Maximum-minimum area from a fixed length rope

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The discussion revolves around maximizing and minimizing the area formed by a fixed-length rope configured into a triangle and a square. The area equations for both shapes are derived, leading to critical points for optimization. The maximum area occurs when the entire length of the rope is used for the square, while the minimum area is calculated based on the triangle's dimensions. Participants identify errors in their calculations and clarify the relationships between the sides of the triangle and square. The final consensus emphasizes the importance of correctly interpreting the area functions and critical points to arrive at accurate solutions.
  • #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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