The Classical Isoperimetric Problem: Worked Examples and Proofs

The classical isoperimetric problem asks a deceptively simple question:

Among closed planar curves having a prescribed perimeter, which curve encloses the largest area?

The answer is the circle. Long before the modern calculus of variations, this geometric extremum appeared in the traditional story of Dido, who sought to enclose as much land as possible with a fixed length of boundary. In the eighteenth century, isoperimetric problems became one of the central classes that motivated the systematic development of variational methods. Euler’s Methodus inveniendi lineas curvas maximi minimive proprietate gaudentes treated such constrained extremal problems as part of the emerging calculus of variations [5, 1, 3].

CV09 introduced the multiplier rule for a functional constraint. This companion article uses that machinery in detail. The main goals are to derive circular arcs from the Euler–Lagrange equation, understand the endpoint geometry in the Dido problem, connect the multiplier to curvature, and finally distinguish a stationary circle from the much stronger global isoperimetric theorem.

A recurring theme will be

(1)

The final exercise goes beyond stationarity and proves the global inequality

(2)

for a sufficiently regular simple closed planar curve, using the periodic Wirtinger inequality.

1 Exercises

Exercise 1: set up the graph-form isoperimetric problem

Let a smooth graph y(x) join

(3)

with y(x) ≥ 0. Its enclosed area above the x-axis is

(4)

and its arc length is

(5)

Suppose S[y] = L is prescribed.

1. Form an augmented functional whose stationary functions are candidates for maximum area.
2. Derive the Euler–Lagrange equation.
3. Show that the stationary graph has constant signed curvature.

Exercise 2: integrate the constant-curvature equation

Continue Exercise 1 and assume the extremal is symmetric about x = 0. Let R > 0 denote the radius associated with the constant curvature.

1. Show that the upper stationary graph can be written

   (6)

2. Introduce an angle α by

   (7)

   Show that

   (8)

3. Interpret the result geometrically.

Exercise 3: the semicircle special case

For the fixed-chord problem of Exercise 2, suppose

(9)

Show that R = a and therefore

(10)

Compute the enclosed area.

Exercise 4: Dido’s straight-shoreline problem

A curve of prescribed length L lies above a straight shoreline. Its two endpoints may slide freely along the shoreline, and together the curve and the shoreline segment enclose an area.

1. Explain why transversality requires the stationary curve to meet the shoreline orthogonally.
2. Combine this endpoint geometry with the constant-curvature result to show that the stationary curve is a semicircle.
3. Find its radius and enclosed area in terms of L.

Exercise 5: closed-curve vector formulation

Let a smooth simple closed curve be represented by

(11)

with periodic endpoint conditions. Its oriented area and perimeter are

(12)

and

(13)

Use a constant multiplier λ to extremize A subject to fixed P.

1. Derive the two Euler–Lagrange equations.
2. Introduce the unit tangent T and show that its derivative with respect to arc length has constant magnitude.
3. Conclude that every regular stationary closed curve has constant curvature and is therefore a circle.

Exercise 6: same perimeter, different areas

For a fixed perimeter P, compute the enclosed area of

1. a circle;
2. a square;
3. a regular n-gon.

Show that the regular n-gon area tends to the circle area as n →∞. Evaluate the area ratio A_square∕A_circle.

Exercise 7: the multiplier as a sensitivity

The globally optimal circle of perimeter P has

(14)

1. Compute dA_∗∕dP.
2. Express the result in terms of the optimal circle radius R.
3. Relate this to the magnitude of the Lagrange multiplier appearing in Exercise 5, noting that its sign depends on the chosen augmented-functional convention and orientation.

Exercise 8: global proof using the periodic Wirtinger inequality

Let r(s) = (x(s),y(s)) be a regular simple closed curve parametrized by arc length s ∈ [0,P]. Translate the origin so that

(15)

Use the following periodic Wirtinger inequality for any mean-zero P-periodic function f:

(16)

Starting from

(17)

prove

(18)

Explain why equality corresponds to a circle.

