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# 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?
Let &quot;L&quot; 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 &pi;L&sup2;.

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

This means the area of the triangle is

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

&pi;L&sup2; + (&radic;3/4)(4 - 2L + L&sup2;) =

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:

L = 1 / (1 + 4&pi;/&radic;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 &quot;must&quot; cut it into 2 pieces, then the second piece can be infinitely small).

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