# prove-that-among-all-closed-figures-having-same-perimeter-circle-has-maximum-area-

Question Number 3417 by RasheedSindhi last updated on 13/Dec/15
$${prove}\:{that}\:{among}\:{all}\:{closed}\:{figures} \\$$$${having}\:{same}\:{perimeter}\:{circle} \\$$$${has}\:{maximum}\:{area}. \\$$
Commented by Filup last updated on 13/Dec/15
$$\mathrm{I}\:\mathrm{dont}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{question} \\$$
Commented by RasheedSindhi last updated on 13/Dec/15
$${Consider}\:{closed}\:{figures}\:{triangle}, \\$$$${quadilateral},{pentagon}…{or}\:{other} \\$$$${closed}\:{curves}\:{or}\:{mixture}\:{of}\:{curves} \\$$$${and}\:{straight}\:{line}\:{segments}\:,{all} \\$$$${having}\:{same}\:{perimeter}.{Among} \\$$$${all}\:{the}\:{areas}\:{these}\:{figures}\:{have} \\$$$${the}\:{largest}\:{is}\:{the}\:{area}\:{of}\:{circle}. \\$$
Commented by 123456 last updated on 13/Dec/15
$$\mathrm{this}\:\mathrm{is}\:\mathrm{isoperemtric}\:\mathrm{inequality} \\$$$$\mathrm{4}\pi\mathrm{A}\leqslant\mathrm{L}^{\mathrm{2}} \\$$
Commented by Filup last updated on 13/Dec/15
$$\mathrm{Circle}\:\mathrm{is}\:\mathrm{essentially}\:\mathrm{an}\:\infty−{sided} \\$$$${polygon} \\$$
Commented by Filup last updated on 13/Dec/15
$$\mathrm{3}−\mathrm{gon}\:\left(\mathrm{triangle}\right)<\mathrm{4}−\mathrm{gon}\:\left(\mathrm{square}\right) \\$$$$\mathrm{4}−\mathrm{gon}<\mathrm{5}−\mathrm{gon} \\$$$$\mathrm{etc} \\$$$$\therefore{n}−{gon}<\infty−{gon}\:\left({circle}\right) \\$$
Commented by Rasheed Soomro last updated on 13/Dec/15
$$\mathcal{W}{hat}\:{is}\:{meant}\:{by}\:\mathrm{isoperemtric}\:\mathrm{inequality}\:? \\$$
Commented by RasheedSindhi last updated on 13/Dec/15
$$\mathcal{T}{h}\propto{nk}\mathcal{S} \\$$
Commented by prakash jain last updated on 13/Dec/15
$$\mathrm{isoperimetric}\:\mathrm{inequality}\:\mathrm{for}\:\mathrm{all}\:\mathrm{closed} \\$$$$\mathrm{plane}\:\mathrm{figure} \\$$$$\mathrm{4}\pi\mathrm{A}\leqslant\mathrm{L}^{\mathrm{2}} \\$$$$\mathrm{A}\:\mathrm{area} \\$$$$\mathrm{L}\:\mathrm{perimeter} \\$$$$\mathrm{equality}\:\mathrm{holding}\:\mathrm{true}\:\mathrm{only}\:\mathrm{for}\:\mathrm{circle}. \\$$