Unit 5Optimization Problems

Concept 4.126Built for May 2027 AP Exam

Optimization Problems

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Optimization is a translation problem before it is a calculus problem. Name the quantity you want to make as large or as small as possible, then write its formula even though the formula has two variables. Find the relationship that ties those variables together; that is the constraint, and it is usually stated in the sentence the problem hopes you will skim. Substitute to get one variable, state the domain the situation allows, and only then differentiate. The calculus that follows is ordinary: set the derivative to zero, check the endpoints, compare. Almost every lost point on this topic is lost in the first two minutes, before any derivative appears.

On the AP Exam

This is the rarest topic in Unit 5 by a wide margin: exactly one released free-response part since 2017 requires building an objective function.

What you’ll see: A context with a fixed budget, a fixed perimeter, or a fixed volume, and a question about the largest or smallest value of some other quantity. On multiple-choice sections the same shape appears far more often, usually with round numbers.
What you’ll do: Write two equations, eliminate one variable, name the domain, differentiate, solve, and justify that the answer is the extremum rather than a point that merely has a zero derivative.
Memory hook: Two equations, one variable, one interval. The calculus is the last step, not the first.
Use it on the exam: For MCQs: Check the domain before you trust a critical point. Distractors are built from solutions that fall outside the geometry of the problem. For FRQs: The justification sentence is worth a point on its own. Name the test you used, whether that is a sign chart or a comparison of endpoint values.

Concept Review

Two Formulas and One Job

Figure 1 — The constraint removes a variable, the objective becomes a function of one variable, and the derivative locates the extremum. The endpoints still need checking.

$\text{constraint} \implies y = f(x) \implies A(x) = x \cdot f(x) \implies A'(x) = 0$

The order is not negotiable. Students who differentiate first end up with a derivative in two variables and no way forward, then spend five minutes trying to repair a step that was never valid. The constraint is the entire content of the problem; the derivative is the routine part.

Optimization: the constraint reduces the objective to one variable, then the derivative locates the extremum

The Four-Step Loop

Every optimization question on the exam runs through the same four steps.

Step	What you write	Why it matters
objective	a formula for the quantity to be optimized	it names the variable you will differentiate
constraint	the equation that links the variables	it removes the second variable
domain	the interval the geometry allows	it removes answers that cannot exist
derivative	solve $A'(x) = 0$ , then compare endpoints	it locates the candidate

The third step is the one that is skipped.

A cut-out box cannot have a negative side, and a rectangle with a fixed perimeter cannot have a side longer than half that perimeter. These limits come from the picture, not from the algebra, and they are often the fastest way to eliminate a distractor.

Endpoints Are Still Candidates

Optimization problems arrive in two flavours, and the justification differs between them.

Closed interval: compare the objective at the critical point and at both endpoints. The largest of those values wins, and the comparison is the justification.
Open interval: a single critical point plus a sign change of the derivative is enough, because there are no endpoints to beat it.

Mixing the two is a common writing error. A sign chart on a closed interval leaves the endpoints unexamined, and an endpoint comparison on an open interval refers to values that do not exist.

What the Justification Sounds Like

Three acceptable forms, in increasing order of strength.

Sign chart: " $A'$ changes from positive to negative at $x = 10$ , so $A(10)$ is the maximum."
Second derivative: " $A''(10) = -2 < 0$ , so the critical point gives a maximum."
Comparison: " $A(0) = A(20) = 0$ and $A(10) = 100$ , so the largest value is $100$ ."

Any of the three earns the point. What does not earn it is the phrase "this is obviously the maximum", which asserts the conclusion the rubric is asking you to prove.

Common Misconception Check

Myth 1: "The answer is where the derivative is zero." Truth: The answer is the best value on the allowed interval. On a closed interval that may be an endpoint.
Myth 2: "Any critical point is the optimum." Truth: A critical point can be a minimum of the objective, or a point outside the physical domain.
Myth 3: "The domain is all real numbers." Truth: Lengths are positive, and the constraint bounds them. The domain is part of the answer.

