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Question Number 56763 by Tawa1 last updated on 23/Mar/19

$$\left(\mathrm{a}\right)\mathrm{Determine}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{largest}\mathrm{rectangle}\mathrm{that}\mathrm{can}\mathrm{be}$$$$\mathrm{inscribed}\mathrm{in}\mathrm{the}\mathrm{circle}{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2.$$$$$$$$\left(\mathrm{b}\right)\mathrm{Name}\mathrm{the}\mathrm{rectangle}\mathrm{so}\mathrm{formed}$$

Answered by kaivan.ahmadi last updated on 23/Mar/19

$$S=4x\sqrt{a^2-x^2}\Rightarrow S^{\prime }=4\sqrt{a^2-x^2}+\frac{-4x^2}{\sqrt{a^2-x^2}}=$$$$\frac{4a^2-4x^2-4x^2}{\sqrt{a^2-x^2}}=0\Rightarrow 8x^2=4a^2\Rightarrow$$$$x^2=\frac{a^2}{2}\Rightarrow x=\frac{a}{\sqrt{2}}\Rightarrow$$$$S=2\sqrt{2}a\sqrt{a^2-\frac{a^2}{2}}=2\sqrt{2}a\sqrt{\frac{a^2}{2}}=2a^2$$$$$$$$$$$$$$

Commented by kaivan.ahmadi last updated on 23/Mar/19

Commented by Tawa1 last updated on 23/Mar/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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