if $b\neq0$	the extreme is at an endpoint or a limit

The pattern is that the location often does not depend on the parameter while the value always does. Recognising that saves time, because the location can be found once and reused.

When There Is No Closed Interval

Verifying Rather Than Assuming

Reporting the Value

Common Misconception Check

Myth 1: "The critical number is the maximum value." Truth: It is the input where the maximum occurs, and the value is the function evaluated there.
Myth 2: "The parameter changes the location." Truth: In most of these models it scales the value only, so the location is independent of it.
Myth 3: "A maximum must be verified at the endpoints." Truth: Only when the domain is closed, and these parameter models usually have an open domain.

Topic-Specific Study Method

Two Stages, Two Lines

Stage	What you write	What it earns
optimize	" $y'=0$ at $x=\sqrt{2}$ "	the location
evaluate	" $y(\sqrt{2})=2c$ "	the extreme as an expression
constrain	" $2c=1.2$ "	the equation the stem implies
solve	" $c=0.6$ "	the answer

① Name which one the stem wants. Underline "value" or "at what time" before starting, because the two paths diverge at the last line.

② Carry the parameter symbolically. Substituting a number early throws away the expression the last step needs.

③ Write the extreme as its own line. " $y(\sqrt{2})=2c$ " on its own line is the step a reader can credit.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find the maximum value of $f(x)=x^{2}-4x+1$ on $[0,5]$ .

Standard Execution:

Locate. $f'(x)=2x-4=0$ gives $x=2$ , which is inside the interval.

Evaluate. $f(0)=1$ , $f(2)=-3$ and $f(5)=6$ .

Compare. The smallest value is $-3$ and the largest is $6$ , which occurs at the right endpoint.

Report. The maximum value on $[0,5]$ is $6$ at $x=5$ , and the minimum value is $-3$ at $x=2$ .

Solution: Differentiating gives $y'=\frac{c(4-2x^{2})}{\sqrt{4-x^{2}}}$ , which vanishes at $x=\sqrt{2}$ independently of $c$ . The maximum value is $y(\sqrt{2})=c\sqrt{2}\cdot\sqrt{2}=2c$ , so $2c=1.2$ and $c=0.6$ . The maximum occurs at $x=\sqrt{2}$ , and the value there is $1.2$ .

Common Traps & Failure Signals:

The location reported (Signal: " $\sqrt{2}$ " as the answer to "what is the largest value"): the input was given instead of the output.
The parameter dropped (Signal: a numeric extreme with no reference to $c$ ): the expression was evaluated too early.
The type unchecked (Signal: a minimum offered as the maximum): no sign change or second derivative was examined.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Stages Merged: You optimized and solved in one motion, so the extreme was never written as an expression and the equation that fixes the parameter was never formed.
Location Reported: You treated the critical number as the answer, which is the address rather than the value the stem asked for.
Type Assumed: You accepted a zero as a maximum.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; One released BC set states a maximum value and asks for the parameter that produces it, and the optimization itself is a single derivative with an interior critical number
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; Counted where the scored answer is an extreme value of a supplied model rather than an extreme on a stated closed interval, which is counted in 5.7
FRQ Years Appeared: 2021 (Q3b)

Practice Examples

MCQ 1

The Value, Not the Address

What is the minimum value of $f(x)=x^{2}-4x+1$ on $[0,5]$ ?

MCQ 2

Locking a Parameter

The maximum value of $y=k\sin x$ on $(0,\pi)$ is $3$ . What is $k$ ?

MCQ 3

Where the Peak Sits

For $y=c\,x\sqrt{4-x^{2}}$ with $c>0$ on $[0,2]$ , at which $x$ does the maximum occur?

FRQ 1 • Optimize, Then Solve

Let $y=c\,x\sqrt{4-x^{2}}$ for $0\le x\le2$ , where $c$ is a positive constant.

(a) Find $\frac{dy}{dx}$ and show that the only critical number in $(0,2)$ is $x=\sqrt{2}$ .

(b) Show that $x=\sqrt{2}$ gives the maximum value of $y$ on $[0,2]$ .

(c) Show that the maximum value of $y$ is $2c$ .

(d) The largest value of $y$ is known to be $1.2$ . Find $c$ and state where the maximum occurs.

Cheatsheet

Maximum and Minimum Values

	What it wants	What to write
"what is the maximum value"	an output	" $y(\sqrt{2})=1.2$ "
"at what time is it greatest"	an input	" $t=3$ "
"the largest value is $1.2$ "	an equation to solve	" $2c=1.2$ "
"justify that it is a maximum"	a sign or second-derivative check	" $y'$ changes from positive to negative"

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 5.20 Implicit Differentiation for Analysis

Unit 5Finding Maximum/Minimum Values

Concept 4.201Built for May 2027 AP Exam

Finding Maximum/Minimum Values

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A model with a constant in it, a stated maximum value, and a request for the constant, or a bare model and a request for the largest value it can take.
What you’ll do: Optimize the model symbolically, express the extreme value in terms of the constant, then solve the resulting equation for that constant.
Memory hook: Optimize first, solve second. The maximum is an expression until the last line turns it into a number.
Use it on the exam: For MCQs: The distractors give the location when the value was asked for, so the noun in the question is the whole test. For FRQs: On 2021 Q3(b) the largest cross-sectional radius is stated as $1.2$ and the task is to recover $c$ , which is exactly the two-stage structure described here.

