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Question Number 5417 by FilupSmith last updated on 14/May/16

$$\mathrm{Prove},\mathrm{or}\mathrm{disprove},\mathrm{that}\mathrm{the}\mathrm{circle}$$$$\mathrm{has}\mathrm{the}\mathrm{largest}Perimiter\mathrm{over}\mathrm{all}$$$$\mathrm{natural}\mathrm{shapes}\mathrm{that}\mathrm{have}\mathrm{area}A$$

Commented by Rasheed Soomro last updated on 14/May/16

$$\mathrm{I}\mathrm{think}\mathrm{that}\mathrm{the}\mathrm{circle}\mathrm{has}\mathrm{the}\mathrm{smallest}$$$$\mathrm{perimeter}\mathrm{among}\mathrm{all}\mathrm{closed}\mathrm{figures}$$$$\mathrm{having}\mathrm{same}\mathrm{area}\mathrm{A}.$$$${\mathrm{Let}}^{\prime }\mathrm{s}\mathrm{compare}\mathrm{the}\mathrm{perimeters}\mathrm{of}\mathrm{a}\mathrm{circle}$$$$\mathrm{and}\mathrm{a}\mathrm{square}\mathrm{who}\mathrm{have}\mathrm{same}\mathrm{area}\mathrm{A}.$$$$\pi {\mathrm{r}}^2={\mathrm{s}}^2\Rightarrow \mathrm{s}=\sqrt{\pi }\mathrm{r}$$$$(\mathrm{r}\mathrm{is}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{and}\mathrm{s}\mathrm{is}\mathrm{measure}$$$$\mathrm{of}\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{square}.)$$$$\mathrm{perimeter}\mathrm{of}\mathrm{circle}\left(\mathrm{circumference}\right)=2\pi \mathrm{r}$$$$\mathrm{perimeter}\mathrm{of}\mathrm{square}=4\mathrm{s}=4\sqrt{\pi }\mathrm{r}$$$$2\pi \mathrm{r}?4\sqrt{\pi }\mathrm{r}$$$$2\pi ?4\sqrt{\pi }$$$$6.283<7.09$$$$\mathrm{circumference}<\mathrm{perimeter}\mathrm{of}\mathrm{square}.$$$$\mathrm{This}\mathrm{disproves}\mathrm{the}\mathrm{claim}.$$

Commented by Rasheed Soomro last updated on 14/May/16

$$\mathrm{The}\mathrm{claim}\mathrm{should}\mathrm{be}\mathrm{for}{}^{\prime }\mathrm{smallest}$$$${\mathrm{perimeter}}^{\prime }\mathrm{instead}\mathrm{of}{}^{\prime }\mathrm{largest}{\mathrm{perimeter}}^{\prime }.$$

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