ISOPERIMETRIC PROBLEMS

6. ISOPERIMETRIC PROBLEMS

Recall that in ordinary calculus we sometimes want to maximize a quantity subject to a condition (for example, find the volume of the largest box you can make with given surface area). Also recall that the method of Lagrange multipliers was useful in such problems (see Chapter 4, Section 9). There are similar problems in the calculus of variations. The original question which gave this class of problems its name was this: Of all the closed plane curves of given perimeter (isoperimetric = same perimeter), which one incloses the largest area? To solve this problem, we must maximize the area, given length l. In other words, we want to maximize an integral subject to the condition that another integral has a given (constant) value; any such problem is called an isoperimetric problem. Let

I=

F (x, y, y ′ ) dx

be the integral we want to make stationary; at the same time,

J=

G(x, y, y ′ ) dx,

with the same integration variable and the same limits, is to have a given constant value. (This means that the allowed varied paths must be paths for which J has the given value.) By using the Lagrange multiplier method, it can be shown that the desired condition is that

(F + λG) dx

should be stationary, that is, that F + λG should satisfy the Euler equation. The Lagrange multiplier λ is a constant. It will appear in the solution y(x) of the Euler

x equation; having found y(x), we can substitute it into 2 x 1 G(x, y, y ′ ) dx = const.

and so find λ if we like. However, for many purposes we do not need to find λ.

492 Calculus of Variations Chapter 9

Example 1. Given two points x 1 and x 2 on the x axis, and an arc length l > x 2 −x 1 , find the shape of the curve of length l joining the given points which, with the x axis, incloses the largest area.

x 1 y dx subject to the condition J = x 1 ds = l. Here F = y and G = 1+y ′2 so

x 2 x We want to maximize I = 2

F + λG = y + λ 1+y ′2 .

We want the Euler equation for F + λG. Since ∂

the Euler equation is

The solution of (6.2) is (Problem 7): (6.3)

(x + c) 2 + (y + c ′ ) 2 =λ 2

We see that the answer to our problem is an arc of a circle passing through the two given points, and the Lagrange multiplier λ is the radius of the circle. The center

and radius of the circle are determined by the given points x 1 and x 2 , and the given arc length l (Problem 7).

PROBLEMS, SECTION 6

In Problems 1 and 2, given the length l of a curve joining two given points, find the equation of the curve so that:

1. The surface of revolution formed by rotating the curve about the x axis has minimum area.

2. The plane area between the curve and a straight line joining the points is a maximum. 3. Given 10 cc of lead, find how to form it into a solid of revolution of height 1 cm and

minimum moment of inertia about its axis. 4. A uniform flexible chain of given length is suspended at given points (x 1 ,y 1 ) and

(x 2 ,y 2 ). Find the curve in which it hangs. Hint: It will hang so that its center of gravity is as low as possible.

5. A curve y = y(x), joining two points x 1 and x 2 on the x axis, is revolved around the x axis to produce a surface and a volume of revolution. Given the surface area, find the shape of the curve y = y(x) to maximize the volume. Hint: You should ′ find a first integral of the Euler equation of the form yf (y, x , λ) = C. Since y = 0 at the endpoints, C = 0. Then either y = 0 for all x, or f = 0. But y ≡ 0 gives zero volume of the solid of revolution, so for maximum volume you want to solve f = 0.

6. In Problem 5, given the volume, find the shape of the curve y = y(x) to minimize the surface area. Hint: See the hint in Problem 5.

7. Integrate (6.2), simplify the result and integrate again to get (6.3) where c and c ′ √ √

3, and l = 4π/3, show that the center and radius of the circle are (0, −1) and λ = radius = 2.

are constants of integration. If x 1 =−

3, x 2 =

Section 7 Variational Notation 493