Topic-Specific Study Method

Two Minutes of Translation, One Minute of Calculus

Borrowed idea: Optimization is a translation problem before it is a calculus problem. What quantity are you trying to make as large or as small as possible? Write that formula first, even though it has two variables. What relationship ties those two variables together? That is the constraint, and it is usually given in the sentence the problem hopes you will skim past. Once the constraint removes one variable, the calculus is ordinary: differentiate, set to zero, compare with the endpoints. The only real difficulty is the first two minutes.

Exam disguise	What you see on the page	What your brain must translate	Required execution
fence against a river	one side needs no fence	perimeter with three sides	objective in one variable, then optimize
open box from a sheet	squares cut from the corners	volume $= x(\text{side} - 2x)^2$	domain from the cut, then optimize
cylinder or cone	a fixed volume or surface	one formula solving for the other variable	substitute, then differentiate
"largest area for a given perimeter"	two quantities linked	square is the answer, but show it	objective, constraint, derivative
a family of curves	a constant in the formula	optimize over the shape, then solve for the constant	differentiate with respect to the shape variable

① Write the objective before reading the question a second time. The quantity to be optimized is almost always named in the first sentence.

② State the domain out loud, in symbols. Half of the wrong answers on this topic are correct algebra applied outside the geometry.

③ Justify with a named test, not with an adjective. "Changes from positive to negative" is a reason; "clearly the largest" is not.

Training & Error Diagnosis

Core Archetype (Simplest Form): A rectangle has perimeter $40$ . What is the largest possible area?

Standard Execution:

Objective. $A = xy$ .

Constraint. $2x + 2y = 40$ , so $y = 20 - x$ .

One variable. $A(x) = x(20 - x)$ on $0 < x < 20$ .

Derivative. $A'(x) = 20 - 2x = 0$ gives $x = 10$ , and $A''(10) = -2 < 0$ , so $A(10) = 100$ is the maximum. The rectangle is a square.

Exam Variation (Context / Parameter Shift): A square sheet of side $12$ has squares of side $x$ cut from each corner, and the flaps are folded up to make an open box. Find the value of $x$ that maximizes the volume, and find that volume.

Solution: The base is $12 - 2x$ on each side and the height is $x$ , so $V(x) = x(12 - 2x)^2$ on $0 < x < 6$ . Expanding is unnecessary: $V'(x) = (12-2x)^2 + x \cdot 2(12-2x)(-2) = (12-2x)(12 - 6x)$ , which is zero at $x = 6$ and at $x = 2$ . Only $x = 2$ lies in the open interval. Since $V' > 0$ on $(0,2)$ and $V' < 0$ on $(2,6)$ , the volume is maximized at $x = 2$ , giving $V(2) = 2 \cdot 64 = 128$ cubic units.

Common Traps & Failure Signals:

Constraint ignored (Signal: a derivative with two variables in it): the substitution step was skipped.
Domain forgotten (Signal: a critical point at $x = 6$ reported for a box whose base would vanish): the geometry was never translated into an interval.
Endpoint unexamined (Signal: a closed-interval problem justified only by a sign chart): the comparison step was omitted.

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

Objective Unwritten: Name the quantity first.
Constraint Blindness: You treated two variables as two independent quantities, so the derivative you computed describes a surface rather than a curve, and every later step is answering a question the problem never asked.
Domain Free Fall: You accepted a solution that the picture rules out, which turns a correct algebraic answer into a physically impossible one.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Exactly one released free-response part since 2017 requires an objective function, and that part asks for the constant in a family of curves rather than for a maximum in context
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; The 2021 spinning-toy question needs the maximum of a family member before it can be solved, and no other exam in the window asks for that step
FRQ Years Appeared: 2021 (Q3b)

Practice Examples

MCQ 1

Largest Area, Fixed Perimeter

A rectangle has perimeter $20$ . What is the largest possible area?

MCQ 2

Smallest Sum of Squares

Two positive numbers have sum $10$ . What is the smallest possible value of the sum of their squares?

MCQ 3

Naming the Domain

Squares of side $x$ are cut from the corners of a $10$ -by- $10$ sheet, and the flaps are folded up to form an open box. What is the domain of the volume function?

FRQ 4

The Open Box

A square sheet of metal measures $12$ inches on each side. Squares of side $x$ inches are cut from each corner, and the remaining flaps are folded up to form an open-top box.

