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Question Number 186302 by ajfour last updated on 03/Feb/23

Commented by ajfour last updated on 03/Feb/23

Can it be = (1/6) ?

$${Can}\:{it}\:{be}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\:? \\ $$

Commented by Frix last updated on 04/Feb/23

The maximum area is reached when the left vertice (with the right angle) is located on the horizontal diameter of the circle. The maximum area is (1/6).

$$\mathrm{The}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{is}\:\mathrm{reached}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{left}\:\mathrm{vertice}\:\left(\mathrm{with}\:\mathrm{the}\:\mathrm{right}\:\mathrm{angle}\right)\:\mathrm{is}\:\mathrm{located} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}. \\ $$$$\mathrm{The}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{6}}. \\ $$

Commented by Frix last updated on 04/Feb/23

Sorry I calculated with r=1 ⇒ maximum area =2 and then forgot a factor...

$$\mathrm{Sorry}\:\mathrm{I}\:\mathrm{calculated}\:\mathrm{with}\:{r}=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{maximum}\:\mathrm{area}\:=\mathrm{2}\:\mathrm{and}\:\mathrm{then}\:\mathrm{forgot}\:\mathrm{a}\:\mathrm{factor}... \\ $$

Answered by ajfour last updated on 04/Feb/23

Commented by ajfour last updated on 04/Feb/23

A=psin θ(pcos θ+(√(R^2 −p^2 sin^2 θ))) here p=2R , hence (A/p^2 )(θ)=sin θcos θ+sin θ(√((1/4)−sin^2 θ)) (d/dθ)((A/(4R^2 )))=cos 2θ+cos θ(√((1/4)−sin^2 θ)) −((sin^2 θcos θ)/( (√((1/4)−sin^2 θ)))) =0 ⇒ cos^2 2θ((1/4)−sin^2 θ) =cos^2 θ(2sin^2 θ−(1/4)) say sin^2 θ=s ⇒ (1−2s)^2 ((1/4)−s)=(1−s)(2s−(1/4)) ⇒ s=(1/5) ⇒ sin θ=(1/( (√5))) , cos θ=(2/( (√5))) Hence (A_(max) /(4R^2 ))=(2/5)+(1/( (√5)))(√((1/4)−(1/5)))=(1/2) and 2R(√3)=1 ⇒ A_(max) =(1/2)(4R^2 )=(1/2)((1/3))=(1/6)

$${A}={p}\mathrm{sin}\:\theta\left({p}\mathrm{cos}\:\theta+\sqrt{{R}^{\mathrm{2}} −{p}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}\right) \\ $$$${here}\:\:{p}=\mathrm{2}{R}\:,\:\:{hence} \\ $$$$\frac{{A}}{{p}^{\mathrm{2}} }\left(\theta\right)=\mathrm{sin}\:\theta\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$\:\:\frac{{d}}{{d}\theta}\left(\frac{{A}}{\mathrm{4}{R}^{\mathrm{2}} }\right)=\mathrm{cos}\:\mathrm{2}\theta+\mathrm{cos}\:\theta\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:\theta}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{sin}\:^{\mathrm{2}} \theta}}\:\:\:\:=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\theta\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{sin}\:^{\mathrm{2}} \theta\right) \\ $$$$\:\:\:\:\:=\mathrm{cos}\:^{\mathrm{2}} \theta\left(\mathrm{2sin}\:^{\mathrm{2}} \theta−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${say}\:\:\:\mathrm{sin}\:^{\mathrm{2}} \theta={s} \\ $$$$\Rightarrow\:\:\left(\mathrm{1}−\mathrm{2}{s}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}−{s}\right)=\left(\mathrm{1}−{s}\right)\left(\mathrm{2}{s}−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\:\:\:\:{s}=\frac{\mathrm{1}}{\mathrm{5}}\:\:\Rightarrow\:\:\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\:,\:\:\mathrm{cos}\:\theta=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$${Hence} \\ $$$$\frac{{A}_{{max}} }{\mathrm{4}{R}^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{5}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${and}\:\:\:\mathrm{2}{R}\sqrt{\mathrm{3}}=\mathrm{1} \\ $$$$\Rightarrow\:\:{A}_{{max}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{R}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented by mr W last updated on 05/Feb/23

nice solution sir!

$${nice}\:{solution}\:{sir}! \\ $$

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