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Question Number 121539 by ajfour last updated on 09/Nov/20
Commented by ajfour last updated on 09/Nov/20
Find (s/R) for maximum blue triangular area. (square remains within semicircle)
$$\:\:\:\:\:\:\:\:{Find}\:\frac{{s}}{{R}}\:\:{for}\:{maximum}\:{blue} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{triangular}\:{area}. \\ $$$$\left({square}\:{remains}\:{within}\:{semicircle}\right) \\ $$
Answered by mr W last updated on 09/Nov/20
area Δ=((sx)/2) (x+s)(2R−x−s)=s^2 (x+s)^2 −2R(x+s)+s^2 =0 x+s=R−(√(R^2 −s^2 )) x=R−s−(√(R^2 −s^2 )) Δ=(1/2)(Rs−s^2 −s(√(R^2 −s^2 ))) (dΔ/ds)=0 R−2s−(√(R^2 −s^2 ))+(s^2 /( (√(R^2 −s^2 ))))=0 let λ=(s/R) ⇒1−2λ−(√(1−λ^2 ))+(λ^2 /( (√(1−λ^2 ))))=0 ⇒1+(λ^2 /( (√(1−λ^2 ))))=2λ+(√(1−λ^2 )) ⇒λ=(s/R)=0.8269
$${area}\:\Delta=\frac{{sx}}{\mathrm{2}} \\ $$$$\left({x}+{s}\right)\left(\mathrm{2}{R}−{x}−{s}\right)={s}^{\mathrm{2}} \\ $$$$\left({x}+{s}\right)^{\mathrm{2}} −\mathrm{2}{R}\left({x}+{s}\right)+{s}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}+{s}={R}−\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} } \\ $$$${x}={R}−{s}−\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} } \\ $$$$\Delta=\frac{\mathrm{1}}{\mathrm{2}}\left({Rs}−{s}^{\mathrm{2}} −{s}\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} }\right) \\ $$$$\frac{{d}\Delta}{{ds}}=\mathrm{0} \\ $$$${R}−\mathrm{2}{s}−\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} }+\frac{{s}^{\mathrm{2}} }{\:\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} }}=\mathrm{0} \\ $$$${let}\:\lambda=\frac{{s}}{{R}} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2}\lambda−\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }+\frac{\lambda^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}+\frac{\lambda^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}=\mathrm{2}\lambda+\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} } \\ $$$$\Rightarrow\lambda=\frac{{s}}{{R}}=\mathrm{0}.\mathrm{8269} \\ $$
Commented by mr W last updated on 09/Nov/20
yes sir, you are right!
$${yes}\:{sir},\:{you}\:{are}\:{right}! \\ $$
Commented by ajfour last updated on 09/Nov/20
Thank you sir. i think you have checked when is the area a maximum for we get two positive values of lambda.
Commented by ajfour last updated on 09/Nov/20
((2△)/R^2 )=λ−λ^2 +λ(√(1−λ^2 )) λ=0.63111 Sir think this is correct.
$$\frac{\mathrm{2}\bigtriangleup}{{R}^{\mathrm{2}} }=\lambda−\lambda^{\mathrm{2}} +\lambda\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} } \\ $$$$\lambda=\mathrm{0}.\mathrm{63111} \\ $$$${Sir}\:{think}\:{this}\:{is}\:{correct}. \\ $$
Commented by ajfour last updated on 09/Nov/20
Answered by ajfour last updated on 09/Nov/20
Commented by ajfour last updated on 09/Nov/20
s=2Rsin θcos θ ; x=2Rcos^2 θ−s 2△=sx let (s/R)=λ = sin 2θ 2△=(Rsin 2θ)(2Rcos^2 θ−Rsin 2θ) ((2△)/R^2 )=sin 2θ(1+cos 2θ−sin 2θ) (d/dθ)(((2△)/R^2 ))=0 ⇒ 2cos 2θ(1+cos 2θ−sin 2θ) =sin 2θ(2cos 2θ+2sin 2θ) ⇒ cos 2θ+cos^2 2θ=2cos 2θsin 2θ +sin^2 2θ with sin 2θ=λ (1−λ^2 )(1−2λ)^2 =(2λ^2 −1)^2 ⇒ 4λ^2 −4λ^4 −4λ+4λ^3 +1−λ^2 =4λ^4 −4λ^2 +1 ⇒ 8λ^3 −4λ^2 −7λ+4=0 λ=0.82694 , 0.63111 I think λ=(s/R)=0.63111 should be the appropriate answer.
