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?
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Expert's answer
2012-12-18T10:33:40-0500
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.
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.
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).
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