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Question Number 41139 by ajfour last updated on 02/Aug/18

Commented by ajfour last updated on 02/Aug/18

Q.41117 alternate solution

$${Q}.\mathrm{41117}\:\:{alternate}\:{solution} \\ $$

Answered by ajfour last updated on 02/Aug/18

AP = 2rsin θ PF =2rsin θcos θ AF = 2rsin^2 θ Area(△APF)=A(θ)=(1/2)×PF×AF A(θ)=2r^2 sin^3 θcos θ (dA/dθ)= 2r^2 (3sin^2 θcos^2 θ−sin^4 θ)=0 ⇒ tan θ = (√3) or θ=60° therefore A_(max) =2r^2 (((√3)/2))^3 ((1/2)) A_(max) =((3(√3))/8) r^2 .

$${AP}\:=\:\mathrm{2}{r}\mathrm{sin}\:\theta \\ $$$${PF}\:=\mathrm{2}{r}\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$${AF}\:=\:\mathrm{2}{r}\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$${Area}\left(\bigtriangleup{APF}\right)={A}\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}×{PF}×{AF} \\ $$$$\:\:\:\:\:\:{A}\left(\theta\right)=\mathrm{2}{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{3}} \theta\mathrm{cos}\:\theta \\ $$$$\:\:\:\frac{{dA}}{{d}\theta}=\:\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{3sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{4}} \theta\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{tan}\:\theta\:=\:\sqrt{\mathrm{3}}\:\:\:{or}\:\:\theta=\mathrm{60}° \\ $$$${therefore}\:{A}_{{max}} =\mathrm{2}{r}^{\mathrm{2}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\boldsymbol{{A}}_{\boldsymbol{{max}}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\:\boldsymbol{{r}}^{\mathrm{2}} \:. \\ $$

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