Concept Review

Value and Location

Extremes With a Parameter

Situation	What the derivative gives	What the last step is
$y=c\,x\sqrt{4-x^{2}}$	the peak sits at $x=\sqrt{2}$ for every $c>0$	set the peak height $2c$ equal to the stated value
$A=k\,x(10-x)$	the peak sits at $x=5$ for every $k>0$	the peak value is $25k$
$C(t)=a+be^{-t}$	no interior critical number if $b\neq0$	the extreme is at an endpoint or a limit

The pattern is that the location often does not depend on the parameter while the value always does. Recognising that saves time, because the location can be found once and reused.

When There Is No Closed Interval

Verifying Rather Than Assuming

Reporting the Value

Common Misconception Check

Myth 1: "The critical number is the maximum value." Truth: It is the input where the maximum occurs, and the value is the function evaluated there.
Myth 2: "The parameter changes the location." Truth: In most of these models it scales the value only, so the location is independent of it.
Myth 3: "A maximum must be verified at the endpoints." Truth: Only when the domain is closed, and these parameter models usually have an open domain.

Topic-Specific Study Method

Two Stages, Two Lines

Stage	What you write	What it earns
optimize	" $y'=0$ at $x=\sqrt{2}$ "	the location
evaluate	" $y(\sqrt{2})=2c$ "	the extreme as an expression
constrain	" $2c=1.2$ "	the equation the stem implies
solve	" $c=0.6$ "	the answer

① Name which one the stem wants. Underline "value" or "at what time" before starting, because the two paths diverge at the last line.

② Carry the parameter symbolically. Substituting a number early throws away the expression the last step needs.

③ Write the extreme as its own line. " $y(\sqrt{2})=2c$ " on its own line is the step a reader can credit.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find the maximum value of $f(x)=x^{2}-4x+1$ on $[0,5]$ .

Standard Execution:

Locate. $f'(x)=2x-4=0$ gives $x=2$ , which is inside the interval.

Evaluate. $f(0)=1$ , $f(2)=-3$ and $f(5)=6$ .

Compare. The smallest value is $-3$ and the largest is $6$ , which occurs at the right endpoint.

Report. The maximum value on $[0,5]$ is $6$ at $x=5$ , and the minimum value is $-3$ at $x=2$ .

Solution: Differentiating gives $y'=\frac{c(4-2x^{2})}{\sqrt{4-x^{2}}}$ , which vanishes at $x=\sqrt{2}$ independently of $c$ . The maximum value is $y(\sqrt{2})=c\sqrt{2}\cdot\sqrt{2}=2c$ , so $2c=1.2$ and $c=0.6$ . The maximum occurs at $x=\sqrt{2}$ , and the value there is $1.2$ .

Common Traps & Failure Signals:

The location reported (Signal: " $\sqrt{2}$ " as the answer to "what is the largest value"): the input was given instead of the output.
The parameter dropped (Signal: a numeric extreme with no reference to $c$ ): the expression was evaluated too early.
The type unchecked (Signal: a minimum offered as the maximum): no sign change or second derivative was examined.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Stages Merged: You optimized and solved in one motion, so the extreme was never written as an expression and the equation that fixes the parameter was never formed.
Location Reported: You treated the critical number as the answer, which is the address rather than the value the stem asked for.
Type Assumed: You accepted a zero as a maximum.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; One released BC set states a maximum value and asks for the parameter that produces it, and the optimization itself is a single derivative with an interior critical number
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; Counted where the scored answer is an extreme value of a supplied model rather than an extreme on a stated closed interval, which is counted in 5.7
FRQ Years Appeared: 2021 (Q3b)

Practice Examples

MCQ 1

The Value, Not the Address

What is the minimum value of $f(x)=x^{2}-4x+1$ on $[0,5]$ ?

MCQ 2

Locking a Parameter

The maximum value of $y=k\sin x$ on $(0,\pi)$ is $3$ . What is $k$ ?

MCQ 3

Where the Peak Sits

For $y=c\,x\sqrt{4-x^{2}}$ with $c>0$ on $[0,2]$ , at which $x$ does the maximum occur?

FRQ 1 • Optimize, Then Solve

Let $y=c\,x\sqrt{4-x^{2}}$ for $0\le x\le2$ , where $c$ is a positive constant.

(a) Find $\frac{dy}{dx}$ and show that the only critical number in $(0,2)$ is $x=\sqrt{2}$ .

(b) Show that $x=\sqrt{2}$ gives the maximum value of $y$ on $[0,2]$ .

(c) Show that the maximum value of $y$ is $2c$ .

(d) The largest value of $y$ is known to be $1.2$ . Find $c$ and state where the maximum occurs.

Cheatsheet

Maximum and Minimum Values

	What it wants	What to write
"what is the maximum value"	an output	" $y(\sqrt{2})=1.2$ "
"at what time is it greatest"	an input	" $t=3$ "
"the largest value is $1.2$ "	an equation to solve	" $2c=1.2$ "
"justify that it is a maximum"	a sign or second-derivative check	" $y'$ changes from positive to negative"

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 5.20 Implicit Differentiation for Analysis