# Answer to Question #20725 in Calculus for Felix

Question #20725

A piece of 2m wire is to be cut into two pieces one of which is to be formed into a circle and the other into an equilateral tiangle. How should the wire be cut so that the total area enclosed is a). a minimum and b) a maximum?

Expert's answer

Let "L" be the length of one piece of wire. Then the other piece is 2 - L.

Make a circle with the first piece. The area is πL².

Make an equilateral triangle with the second piece. The base is (2-L). The height (h) splits the triangle into two smaller 30-60-90 triangles, such that

sin(60) = h / (2-L),

so

h = sin(60)(2-L),

or

(√3/2)(2-L).

This means the area of the triangle is

(1/2)(2-L)(√3/2)(2-L).

Add these two areas up. Simplify. Take the derivative with respect to L and set it equal to 0. Solve for L.

πL² + (1/2)(2-L)(√3/2)(2-L) =

πL² + (2-L)(√3/4)(2-L) =

πL² + (√3/4)(4 - 2L + L²) =

πL² + (√3/4)4 - 2L·√3/4 + √3L²/4 =

√3 - √3L/2 + (√3/4+π)L².

The graph of this would a parabola that opens upward, so the vertex point we found here should a MINIMUM. So we get the smallest area when we cut one piece of this length, and use that to make the square.

Derivative is:

-√3/2 + 2(√3/4+π)L = 0,

2(√3/4+π)L = √3/2,

(√3/4+π)L = √3/4,

L = √3/4 / (√3/4+π),

L = 1 / (1 + 4π/√3).

As for the maximum area, if you think back to the graph, this would be when we can take L as large as possible. The largest this can be is L = 2, which means you maximize the area by not cutting it at all and making a square of the whole thing (if you "must" cut it into 2 pieces, then the second piece can be infinitely small).

Make a circle with the first piece. The area is πL².

Make an equilateral triangle with the second piece. The base is (2-L). The height (h) splits the triangle into two smaller 30-60-90 triangles, such that

sin(60) = h / (2-L),

so

h = sin(60)(2-L),

or

(√3/2)(2-L).

This means the area of the triangle is

(1/2)(2-L)(√3/2)(2-L).

Add these two areas up. Simplify. Take the derivative with respect to L and set it equal to 0. Solve for L.

πL² + (1/2)(2-L)(√3/2)(2-L) =

πL² + (2-L)(√3/4)(2-L) =

πL² + (√3/4)(4 - 2L + L²) =

πL² + (√3/4)4 - 2L·√3/4 + √3L²/4 =

√3 - √3L/2 + (√3/4+π)L².

The graph of this would a parabola that opens upward, so the vertex point we found here should a MINIMUM. So we get the smallest area when we cut one piece of this length, and use that to make the square.

Derivative is:

-√3/2 + 2(√3/4+π)L = 0,

2(√3/4+π)L = √3/2,

(√3/4+π)L = √3/4,

L = √3/4 / (√3/4+π),

L = 1 / (1 + 4π/√3).

As for the maximum area, if you think back to the graph, this would be when we can take L as large as possible. The largest this can be is L = 2, which means you maximize the area by not cutting it at all and making a square of the whole thing (if you "must" cut it into 2 pieces, then the second piece can be infinitely small).

## Comments

## Leave a comment