CSUN MATH 150A Optimization Study Notes: Maximum Area for
a Given Perimeter
Topic: Maximizing Functions Subject to Constraints – The Calculus of Optimization
Context: Applications of the Derivative (Module 4)
Part I: Student Insight – The Necessity of Calculus
My initial encounter with the problem of finding the maximum area for a fixed
perimeter was met with an intuitive answer: it must be a square. Geometrically, a
square is the most "balanced" shape, and it simply feels right that it would enclose
the most space. However, as a student of MATH 150A, the most powerful learning
realization is that intuition is not proof, and proof is the domain of Calculus.
The beauty of this optimization problem lies in its transition from a simple
geometric observation to a rigorous mathematical demonstration using the power
of the first derivative. We are no longer guessing; we are proving that the derivative
of the area function must be zero at the point where the geometric shape transitions
from increasing its area to decreasing it. This critical point, where the instantaneous
rate of change is zero, is the heart of optimization.
This problem is a foundational microcosm of all real-world calculus applications.
Every time an engineer minimizes material cost, a physicist maximizes efficiency, or
an economist maximizes profit, they are performing the same core algebraic steps:
identifying the objective function, defining the constraint, substituting to create a
single-variable model, and finding the roots of the first derivative. The maximum
area problem is the perfect introduction to this powerful, universal methodology,
teaching us not just what the answer is, but why it must be so, with the unwavering
certainty provided by differential calculus. The structure—Primary Equation,
Constraint Equation, Simplification, Differentiation, and Justification—must become
second nature.
Part II: Theoretical Foundation – The Calculus of Extreme Values
Before solving the core problem, it is essential to establish the theoretical
scaffolding that validates our procedure. Optimization problems rely entirely on the
theorems governing extreme values of a function.
1. The Extreme Value Theorem (EVT) and the Principle of Critical Points
The Extreme Value Theorem (EVT) is the guarantee that an absolute maximum and
minimum exist for a continuous function on a closed interval
[a , b]
. While our
function might seem open-ended, the physical constraints of geometry will always
impose a natural domain, making the EVT relevant.
The EVT and Critical Points: If a function
f(x)
has an absolute extremum on an
interval, that extremum must occur at one of three types of points:
Critical Points: Where the first derivative
f′(x)
is equal to zero or is undefined.
Endpoints: The boundaries of the domain (where
x=a
or
x=b
).
Our methodology is therefore robust: we find all critical points and compare the
functions value at these points and the endpoints to definitively identify the
absolute maximum.
2. The First Derivative Test (FDT) for Local Extrema
The First Derivative Test is the method used to classify a critical point as a local
maximum, local minimum, or neither.
If
f′(x)
changes sign from positive (
+¿
) to negative (
−
), the function
f(x)
has a local
maximum at
x=c
. The function was increasing and is now decreasing.
If
f′(x)
changes sign from negative (
−
) to positive (
+¿
), the function
f(x)
has a local
minimum at
x=c
.
If
f′(x)
does not change sign, the point is an inflection point.
For optimization problems, the FDT provides the necessary local confirmation
before comparing with the endpoints.
3. The Second Derivative Test (SDT) for Concavity and Classification
The Second Derivative Test provides a quicker, though sometimes inconclusive,
method for classifying critical points by analyzing the functions concavity.
If
f′(c)=0
and
f″(c)<0
, the function is concave down at
c
, meaning
c
is a local
maximum.
If
f′(c)=0
and
f″(c)>0
, the function is concave up at
c
, meaning
c
is a local
minimum.
If
f″(c)=0
, the test is inconclusive, and one must resort to the FDT.
In the case of the maximum area problem, the SDT provides a highly efficient and
elegant proof of concavity, confirming the critical point is indeed a maximum.
Part III: The Core Optimization Problem – Maximum Area of a Rectangle
We now apply the rigorous four-step calculus method to the problem: Given a fixed
perimeter
P
, find the dimensions
L
(length) and
W
(width) of a rectangle that
maximize its area
A
.
1. Step 1: Define the Primary and Secondary Equations
We need two equations to describe the system: the function we wish to maximize
(Primary) and the fixed relationship between the variables (Secondary/Constraint).
Primary Equation (Objective Function): The function we want to maximize is the
area
A
.
A=L⋅W
Secondary Equation (Constraint): The perimeter
P
is fixed.
P=2L+2W
Here,
P
is a constant value, even though we do not specify a number (e.g.,
P=100
meters).
