PART 10：例題-摺紙問題1

假設一片長方形紙片，由左下角上折使之切齊上端(如下圖)
(1) 欲三角形A面積最大，則 \(x = ?\)

SOL：

\(A\) 之一股為 \(a - x\) ，斜邊長為 \(x\) ，故另一股為 \(y = \sqrt {{x^2} - {{(a - x)}^2}}  = \sqrt {2ax - {a^2}} \)

\(A\) 之面積設為 \(f(x) = \frac{1}{2}(a - x)\sqrt {2ax - {a^2}} \) ， \(x \in\left[ {\;\frac{a}{2}\;\;,\;\;a\;} \right]\) 

\(f'(x) =  - \frac{1}{2}\sqrt {2ax - {a^2}}  + \frac{1}{2}(a - x)a{\left( {2ax - {a^2}} \right)^{ - \frac{1}{2}}}\) 

\( = \frac{1}{2}{\left( {2ax - {a^2}} \right)^{ - \frac{1}{2}}}\left[ { - \left( {2ax - {a^2}} \right) + (a - x)a} \right]\) 

\( = \frac{1}{2}{\left( {2ax - {a^2}} \right)^{ - \frac{1}{2}}}\left[ { - 3ax + 2{a^2}} \right]\)  \( = \frac{a}{2}{\left( {2ax - {a^2}} \right)^{ - \frac{1}{2}}}\left[ { - 3x + 2a} \right]\) ，

臨界值:  \(x = \;\frac{a}{2}\) ， \(x = \;\frac{{2a}}{3}\)

\(f(\frac{a}{2})=\frac{1}{2}(\frac{a}{2})\sqrt{2a\cdot \frac{a}{2}-{{a}^{2}}}=0\)

\(f(\frac{2a}{3})=\frac{1}{2}(\frac{2a}{3})\sqrt{2a\cdot \frac{2a}{3}-{{a}^{2}}}=\frac{a}{6}\cdot \frac{a}{\sqrt{3}}\) \(=\frac{{{a}^{2}}}{6\sqrt{3}}=\frac{\sqrt{3}{{a}^{2}}}{18}\)

當 \(x=\frac{2a}{3}\) 時，三角形A面積最大