$${s}=\mathrm{2}{R}\mathrm{sin}\:\theta\mathrm{cos}\:\theta\:\:;\:\:\:{x}=\mathrm{2}{R}\mathrm{cos}\:^{\mathrm{2}} \theta−{s} \\ $$$$\mathrm{2}\bigtriangleup={sx} \\ $$$${let}\:\:\:\frac{{s}}{{R}}=\lambda\:=\:\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\mathrm{2}\bigtriangleup=\left({R}\mathrm{sin}\:\mathrm{2}\theta\right)\left(\mathrm{2}{R}\mathrm{cos}\:^{\mathrm{2}} \theta−{R}\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$$\frac{\mathrm{2}\bigtriangleup}{{R}^{\mathrm{2}} }=\mathrm{sin}\:\mathrm{2}\theta\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta−\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$$\frac{{d}}{{d}\theta}\left(\frac{\mathrm{2}\bigtriangleup}{{R}^{\mathrm{2}} }\right)=\mathrm{0}\:\:\:\Rightarrow \\ $$$$\mathrm{2cos}\:\mathrm{2}\theta\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta−\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{sin}\:\mathrm{2}\theta\left(\mathrm{2cos}\:\mathrm{2}\theta+\mathrm{2sin}\:\mathrm{2}\theta\right) \\ $$$$\Rightarrow\:\:\mathrm{cos}\:\mathrm{2}\theta+\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\theta=\mathrm{2cos}\:\mathrm{2}\theta\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\theta \\ $$$${with}\:\:\mathrm{sin}\:\mathrm{2}\theta=\lambda \\ $$$$\:\:\:\:\:\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{2}\lambda\right)^{\mathrm{2}} =\left(\mathrm{2}\lambda^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{4}\lambda^{\mathrm{2}} −\mathrm{4}\lambda^{\mathrm{4}} −\mathrm{4}\lambda+\mathrm{4}\lambda^{\mathrm{3}} +\mathrm{1}−\lambda^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\lambda^{\mathrm{4}} −\mathrm{4}\lambda^{\mathrm{2}} +\mathrm{1} \\ $$$$\Rightarrow\:\:\mathrm{8}\lambda^{\mathrm{3}} −\mathrm{4}\lambda^{\mathrm{2}} −\mathrm{7}\lambda+\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\:\lambda=\mathrm{0}.\mathrm{82694}\:\:\:,\:\:\:\mathrm{0}.\mathrm{63111} \\ $$$${I}\:{think}\:\:\lambda=\frac{{s}}{{R}}=\mathrm{0}.\mathrm{63111}\:\:{should}\: \\ $$$${be}\:{the}\:\:{appropriate}\:{answer}. \\ $$$$ \\ $$
Answered by MJS_new last updated on 09/Nov/20
let R=1 semicircle: y=(√(1−x^2 )) s= (√(1−p^2 )) base of triangle ∣p−1+(√(1−p^2 ))∣ 2×area of triangle is ∣p−1+(√(1−p^2 ))∣(√(1−p^2 )) its maximum is at p≈−.775694 ⇒ s≈.631109
$$\mathrm{let}\:{R}=\mathrm{1} \\ $$$$\mathrm{semicircle}:\:{y}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${s}=\:\sqrt{\mathrm{1}−{p}^{\mathrm{2}} } \\ $$$$\mathrm{base}\:\mathrm{of}\:\mathrm{triangle}\:\mid{p}−\mathrm{1}+\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }\mid \\ $$$$\mathrm{2}×\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{is}\:\mid{p}−\mathrm{1}+\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }\mid\sqrt{\mathrm{1}−{p}^{\mathrm{2}} } \\ $$$$\mathrm{its}\:\mathrm{maximum}\:\mathrm{is}\:\mathrm{at}\:{p}\approx−.\mathrm{775694} \\ $$$$\Rightarrow\:{s}\approx.\mathrm{631109} \\ $$
Commented by MJS_new last updated on 09/Nov/20
please check...
$$\mathrm{please}\:\mathrm{check}… \\ $$
Commented by ajfour last updated on 09/Nov/20
Commented by ajfour last updated on 09/Nov/20
s=(√(1−p^2 )) , R=1 base of triangle x=1+p−(√(1−p^2 )) 2△=(1+p−(√(1−p^2 )) )(√(1−p^2 )) = (1+p)(√(1−p^2 )) −1+p^2 ((d(2△))/dp)=(√(1−p^2 ))−((p(1+p))/( (√(1−p^2 ))))+2p = 0 ⇒ 1−p^2 −p−p^2 +2p(√(1−p^2 )) = 0 ⇒ (2p^2 +p−1)^2 =4p^2 (1−p^2 ) ⇒ 8p^4 +4p^3 −7p^2 −2p+1=0 max (2△)≈ 0.72236 at p≈ 0.77569 s=0.63111 If this is what you mean Sir MjS.
$${s}=\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }\:\:\:\:\:,\:\:{R}=\mathrm{1} \\ $$$${base}\:{of}\:\:{triangle}\:\:\boldsymbol{{x}}=\mathrm{1}+{p}−\sqrt{\mathrm{1}−{p}^{\mathrm{2}} } \\ $$$$\mathrm{2}\bigtriangleup=\left(\mathrm{1}+{p}−\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }\:\right)\sqrt{\mathrm{1}−{p}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\:\left(\mathrm{1}+{p}\right)\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }\:−\mathrm{1}+{p}^{\mathrm{2}} \\ $$$$\frac{{d}\left(\mathrm{2}\bigtriangleup\right)}{{dp}}=\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }−\frac{{p}\left(\mathrm{1}+{p}\right)}{\:\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }}+\mathrm{2}{p}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{1}−{p}^{\mathrm{2}} −{p}−{p}^{\mathrm{2}} +\mathrm{2}{p}\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\left(\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}{p}^{\mathrm{2}} \left(\mathrm{1}−{p}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:\mathrm{8}{p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{3}} −\mathrm{7}{p}^{\mathrm{2}} −\mathrm{2}{p}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:{max}\:\left(\mathrm{2}\bigtriangleup\right)\approx\:\mathrm{0}.\mathrm{72236} \\ $$$$\:\:\:\:\:\:\:{at}\:\:{p}\approx\:\mathrm{0}.\mathrm{77569} \\ $$$$\:\:\:\:\:\:\:{s}=\mathrm{0}.\mathrm{63111} \\ $$$${If}\:{this}\:{is}\:{what}\:{you}\:{mean}\:{Sir}\:{MjS}. \\ $$
Commented by ajfour last updated on 09/Nov/20
Commented by MJS_new last updated on 09/Nov/20
yes thank you, typo while approximating the zero. I have corrected it
$$\mathrm{yes}\:\mathrm{thank}\:\mathrm{you},\:\mathrm{typo}\:\mathrm{while}\:\mathrm{approximating}\:\mathrm{the} \\ $$$$\mathrm{zero}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{corrected}\:\mathrm{it} \\ $$

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