(a) Write an expression for the volume $V(x)$ of the box, and state the domain of $V$ that the geometry allows.

(b) Find $V'(x)$ , and find all values of $x$ at which $V'(x) = 0$ .

(c) Determine which of the values found in part (b) gives the maximum volume, and justify your answer.

(d) Find the maximum volume of the box, and state it with units.

Cheatsheet

Core Reference

	Constraint	Objective after substitution
fixed perimeter, largest area	$2x + 2y = P$	$A(x) = x\left(\frac{P}{2} - x\right)$
fixed area, least perimeter	$xy = A$	$P(x) = 2x + \frac{2A}{x}$
open box from a square sheet	corners of side $x$ removed	$V(x) = x(s-2x)^2$
fixed sum, extremal sum of squares	$x + y = S$	$Q(x) = x^2 + (S-x)^2$
fence with one side on a river	only three sides fenced	$A(x) = x(L - 2x)$

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Unit 5Optimization Problems

Concept 4.126Built for May 2027 AP Exam

Optimization Problems

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

This is the rarest topic in Unit 5 by a wide margin: exactly one released free-response part since 2017 requires building an objective function.

What you’ll see: A context with a fixed budget, a fixed perimeter, or a fixed volume, and a question about the largest or smallest value of some other quantity. On multiple-choice sections the same shape appears far more often, usually with round numbers.
What you’ll do: Write two equations, eliminate one variable, name the domain, differentiate, solve, and justify that the answer is the extremum rather than a point that merely has a zero derivative.
Memory hook: Two equations, one variable, one interval. The calculus is the last step, not the first.
Use it on the exam: For MCQs: Check the domain before you trust a critical point. Distractors are built from solutions that fall outside the geometry of the problem. For FRQs: The justification sentence is worth a point on its own. Name the test you used, whether that is a sign chart or a comparison of endpoint values.

Concept Review

Two Formulas and One Job

Figure 1 — The constraint removes a variable, the objective becomes a function of one variable, and the derivative locates the extremum. The endpoints still need checking.

$\text{constraint} \implies y = f(x) \implies A(x) = x \cdot f(x) \implies A'(x) = 0$

Optimization: the constraint reduces the objective to one variable, then the derivative locates the extremum

The Four-Step Loop

Every optimization question on the exam runs through the same four steps.

Step	What you write	Why it matters
objective	a formula for the quantity to be optimized	it names the variable you will differentiate
constraint	the equation that links the variables	it removes the second variable
domain	the interval the geometry allows	it removes answers that cannot exist
derivative	solve $A'(x) = 0$ , then compare endpoints	it locates the candidate

The third step is the one that is skipped.

Endpoints Are Still Candidates

Optimization problems arrive in two flavours, and the justification differs between them.

Closed interval: compare the objective at the critical point and at both endpoints. The largest of those values wins, and the comparison is the justification.
Open interval: a single critical point plus a sign change of the derivative is enough, because there are no endpoints to beat it.

Mixing the two is a common writing error. A sign chart on a closed interval leaves the endpoints unexamined, and an endpoint comparison on an open interval refers to values that do not exist.

What the Justification Sounds Like

Three acceptable forms, in increasing order of strength.

Sign chart: " $A'$ changes from positive to negative at $x = 10$ , so $A(10)$ is the maximum."
Second derivative: " $A''(10) = -2 < 0$ , so the critical point gives a maximum."
Comparison: " $A(0) = A(20) = 0$ and $A(10) = 100$ , so the largest value is $100$ ."

Any of the three earns the point. What does not earn it is the phrase "this is obviously the maximum", which asserts the conclusion the rubric is asking you to prove.

Common Misconception Check

Myth 1: "The answer is where the derivative is zero." Truth: The answer is the best value on the allowed interval. On a closed interval that may be an endpoint.
Myth 2: "Any critical point is the optimum." Truth: A critical point can be a minimum of the objective, or a point outside the physical domain.
Myth 3: "The domain is all real numbers." Truth: Lengths are positive, and the constraint bounds them. The domain is part of the answer.