2. Step 2: Substitution and Domain
We must convert the two-variable objective function
A(L , W )
into a single-variable
function,
A(L)
or
A(W)
, using the constraint. We choose to solve the constraint for
W
:
2W=P −2LW =P
2− L
Now, substitute this expression for
W
into the Primary Area equation:
A(L)=L⋅
(
P
2− L
)
A(L)= P
2L− L2
Domain Analysis: The physical dimensions
L
and
W
must be non-negative.
L ≥0
W=P
2− L≥ 0⇒L ≤ P
2
Therefore, the domain for the variable
L
is the closed interval:
D:
[
0,P
2
]
This finite, closed domain guarantees by the EVT that an absolute maximum area
exists.
3. Step 3: Differentiation and Finding Critical Points
We now find the derivative of the single-variable area function
A(L)
with respect to
L
. Recall that
P
is treated as a constant.
A(L)= P
2L − L2A′(L)= d
dL
(
P
2L− L2
)
A′(L)= P
2−2L
To find the critical points, we set the first derivative equal to zero and solve for
L
:
A′(L)=0P
2−2L=02 L=P
2L=P
4
The only critical point is
L=P
4
. Since
P
is a positive constant, this critical point lies
strictly within our domain
D
.
4. Step 4: Classification and Conclusion
We use the Second Derivative Test (SDT) to classify this critical point
L=P
4
.
First, we calculate the second derivative
A″(L)
:
A″(L)= d
dL
(
P
2−2L
)
A″(L)=−2
Since the second derivative
A″(L)
is
−2
, which is a negative constant for all values of
L
, we conclude that the area function
A(L)
is concave down everywhere.
By the SDT, since
A′(P
4)=0
and
A″(P
4)=−2<0
, the critical point
L=P
4
corresponds
to a local maximum. Since the function is concave down everywhere, this local
maximum is also the absolute maximum over the entire domain.
Finding the Dimensions: Now we substitute
L=P
4
back into our expression for
W
:
W=P
2− LW =P
2−P
4W=2P
4−P
4W=P
4
Since
L=P
4
and
W=P
4
, we have
L=W
.
The Conclusion of the Proof: For a fixed perimeter
P
, the rectangle that yields the
maximum area is a square with side length
P
4
.
Checking the Endpoints: As a final step required by the EVT, we check the endpoints
of the domain
D:[0,P
2]
.
At
L=0
(Endpoint 1):
A(0)= P
2(0)−¿
.
At
L=P
2
(Endpoint 2):
W=P
2−P
2=0
.
A(P
2)= P
2(P
2)−¿
.
The area at the critical point
L=P
4
is
A(P
4)= P
2(P
4)−¿
.
Since
P2
16 >0
, the maximum area occurs at the critical point, confirming our rigorous
derivation.
Part IV: Related Optimization Examples and Variations
To deepen the understanding of this optimization method, we explore variations
that demonstrate the power and adaptability of the calculus approach.
1. Variation A: Minimizing Perimeter for a Fixed Area
A
This is the inverse problem: Given a fixed area
A
, what dimensions minimize the
perimeter
P
?
Primary Equation (Objective Function): Minimize the perimeter
P
.
P=2L+2W
Secondary Equation (Constraint): The area
A
is fixed.
A=L⋅W
Solve the constraint for
W
:
W=A
L
.
Substitution: Substitute
W
into the Primary Equation to get
P(L)
.
P(L)=2L+2
(
A
L
)
P(L)=2L+2A L−1
Domain Analysis: Since
A
is fixed and
L
must be positive, the domain for
L
is
(0, ∞ )
.
We cannot use the EVT directly, but we will rely on the FDT/SDT and the nature of
the function (since it is a single local minimum, it must be the global minimum).
Differentiation and Critical Points:
P′(L)= d
dL (2L+2A L−1)P′(L)=2−2A L−2
Set the derivative to zero:
2−2A
L2=02=2A
L2L2=AL=
√
A
Since length must be positive, the only critical point is
L=
√
A
.
Classification using SDT:
P″(L)= d
dL (2−2A L−2)P″(L)=−2A(−2)L−3P″(L)= 4A
L3
Substitute the critical point
L=
√
A
:
P″(
√
A)= 4A
¿¿
Since
A
(area) is positive,
P″(
√
A)>0
. By the SDT, the critical point is a local
minimum.
