Maximum-minimum area from a fixed length rope

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SUMMARY

The forum discussion focuses on maximizing and minimizing the area formed by a fixed length rope, specifically when creating an equilateral triangle and a square. The key equations derived include the area of the triangle, \( A_t = \frac{\sqrt{3}}{4}a^2 \), and the total area \( A = (L - 3a)^2 + \frac{\sqrt{3}}{4}a^2 \). The critical points for area optimization were calculated, leading to the conclusion that the maximum area occurs when the entire length is used for the square, yielding \( a = \frac{3\sqrt{3}L}{4 + 3\sqrt{3}} \) for the triangle's side. The discussion emphasizes the importance of correctly identifying maximum and minimum areas based on the configuration of shapes.

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  • #31
And from my answer to the book's, without the L:
$$a=\frac{9}{4\sqrt{3}+9}=\frac{9}{\sqrt{3}(4+3\sqrt{3})}=\frac{3\sqrt{3}\sqrt{3}}{\sqrt{3}(4+3\sqrt{3})}=\frac{9}{4+3\sqrt{3}}$$
Thank you kuruman, Mark, Ray and LCKurtz
 

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