Topic-Specific Study Method

Two Minutes of Translation, One Minute of Calculus

Exam disguise	What you see on the page	What your brain must translate	Required execution
fence against a river	one side needs no fence	perimeter with three sides	objective in one variable, then optimize
open box from a sheet	squares cut from the corners	volume $= x(\text{side} - 2x)^2$	domain from the cut, then optimize
cylinder or cone	a fixed volume or surface	one formula solving for the other variable	substitute, then differentiate
"largest area for a given perimeter"	two quantities linked	square is the answer, but show it	objective, constraint, derivative
a family of curves	a constant in the formula	optimize over the shape, then solve for the constant	differentiate with respect to the shape variable

① Write the objective before reading the question a second time. The quantity to be optimized is almost always named in the first sentence.

② State the domain out loud, in symbols. Half of the wrong answers on this topic are correct algebra applied outside the geometry.

③ Justify with a named test, not with an adjective. "Changes from positive to negative" is a reason; "clearly the largest" is not.

Training & Error Diagnosis

Core Archetype (Simplest Form): A rectangle has perimeter $40$ . What is the largest possible area?

Standard Execution:

Objective. $A = xy$ .

Constraint. $2x + 2y = 40$ , so $y = 20 - x$ .

One variable. $A(x) = x(20 - x)$ on $0 < x < 20$ .

Derivative. $A'(x) = 20 - 2x = 0$ gives $x = 10$ , and $A''(10) = -2 < 0$ , so $A(10) = 100$ is the maximum. The rectangle is a square.

Solution: The base is $12 - 2x$ on each side and the height is $x$ , so $V(x) = x(12 - 2x)^2$ on $0 < x < 6$ . Expanding is unnecessary: $V'(x) = (12-2x)^2 + x \cdot 2(12-2x)(-2) = (12-2x)(12 - 6x)$ , which is zero at $x = 6$ and at $x = 2$ . Only $x = 2$ lies in the open interval. Since $V' > 0$ on $(0,2)$ and $V' < 0$ on $(2,6)$ , the volume is maximized at $x = 2$ , giving $V(2) = 2 \cdot 64 = 128$ cubic units.

Common Traps & Failure Signals:

Constraint ignored (Signal: a derivative with two variables in it): the substitution step was skipped.
Domain forgotten (Signal: a critical point at $x = 6$ reported for a box whose base would vanish): the geometry was never translated into an interval.
Endpoint unexamined (Signal: a closed-interval problem justified only by a sign chart): the comparison step was omitted.

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

Objective Unwritten: Name the quantity first.
Constraint Blindness: You treated two variables as two independent quantities, so the derivative you computed describes a surface rather than a curve, and every later step is answering a question the problem never asked.
Domain Free Fall: You accepted a solution that the picture rules out, which turns a correct algebraic answer into a physically impossible one.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Exactly one released free-response part since 2017 requires an objective function, and that part asks for the constant in a family of curves rather than for a maximum in context
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; The 2021 spinning-toy question needs the maximum of a family member before it can be solved, and no other exam in the window asks for that step
FRQ Years Appeared: 2021 (Q3b)

Practice Examples

MCQ 1

Largest Area, Fixed Perimeter

A rectangle has perimeter $20$ . What is the largest possible area?

MCQ 2

Smallest Sum of Squares

Two positive numbers have sum $10$ . What is the smallest possible value of the sum of their squares?

MCQ 3

Naming the Domain

Squares of side $x$ are cut from the corners of a $10$ -by- $10$ sheet, and the flaps are folded up to form an open box. What is the domain of the volume function?

FRQ 4

The Open Box

A square sheet of metal measures $12$ inches on each side. Squares of side $x$ inches are cut from each corner, and the remaining flaps are folded up to form an open-top box.

(a) Write an expression for the volume $V(x)$ of the box, and state the domain of $V$ that the geometry allows.

(b) Find $V'(x)$ , and find all values of $x$ at which $V'(x) = 0$ .

(c) Determine which of the values found in part (b) gives the maximum volume, and justify your answer.

(d) Find the maximum volume of the box, and state it with units.