Conclusion: The dimensions that minimize the perimeter for a fixed area
A
occur
when
L=
√
A
. Since
W=A
L=A
√
A=
√
A
, we again prove that
L=W
. The geometric
optimal shape for both maximizing area (fixed perimeter) and minimizing perimeter
(fixed area) is the square.
2. Variation B: Maximizing Area with a Fixed Perimeter and a River Boundary
(The Fence Problem)
A farmer wishes to enclose a rectangular field adjacent to a straight river, where no
fencing is needed along the river side. Given a fixed length of fence
F
(the perimeter
constraint), find the dimensions that maximize the enclosed area
A
.
Let
L
be the length parallel to the river and
W
be the width perpendicular to the
river (there are two of these).
Primary Equation (Objective Function): Maximize the area
A
.
A=L⋅W
Secondary Equation (Constraint): The total fence length
F
is fixed.
F=L+2W
Solve the constraint for
L
:
L=F − 2W
. (Choosing
L
simplifies the subsequent
derivative).
Substitution: Substitute
L
into the Primary Equation to get
A(W)
.
A(W)=(F − 2W)⋅WA (W)=FW −2W2
Domain Analysis: The widths and length must be non-negative.
W ≥ 0
L=F − 2W ≥ 0⇒2W ≤ F ⇒W ≤ F
2
The domain for the variable
W
is the closed interval:
D:[0,F
2]
.
Differentiation and Critical Points:
A′(W)= d
dW (FW −2W2)A′(W)=F −4W
Set the derivative to zero:
F − 4W=04 W=FW =F
4
The critical point is
W=F
4
.
Classification and Conclusion:
Calculate the second derivative
A″(W)
:
A″(W)= d
dW (F − 4W)A″(W)=−4
Since
A″(W)=−4
(a negative constant), the function
A(W)
is concave down, and the
critical point
W=F
4
is the absolute maximum.
Finding the Dimensions: Substitute
W=F
4
back into the expression for
L
:
L=F − 2WL=F − 2
(
F
4
)
L=F − F
2L=F
2
Conclusion of Variation B: The dimensions that maximize the area with a river
boundary are
L=F
2
and
W=F
4
. In this case, the length parallel to the river (
L
) is
twice the width perpendicular to the river (
W
). This is no longer a square,
demonstrating how changing the constraint equation dramatically alters the
geometric result, while the calculus methodology remains perfectly consistent.
Part V: Generalization and Advanced Mathematical Context
The problem of maximizing area for a fixed perimeter is part of a much larger and
historically significant branch of mathematics called the Isoperimetric Problem.
Discussing this provides advanced context for the MATH 150A student.
1. The Isoperimetric Theorem and The General Solution
The maximum area problem we solved is a constrained version of the general
Isoperimetric Problem: among all closed plane curves of a given length
P
, which one
encloses the maximum area?
The Isoperimetric Theorem: The closed curve that encloses the maximum area for a
fixed perimeter is the circle.
Our rectangular problem is limited by a stronger constraint: the shape must be a
rectangle. The calculus proof confirms that the most circle-like rectangle—the one
with the minimum aspect ratio, the square—is the optimal solution within that
constraint set. The square is the quadrilateral of maximum area for a fixed
perimeter, just as the circle is the closed curve of maximum area. This link between
the local calculus solution and the global geometric principle is conceptually
powerful.
2. The Power of Multivariate Optimization: Lagrange Multipliers
While the substitution method is mandatory for single-variable calculus (MATH
150A), the problem is intrinsically multivariate and can be solved more elegantly
using Lagrange Multipliers, a topic typically covered in Calculus III (MATH 252 or
equivalent) but valuable for insight here.
The method solves the optimization problem without substitution by introducing a
new variable,
λ
(lambda), the Lagrange Multiplier.
Objective Function to Maximize:
f(L ,W )=L⋅W
Constraint Function:
g(L , W )=2L+2W − P=0
The core principle states that at the maximum (or minimum), the gradient of the
objective function (
∇f
) must be parallel to the gradient of the constraint function (
∇g
). Mathematically:
∇f=λ∇g
This expands into a system of three equations:
∂ f
∂ L =λ∂ g
∂ L ⇒W=λ(2)
∂ f
∂W =λ∂ g
∂W ⇒L=λ(2)
2L+2W=P
From the first two equations,
W=2λ
and
L=2λ
. Therefore,
L=W
.
