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