Cheatsheet

Core Reference

	Constraint	Objective after substitution
fixed perimeter, largest area	$2x + 2y = P$	$A(x) = x\left(\frac{P}{2} - x\right)$
fixed area, least perimeter	$xy = A$	$P(x) = 2x + \frac{2A}{x}$
open box from a square sheet	corners of side $x$ removed	$V(x) = x(s-2x)^2$
fixed sum, extremal sum of squares	$x + y = S$	$Q(x) = x^2 + (S-x)^2$
fence with one side on a river	only three sides fenced	$A(x) = x(L - 2x)$

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Unit 5Optimization Problems

Concept 4.126Built for May 2027 AP Exam

Optimization Problems

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

This is the rarest topic in Unit 5 by a wide margin: exactly one released free-response part since 2017 requires building an objective function.

What you’ll see: A context with a fixed budget, a fixed perimeter, or a fixed volume, and a question about the largest or smallest value of some other quantity. On multiple-choice sections the same shape appears far more often, usually with round numbers.
What you’ll do: Write two equations, eliminate one variable, name the domain, differentiate, solve, and justify that the answer is the extremum rather than a point that merely has a zero derivative.
Memory hook: Two equations, one variable, one interval. The calculus is the last step, not the first.
Use it on the exam: For MCQs: Check the domain before you trust a critical point. Distractors are built from solutions that fall outside the geometry of the problem. For FRQs: The justification sentence is worth a point on its own. Name the test you used, whether that is a sign chart or a comparison of endpoint values.

Concept Review

Two Formulas and One Job

Figure 1 — The constraint removes a variable, the objective becomes a function of one variable, and the derivative locates the extremum. The endpoints still need checking.

$\text{constraint} \implies y = f(x) \implies A(x) = x \cdot f(x) \implies A'(x) = 0$

Optimization: the constraint reduces the objective to one variable, then the derivative locates the extremum

The Four-Step Loop

Every optimization question on the exam runs through the same four steps.

Step	What you write	Why it matters
objective	a formula for the quantity to be optimized	it names the variable you will differentiate
constraint	the equation that links the variables	it removes the second variable
domain	the interval the geometry allows	it removes answers that cannot exist
derivative	solve $A'(x) = 0$ , then compare endpoints	it locates the candidate

The third step is the one that is skipped.

Endpoints Are Still Candidates

Optimization problems arrive in two flavours, and the justification differs between them.

Closed interval: compare the objective at the critical point and at both endpoints. The largest of those values wins, and the comparison is the justification.
Open interval: a single critical point plus a sign change of the derivative is enough, because there are no endpoints to beat it.

Mixing the two is a common writing error. A sign chart on a closed interval leaves the endpoints unexamined, and an endpoint comparison on an open interval refers to values that do not exist.

What the Justification Sounds Like

Three acceptable forms, in increasing order of strength.

Sign chart: " $A'$ changes from positive to negative at $x = 10$ , so $A(10)$ is the maximum."
Second derivative: " $A''(10) = -2 < 0$ , so the critical point gives a maximum."
Comparison: " $A(0) = A(20) = 0$ and $A(10) = 100$ , so the largest value is $100$ ."

Any of the three earns the point. What does not earn it is the phrase "this is obviously the maximum", which asserts the conclusion the rubric is asking you to prove.

Common Misconception Check

Myth 1: "The answer is where the derivative is zero." Truth: The answer is the best value on the allowed interval. On a closed interval that may be an endpoint.
Myth 2: "Any critical point is the optimum." Truth: A critical point can be a minimum of the objective, or a point outside the physical domain.
Myth 3: "The domain is all real numbers." Truth: Lengths are positive, and the constraint bounds them. The domain is part of the answer.

Topic-Specific Study Method

Two Minutes of Translation, One Minute of Calculus

Exam disguise	What you see on the page	What your brain must translate	Required execution
fence against a river	one side needs no fence	perimeter with three sides	objective in one variable, then optimize
open box from a sheet	squares cut from the corners	volume $= x(\text{side} - 2x)^2$	domain from the cut, then optimize
cylinder or cone	a fixed volume or surface	one formula solving for the other variable	substitute, then differentiate
"largest area for a given perimeter"	two quantities linked	square is the answer, but show it	objective, constraint, derivative
a family of curves	a constant in the formula	optimize over the shape, then solve for the constant	differentiate with respect to the shape variable

① Write the objective before reading the question a second time. The quantity to be optimized is almost always named in the first sentence.

