This document discusses the stick and rope problem of finding a smooth function that maximizes the area under the graph subject to the constraint that the length of the graph is a given fixed value. The problem is analyzed for the case where both ends of the rope are fixed at zero. It is shown that when the fixed length is between 1 and π/2, the optimal solution is a segment of a circle with its center on the vertical line at t=1/2. The proof uses Lagrange multipliers to derive an equation that the optimal function must satisfy, showing it is the equation of a circle. Boundary conditions then determine the circle's parameters. Special cases for longer rope lengths are also discussed

1. On The Stick and Rope Problem
Iwan Pranoto
Institute of Technology Bandung
pranoto@math.itb.ac.id
November 19, 2012

2. Problem
Isoperimetric Problem in Halfplane?
Finding a smooth function x on I = [0, 1] satisfying
x(0) = x(1) = 0 and the area formed by the graph of x and the
interval I is maximum, with a constrain: the length of the graph of
x must be , where is a positive number given but fixed.

3. Stick and Rope
Figure: Free Ends, Left; Fixed Ends, Right
In this paper, we consider the right situation.

4. Figure: Half a Bubble

5. We assume x(t) ≥ 0 for all t. Therefore, the area A of the region
bounded by the graph of x and I is to be maximized
1
A[x] = x(t) dt,
0
and at the same time the length L[x] = , where
1
L[x] = 1 + (x)2 dt.
˙ (1)
0

6. Results
Theorem
If 1 < ≤ π , the optimal solution of the problem is a segment of a
2
circle whose center is on the vertical line t = 1 .
2

7. Proof
Using Lagrange method, define a new form
1
¯
A[x, x, λ] =
˙ x(t) − λ 1 + (x)2 −
˙ dt.
0
After substituting this into the Euler-Lagrange equation, we obtain
∂
x(t) − λ 1 + (x)2 −
˙
∂x
d ∂
= x(t) − λ 1 + (x)2 −
˙ . (2)
dt ∂x
˙

8. This leads to
d λx
˙
1= .
dt 1 + (x)2
˙
After integrating both sides with respect to t, we obtain
λx
˙
t= + C1
1 + (x)2
˙
for some constant C1 . Thus, x must satisfy
(λx)2
˙
(t − C1 )2 = . (3)
1 + (x)2
˙

9. Since x vanishes only if t = C1 , one obtains
˙
1 + (x)2
˙ λ2
2
= .
x
˙ (t − C1 )2
Thus,
1 λ2 − (t − C1 )2
= ,
x2
˙ (t − C1 )2
or
(t − C1 )
x =±
˙ .
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)

10. 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 = 1 .2
Thus, the circle’s center must lie on the vertical line t = 1 .
2
One may compute the other constants, the center’s ordinate −|C2 |
and radius |λ|, geometrically.

11. Special Cases
Figure: Left = π/2; Right = π/3

12. How if > π/2?
We consider = 2 and a special kind of functions:
m
1 1
x(t) = a |t − |m − ,
2 2
where m is any natural number greater than 1 and a is some
constant to be chosen, depending on m.
Figure: Table, Left; Numerical Results, Right

13. How About A Rectangle and A Semicircle?
Figure: The Area = (1 − π ) ≈ 0.60730091830 · · ·
8

14. The End