on the stick and rope problem - draft 1

On The Stick and Rope Problem

Iwan Pranoto

Institute of Technology Bandung

pranoto@math.itb.ac.id

November 19, 2012

Problem

Isoperimetric Problem in Halfplane?

Finding a smooth function x on I = [0, 1] satisfyingx(0) = x(1) = 0 and the area formed by the graph of x and theinterval I is maximum, with a constrain: the length of the graph ofx must be `, where ` is a positive number given but fixed.

Stick and Rope

Figure: Free Ends, Left; Fixed Ends, Right

In this paper, we consider the right situation.

Figure: Half a Bubble

We assume x(t) ≥ 0 for all t. Therefore, the area A of the regionbounded by the graph of x and I is to be maximized

A[x ] =

∫ 1

0x(t) dt,

and at the same time the length L[x ] = `, where

L[x ] =

∫ 1

0

√1 + (x)2 dt. (1)

Results

TheoremIf 1 < ` ≤ π

2 , the optimal solution of the problem is a segment of acircle whose center is on the vertical line t = 1

2 .

Proof

Using Lagrange method, define a new form

A[x , x , λ] =

∫ 1

0

[x(t)− λ

(√1 + (x)2 − `

)]dt.

After substituting this into the Euler-Lagrange equation, we obtain

∂

∂x

[x(t)− λ

(√1 + (x)2 − `

)]

=d

dt

(∂

∂x

[x(t)− λ

(√1 + (x)2 − `

)]). (2)

This leads to

1 =d

dt

(λx√

1 + (x)2

).

After integrating both sides with respect to t, we obtain

t =λx√

1 + (x)2+ C1

for some constant C1. Thus, x must satisfy

(t − C1)2 =(λx)2

1 + (x)2. (3)

Since x vanishes only if t = C1, one obtains

1 + (x)2

x2=

λ2

(t − C1)2.

Thus,1

x2=λ2 − (t − C1)2

(t − C1)2,

or

x = ± (t − C1)√λ2 − (t − C1)2

.

Therefore, x must satisfy

x = ±√λ2 − (t − C1)2 + C2

for some constant C2, or

(x − C2)2 = λ2 − (t − C1)2. (4)

This means, x must satisfy the equation of a circle whose center is(C1,C2)

(t − C1)2 + (x − C2)2 = λ2. (5)

After using the conditions x(0) = x(1) = 0, one obtains C1 = 12 .

Thus, the circle’s center must lie on the vertical line t = 12 .

One may compute the other constants, the center’s ordinate −|C2|and radius |λ|, geometrically.

�

Special Cases

Figure: Left ` = π/2; Right ` = π/3

How if ` > π/2?

We consider ` = 2 and a special kind of functions:

x(t) = a

(|t − 1

2|m −

(1

2

)m),

where m is any natural number greater than 1 and a is someconstant to be chosen, depending on m.

Figure: Table, Left; Numerical Results, Right

How About A Rectangle and A Semicircle?

Figure: The Area = (1− π8 ) ≈ 0.60730091830 · · ·

The End