Find maximum possible area of a closed curve with perimeter

My approach : \(\displaystyle \max A = \frac{1}{2} \int_{0}^{\alpha } \max y^2 (\alpha ) d\alpha=\frac{1}{2}\int_{0}^{\alpha} c^2 d\alpha =\frac{1}{2}c^2 \alpha\) which is the circle's sector area and \(\displaystyle \frac{dy^2 (\alpha ) }{d\alpha}=0\).

So \(\displaystyle r=\frac{L}{2\pi}\) and \(\displaystyle A=r^2 \pi = \frac{L^2 \pi }{4\pi^2 }=\frac{L^2 }{4\pi }\).

**L**.My approach : \(\displaystyle \max A = \frac{1}{2} \int_{0}^{\alpha } \max y^2 (\alpha ) d\alpha=\frac{1}{2}\int_{0}^{\alpha} c^2 d\alpha =\frac{1}{2}c^2 \alpha\) which is the circle's sector area and \(\displaystyle \frac{dy^2 (\alpha ) }{d\alpha}=0\).

So \(\displaystyle r=\frac{L}{2\pi}\) and \(\displaystyle A=r^2 \pi = \frac{L^2 \pi }{4\pi^2 }=\frac{L^2 }{4\pi }\).

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