HOMEWORK 1: A wire of length L is cut into two pieces: one to be

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Math 4580, Winter/Spring 2000, Georgia Institute of Technology

HOMEWORK 1: A wire of length L is cut into two pieces: one to be used to form the perimater of a square, and the other a circle. How should the wire be cut in order that the sum of the areas be as small [and also as large] as possible? Set this up as an optimization problem. What is the objective function and what are the constraints? Do the problem in two different ways: as a one variable problem, and also as a two variable problem using the Lagrange multiplier principle.

FIRST METHOD: Let the piece that goes into the square have length x and let the piece that goes into the circle have length y. Then x+y=L. The quantity to maximize or minimize is

A = (x/4)

2

+ y

2

4 p subject to 0 £ x, 0 £ y. Substituting y=L-x and differentiating A = x

16

2

+

(L-x)

4 p

2

, we see that dA dx

= 0 when x =

4 p +4

L

»

.5601 L and the corresponding value of A is computed to be

»

.03501 L

2

. Since A = x

2

16

+

(L-x)

2

is a quadratic in x

4 p and the coefficient of x

2

is positive, this value for x must be the minimizer.

The maximizer must be one of the endpoints, x=0 or x=L. The value of A is L

2

/4 p

»

.07958L

2

at x=0 and L

2

/16 = .0625L

2

at x=L. The maximum occurs therefore when x=0, when there is no square and all the wire is put into the circumference of the circle.

SECOND METHOD: We wish to minimize A(x,y) over the line segment defined by the constraints x+y=L, 0 £ x, 0 £ y. The min or max can only occur at an endpoint of this line segment, or at an interior point of the line segment where the gradient Ñ A(x,y) = (x/8, y/2 p ) is perpendicular to the line segment

(that is to say, a multiple of (1,1) which is the gradient of the constraint function x+y-L). Solving the system:

(x/8, y/2 p ) = l (1,1)

4 yields y = p x/4. Thus x + p x/4 = L, or x =

4+ p

L. Now we proceed as before.

REMARK: There is a famous theorem known as the Isoperimetric Inequality that states that of all plane figures with a fixed perimeter, the one with the largest area is a circle. Thus we should have expected the maximum to occur when all the wire went into the circle.

copyright Jonathan Spingarn, January 2000

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