Answer to Question #20725 in Calculus for Felix
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),
h = sin(60)(2-L),
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.
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.
-√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).
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