2 Solutions

Solution 1: graph formulation and constant curvature

The objective is area,

(19)

while the constraint is

(20)

Using a constant multiplier λ, define the augmented integrand

(21)

The sign of λ is conventional; replacing λ by −λ changes no physics or geometry.

We have

(22)

and

(23)

The Euler–Lagrange equation gives

(24)

Because λ is constant,

(25)

Differentiate explicitly:

	= −	(26)
	= .	(27)

Hence

(28)

The left-hand side is the signed curvature of the graph. Therefore every smooth stationary graph has constant signed curvature. A nonzero constant-curvature plane curve is an arc of a circle.

This conclusion is a necessary stationarity condition. By itself, it does not yet prove that the circle is the global maximizer.

Solution 2: integrating to obtain the circular arc

For an upper concave-down arc, it is convenient to write

(29)

Integrating once gives

(30)

Symmetry about x = 0 implies y′(0) = 0, hence C₁ = 0:

(31)

Square and solve for y′:

(32)

so

(33)

The upper arc has negative slope for x > 0, therefore

(34)

Integrating,

(35)

The endpoint condition y(a) = 0 gives

(36)

hence

(37)

This is a circle of radius R whose center lies below the chord.

Now define α by

(38)

The circular arc subtends the central angle 2α, so its length is

(39)

Equivalently,

(40)

Thus the prescribed chord and prescribed length together determine the radius of the stationary circular arc.

For the graph formulation, the minor-arc family runs from the straight-line limit L → 2a to the semicircle L = πa.

Solution 3: the semicircle

If

(41)

then the circular arc must satisfy

(42)

and

(43)

The semicircle corresponds to

(44)

so

(45)

The stationary graph therefore becomes

(46)

Its area is the area of a semicircle:

(47)

Notice the distinction between the semicircle as a stationary solution and the statement that it gives the largest area among all admissible curves. The latter is a global theorem; Exercise 8 supplies a global argument for the closed-curve version.

Solution 4: Dido’s shoreline problem

The Dido-type problem differs from the fixed-chord problem because the endpoints may slide along the straight shoreline. Endpoint freedom adds transversality information.

In the graph formulation the shoreline is y = 0. At a sliding endpoint,

(48)

while δx is free. The moving-endpoint boundary term from CV05 therefore requires

(49)

For

(50)

we obtain

H − y′Hy′	= y + λ − λ	(51)
	= y + .	(52)

At the shoreline y = 0, transversality gives

(53)

For a nontrivial finite multiplier this requires

(54)

Thus the tangent to the stationary arc is vertical at the shoreline: the curve meets the straight boundary orthogonally.

A constant-curvature arc that meets the same straight line orthogonally at both endpoints must be a semicircle. If its radius is R, then its curved-boundary length is

(55)

so

(56)

The enclosed area is therefore

(57)

This is the familiar “Dido problem” version of isoperimetric geometry [2, 6].

Solution 5: closed curves and the vector Euler–Lagrange equations

Define

(58)

and use the augmented integrand

(59)

For x,

(60)

and

(61)

Hence

0	= Fx −Fẋ	(62)
	= ẏ −,	(63)

so

(64)

Similarly, for y,

(65)

and

(66)

which gives

(67)

Introduce the unit tangent

(68)

Because

(69)

we have

(70)

The two equations become

(71)

Therefore

(72)

But

(73)

so the curvature magnitude is constant. A regular closed plane curve with nonzero constant curvature is a circle. Thus the vector formulation reaches the same stationary geometry without privileging a graph coordinate.

This formulation also shows why the multiplier has units of length: its magnitude is the stationary radius.

Solution 6: comparing equal-perimeter shapes

For a circle of perimeter P,

(74)

so

(75)

Hence

(76)

For a square, the side length is P∕4, giving

(77)

Therefore

(78)

For a regular n-gon of side length s = P∕n, the apothem is

(79)

The area is one half perimeter times apothem:

(80)

As n →∞,

(81)

so

(82)

These comparisons are illustrative, not a proof against all possible curves.

