Question Number 41139 by ajfour last updated on 02/Aug/18
Commented by ajfour last updated on 02/Aug/18
$${Q}.\mathrm{41117}\:\:{alternate}\:{solution} \\ $$
Answered by ajfour last updated on 02/Aug/18
$${AP}\:=\:\mathrm{2}{r}\mathrm{sin}\:\theta \\ $$$${PF}\:=\mathrm{2}{r}\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$${AF}\:=\:\mathrm{2}{r}\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$${Area}\left(\bigtriangleup{APF}\right)={A}\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}×{PF}×{AF} \\ $$$$\:\:\:\:\:\:{A}\left(\theta\right)=\mathrm{2}{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{3}} \theta\mathrm{cos}\:\theta \\ $$$$\:\:\:\frac{{dA}}{{d}\theta}=\:\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{3sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{4}} \theta\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{tan}\:\theta\:=\:\sqrt{\mathrm{3}}\:\:\:{or}\:\:\theta=\mathrm{60}° \\ $$$${therefore}\:{A}_{{max}} =\mathrm{2}{r}^{\mathrm{2}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\boldsymbol{{A}}_{\boldsymbol{{max}}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\:\boldsymbol{{r}}^{\mathrm{2}} \:. \\ $$