② State the domain out loud, in symbols. Half of the wrong answers on this topic are correct algebra applied outside the geometry.

③ Justify with a named test, not with an adjective. "Changes from positive to negative" is a reason; "clearly the largest" is not.

Training & Error Diagnosis

Core Archetype (Simplest Form): A rectangle has perimeter $40$ . What is the largest possible area?

Standard Execution:

Objective. $A = xy$ .

Constraint. $2x + 2y = 40$ , so $y = 20 - x$ .

One variable. $A(x) = x(20 - x)$ on $0 < x < 20$ .

Derivative. $A'(x) = 20 - 2x = 0$ gives $x = 10$ , and $A''(10) = -2 < 0$ , so $A(10) = 100$ is the maximum. The rectangle is a square.

Solution: The base is $12 - 2x$ on each side and the height is $x$ , so $V(x) = x(12 - 2x)^2$ on $0 < x < 6$ . Expanding is unnecessary: $V'(x) = (12-2x)^2 + x \cdot 2(12-2x)(-2) = (12-2x)(12 - 6x)$ , which is zero at $x = 6$ and at $x = 2$ . Only $x = 2$ lies in the open interval. Since $V' > 0$ on $(0,2)$ and $V' < 0$ on $(2,6)$ , the volume is maximized at $x = 2$ , giving $V(2) = 2 \cdot 64 = 128$ cubic units.

Common Traps & Failure Signals:

Constraint ignored (Signal: a derivative with two variables in it): the substitution step was skipped.
Domain forgotten (Signal: a critical point at $x = 6$ reported for a box whose base would vanish): the geometry was never translated into an interval.
Endpoint unexamined (Signal: a closed-interval problem justified only by a sign chart): the comparison step was omitted.

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

Objective Unwritten: Name the quantity first.
Constraint Blindness: You treated two variables as two independent quantities, so the derivative you computed describes a surface rather than a curve, and every later step is answering a question the problem never asked.
Domain Free Fall: You accepted a solution that the picture rules out, which turns a correct algebraic answer into a physically impossible one.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Exactly one released free-response part since 2017 requires an objective function, and that part asks for the constant in a family of curves rather than for a maximum in context
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; The 2021 spinning-toy question needs the maximum of a family member before it can be solved, and no other exam in the window asks for that step
FRQ Years Appeared: 2021 (Q3b)

Practice Examples

MCQ 1

Largest Area, Fixed Perimeter

A rectangle has perimeter $20$ . What is the largest possible area?

MCQ 2

Smallest Sum of Squares

Two positive numbers have sum $10$ . What is the smallest possible value of the sum of their squares?

MCQ 3

Naming the Domain

Squares of side $x$ are cut from the corners of a $10$ -by- $10$ sheet, and the flaps are folded up to form an open box. What is the domain of the volume function?

FRQ 4

The Open Box

A square sheet of metal measures $12$ inches on each side. Squares of side $x$ inches are cut from each corner, and the remaining flaps are folded up to form an open-top box.

(a) Write an expression for the volume $V(x)$ of the box, and state the domain of $V$ that the geometry allows.

(b) Find $V'(x)$ , and find all values of $x$ at which $V'(x) = 0$ .

(c) Determine which of the values found in part (b) gives the maximum volume, and justify your answer.

(d) Find the maximum volume of the box, and state it with units.

Cheatsheet

Core Reference

	Constraint	Objective after substitution
fixed perimeter, largest area	$2x + 2y = P$	$A(x) = x\left(\frac{P}{2} - x\right)$
fixed area, least perimeter	$xy = A$	$P(x) = 2x + \frac{2A}{x}$
open box from a square sheet	corners of side $x$ removed	$V(x) = x(s-2x)^2$
fixed sum, extremal sum of squares	$x + y = S$	$Q(x) = x^2 + (S-x)^2$
fence with one side on a river	only three sides fenced	$A(x) = x(L - 2x)$

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