Solution 7: multiplier sensitivity

For the globally optimal circle,

(83)

Differentiate:

(84)

Since

(85)

we have

(86)

Thus an infinitesimal increase dP in the available perimeter increases the optimal area by approximately

(87)

In Exercise 5, the stationary radius satisfies

(88)

The sign depends on whether the augmented functional is written as A + λ(P − P₀) or A − μ(P − P₀) and on the orientation of the curve. The magnitude has the sensitivity interpretation expected from the multiplier discussion in CV09.

Solution 8: the global isoperimetric inequality

Now we prove that the circle is not merely stationary but globally optimal under the stated smoothness assumptions. Parametrize the closed curve by arc length s, so

(89)

Therefore

(90)

The oriented area is

(91)

Introduce the quarter-turn operator

(92)

Then

(93)

Since J preserves Euclidean length,

(94)

Cauchy–Schwarz gives

|A|	≤1∕21∕2	(95)
	= 1∕2P1∕2.	(96)

Translate the origin so that both coordinate functions have zero mean. The periodic Wirtinger inequality gives

(97)

and

(98)

Add them:

∫ 0P |r|2 ds	≤2 ∫ 0P |r′|2 ds	(99)
	= 2P	(100)
	= .	(101)

Substitute into the Cauchy–Schwarz bound:

|A|	≤1∕2P1∕2	(102)
	= .	(103)

To understand equality, both inequalities must be sharp. Equality in the periodic Wirtinger inequality requires x and y to contain only the first periodic harmonic after centering. Equality in Cauchy–Schwarz requires the position vector and the quarter-turn of the unit tangent to be linearly dependent with constant proportionality. Together with unit-speed parametrization, these conditions reduce to

(104)

where

(105)

Thus equality is attained by a circle, and the stationary circle found by the multiplier calculation is indeed the global maximizer. This classical proof strategy and related variants are discussed in standard treatments of the isoperimetric inequality [4, 6].

3 What the worked examples establish

The examples reveal three logically distinct levels of conclusion.

1. Multiplier stationarity: constrained Euler–Lagrange gives constant curvature.
2. Endpoint geometry: transversality can select which circular arc is admissible, as in the Dido semicircle.
3. Global optimality: an inequality argument is required to prove that no other admissible closed curve encloses more area.

It is important not to collapse these three statements into one. Euler–Lagrange supplies a necessary condition. The isoperimetric inequality supplies the global comparison.

4 Common mistakes

• Calling the circle a maximum immediately after Euler–Lagrange. Constant curvature identifies stationary candidates; it does not by itself prove global maximality.
• Forgetting the original perimeter constraint. The differential equation gives a family of circles; the prescribed length fixes the radius or remaining parameter.
• Using a variable multiplier for a single scalar perimeter constraint. The isoperimetric multiplier is constant.
• Ignoring endpoint freedom in the Dido problem. Sliding endpoints contribute transversality conditions and force orthogonal contact with the shoreline.
• Confusing the graph formulation with the full closed-curve problem. A single graph cannot represent every closed curve or every major circular arc.
• Dropping orientation signs. The signed curvature and the multiplier may change sign if the curve orientation or augmented-functional convention changes; the radius is positive and corresponds to the magnitude.
• Using the area formula without regularity assumptions. The global proof assumes a sufficiently regular simple closed curve so that arc-length parametrization and the line-integral area formula are valid.

5 Bridge to later variational theory

The classical isoperimetric problem anticipates several ideas that recur later in the PhysicsLibrary CV sequence.

The multiplier behaves like a conjugate sensitivity variable. The closed-curve derivation is naturally vector-valued, linking directly to CV07. Endpoint orthogonality uses the transversality ideas of CV05. The constant-curvature result is a geometric first integral in spirit, linking to CV06. Finally, the global proof requires an inequality rather than only stationarity, foreshadowing the second-variation and sufficiency theory developed in CV11–CV13.

References

References