Question Number 56763 by Tawa1 last updated on 23/Mar/19
$$\left(\mathrm{a}\right)\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{rectangle}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{inscribed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{circle}\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:\:=\:\:\mathrm{a}^{\mathrm{2}} \:. \\ $$$$ \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Name}\:\mathrm{the}\:\mathrm{rectangle}\:\mathrm{so}\:\mathrm{formed} \\ $$
Answered by kaivan.ahmadi last updated on 23/Mar/19
$${S}=\mathrm{4}{x}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\Rightarrow{S}'=\mathrm{4}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }+\frac{−\mathrm{4}{x}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}= \\ $$$$\frac{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}=\mathrm{0}\Rightarrow\mathrm{8}{x}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} \Rightarrow \\ $$$${x}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow{x}=\frac{{a}}{\sqrt{\mathrm{2}}}\Rightarrow \\ $$$${S}=\mathrm{2}\sqrt{\mathrm{2}}{a}\sqrt{{a}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{2}}}=\mathrm{2}\sqrt{\mathrm{2}}{a}\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{2}}}=\mathrm{2}{a}^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 23/Mar/19
Commented by Tawa1 last updated on 23/Mar/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$