Substituting
L=W
into the constraint (Equation 3):
2L+2L=P4L=PL=P
4
Conclusion (Lagrange): The result
L=W=P
4
is confirmed without needing the
cumbersome substitution step. This demonstrates how advanced calculus simplifies
the process while resting on the same geometric foundation: the critical point
occurs where the level curves of the objective function are tangent to the constraint
curve. The variable
λ
itself has a powerful interpretation as the rate of change of the
optimal value with respect to a small change in the constraint—in this case,
λ
is the
instantaneous change in maximum area per unit change in perimeter.
Part VI: Pitfalls, Domain Issues, and Real-World Constraints
A complete calculus analysis must acknowledge the subtleties and potential pitfalls,
especially in applied problems.
1. The Importance of the Closed Interval and Endpoints
If we had ignored the domain analysis in Part III, we would have found only one
critical point,
L=P
4
. While this point is the absolute maximum, without checking the
endpoints
L=0
and
L=P
2
, we fail to fully satisfy the conditions of the Extreme Value
Theorem.
In most physical optimization problems, the endpoints represent the most trivial or
degenerate solutions (e.g., zero area), but their explicit inclusion in the verification
process is non-negotiable for a rigorous mathematical proof. Had the area function
A(L)
been more complex (e.g.,
A(L)=L3−3L2+5
on the interval
[0,5]
), the absolute
maximum could have occurred at an endpoint, invalidating the critical point
solution. The optimization procedure is a two-step process: find the critical points
and compare against the boundaries.
2. Error Analysis and Practical Application
In a real-world scenario, measurements of the perimeter
P
will contain some small
error,
Δ P
. The concept of the differential from Calculus I allows us to estimate the
resulting error in the maximum area,
Δ A
.
The maximum area is
A(P)= P2
16
. The error in area
dA
is estimated by the
differential:
dA=A′(P)dP
First, find the derivative of the area function with respect to the constraint variable
P
:
A′(P)= d
dP
(
P2
16
)
=2P
16 =P
8
If the measured perimeter
P
has an error of
dP=Δ P
, the corresponding error in the
calculated maximum area is:
Δ A ≈ P
8⋅Δ P
This estimation allows engineers to quantify the uncertainty in their optimized
solution based on the precision of their initial measurements, demonstrating the
critical link between differential calculus and applied physics.
3. Non-Continuity and Physical Jumps
The calculus method assumes continuity and differentiability of the objective
function. While area and perimeter are always continuous, real-world constraints
often introduce discontinuities. For example:
Discrete Units: If a fence must be bought in discrete 1-meter lengths, the variable
L
can only take integer values. The continuous solution
L=P
4
might not be attainable.
The optimal solution is the integer value closest to
P
4
.
Cost Jumps: If the cost of fencing materials jumps sharply at a certain length (e.g.,
L>50
meters), the constraint might become piecewise, potentially causing a
maximum to occur at the point of discontinuity, where the derivative is undefined.
In such cases, the calculus solution provides the ideal, theoretical optimum, but the
practical solution requires checking the values adjacent to the discontinuity or the
nearest feasible discrete points.
Part VII: Summary of the Optimization Toolkit
The successful execution of the maximum area problem provides the fundamental
framework for solving all optimization problems in MATH 150A. The process is
defined by the following sequence of transformations and calculations:
Modeling (The Setup):
Identify the physical variables (
L ,W
).
Write the Primary Equation (the function to maximize/minimize).
Write the Secondary Equation (the constraint that links the variables).
Reduction (The Calculus Pre-work):
Solve the Secondary Equation for one variable.
Substitute this expression into the Primary Equation to obtain a single-variable
function,
f(x)
.
Determine the Domain of
f(x)
based on the physical constraints.
Analysis (The Calculus Core):
Calculate the first derivative,
f′(x)
.
Set
f′(x)=0
to find the Critical Points.
Verification (The Conclusion):
Use the First Derivative Test (FDT) or Second Derivative Test (SDT) to classify the
critical points.
Compare the functions value at the critical points and the Endpoints of the domain
to identify the absolute maximum or minimum.
Translate the mathematical result back into the physical context (e.g., "The
dimensions are
L=W=P
4
").
The problem of maximizing the area of a rectangle for a given perimeter is a perfect
illustration of this toolkit, resulting in the elegant and geometrically intuitive
conclusion that the optimal rectangle is a square. This conclusion, however, is now
substantiated not by mere intuition, but by the irrefutable power of differential
calculus.
