Question Number 81193 by ajfour last updated on 10/Feb/20
Commented by ajfour last updated on 10/Feb/20
$${If}\:{both}\:{coloured}\:{regions}\:{have}\:{equal} \\ $$$$\left({areas}\right)_{{maximum}\:} ,\:{determine}\:\alpha. \\ $$
Commented by jagoll last updated on 10/Feb/20
$${waw} \\ $$
Commented by ajfour last updated on 14/Feb/20
Answered by mr W last updated on 10/Feb/20
Commented by mr W last updated on 10/Feb/20
![R=radius of semi−circle R^2 =a^2 +((a/(tan α))+b−R)^2 b^2 +2((a/(tan α))−R)b+(a^2 /(sin^2 α))−((2aR)/(tan α))=0 b=R−(a/(tan α))+(√((R−(a/(tan α)))^2 −(a^2 /(sin^2 α))+((2aR)/(tan α)))) ⇒b=R+(√(R^2 −a^2 ))−(a/(tan α)) d=2((a/(sin α))−R cos α) (c+R sin α)^2 +((d/2))^2 =R^2 (c+R sin α)^2 +((a/(sin α))−R cos α)^2 =R^2 c^2 +2R sin α c+((a(a−R sin 2α))/(sin^2 α))=0 ⇒c=((√(R^2 sin^4 α−a(a−R sin 2α)))/(sin α))−R sin α Area=ab=cd let λ=(a/R) P=((Area)/R^2 )=λ(1+(√(1−λ^2 ))−(λ/(tan α)))=2((λ/(sin α))−cos α)[((√(sin^4 α−λ(λ−sin 2α)))/(sin α))−sin α] now it is to find the maximum of P=λ(1+(√(1−λ^2 ))−(λ/(tan α))) under the condition λ(1+(√(1−λ^2 ))−(λ/(tan α)))=(((2λ−sin 2α)[(√(sin^4 α−λ(λ−sin 2α)))−sin^2 α])/(sin^2 α)) P_(max) ≈0.3795 at α≈27.01°, λ≈0.5943](https://www.tinkutara.com/question/Q81238.png)
$${R}={radius}\:{of}\:{semi}−{circle} \\ $$$${R}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{tan}\:\alpha}+{b}−{R}\right)^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} +\mathrm{2}\left(\frac{{a}}{\mathrm{tan}\:\alpha}−{R}\right){b}+\frac{{a}^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \:\alpha}−\frac{\mathrm{2}{aR}}{\mathrm{tan}\:\alpha}=\mathrm{0} \\ $$$${b}={R}−\frac{{a}}{\mathrm{tan}\:\alpha}+\sqrt{\left({R}−\frac{{a}}{\mathrm{tan}\:\alpha}\right)^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \:\alpha}+\frac{\mathrm{2}{aR}}{\mathrm{tan}\:\alpha}} \\ $$$$\Rightarrow{b}={R}+\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }−\frac{{a}}{\mathrm{tan}\:\alpha} \\ $$$${d}=\mathrm{2}\left(\frac{{a}}{\mathrm{sin}\:\alpha}−{R}\:\mathrm{cos}\:\alpha\right) \\ $$$$\left({c}+{R}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} +\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\left({c}+{R}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{sin}\:\alpha}−{R}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} +\mathrm{2}{R}\:\mathrm{sin}\:\alpha\:{c}+\frac{{a}\left({a}−{R}\:\mathrm{sin}\:\mathrm{2}\alpha\right)}{\mathrm{sin}^{\mathrm{2}} \:\alpha}=\mathrm{0} \\ $$$$\Rightarrow{c}=\frac{\sqrt{{R}^{\mathrm{2}} \mathrm{sin}^{\mathrm{4}} \:\alpha−{a}\left({a}−{R}\:\mathrm{sin}\:\mathrm{2}\alpha\right)}}{\mathrm{sin}\:\alpha}−{R}\:\mathrm{sin}\:\alpha \\ $$$${Area}={ab}={cd} \\ $$$${let}\:\lambda=\frac{{a}}{{R}} \\ $$$${P}=\frac{{Area}}{{R}^{\mathrm{2}} }=\lambda\left(\mathrm{1}+\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }−\frac{\lambda}{\mathrm{tan}\:\alpha}\right)=\mathrm{2}\left(\frac{\lambda}{\mathrm{sin}\:\alpha}−\mathrm{cos}\:\alpha\right)\left[\frac{\sqrt{\mathrm{sin}^{\mathrm{4}} \:\alpha−\lambda\left(\lambda−\mathrm{sin}\:\mathrm{2}\alpha\right)}}{\mathrm{sin}\:\alpha}−\mathrm{sin}\:\alpha\right] \\ $$$$ \\ $$$${now}\:{it}\:{is}\:{to}\:{find}\:{the}\:{maximum}\:{of} \\ $$$${P}=\lambda\left(\mathrm{1}+\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }−\frac{\lambda}{\mathrm{tan}\:\alpha}\right) \\ $$$${under}\:{the}\:{condition} \\ $$$$\lambda\left(\mathrm{1}+\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }−\frac{\lambda}{\mathrm{tan}\:\alpha}\right)=\frac{\left(\mathrm{2}\lambda−\mathrm{sin}\:\mathrm{2}\alpha\right)\left[\sqrt{\mathrm{sin}^{\mathrm{4}} \:\alpha−\lambda\left(\lambda−\mathrm{sin}\:\mathrm{2}\alpha\right)}−\mathrm{sin}^{\mathrm{2}} \:\alpha\right]}{\mathrm{sin}^{\mathrm{2}} \:\alpha} \\ $$$$ \\ $$$${P}_{{max}} \approx\mathrm{0}.\mathrm{3795}\:{at}\:\alpha\approx\mathrm{27}.\mathrm{01}°,\:\lambda\approx\mathrm{0}.\mathrm{5943} \\ $$
Commented by mr W last updated on 10/Feb/20
Commented by mr W last updated on 10/Feb/20
Commented by mr W last updated on 10/Feb/20
![Explanation of method i used: red curve shows the condition λ(1+(√(1−λ^2 ))−(λ/(tan α)))=(((2λ−sin 2α)[(√(sin^4 α−λ(λ−sin 2α)))−sin^2 α])/(sin^2 α)) the green curves represent the equation F(λ,α)=λ(1+(√(1−λ^2 ))−(λ/(tan α)))−P=0 with different values of P: curve 1: with P which fulfills the condition curve 3: with P which doesn′t fulfill the condition curve 2: with P_(max) which fulfills the condition](https://www.tinkutara.com/question/Q81250.png)
$${Explanation}\:{of}\:{method}\:{i}\:{used}: \\ $$$${red}\:{curve}\:{shows}\:{the}\:{condition} \\ $$$$\lambda\left(\mathrm{1}+\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }−\frac{\lambda}{\mathrm{tan}\:\alpha}\right)=\frac{\left(\mathrm{2}\lambda−\mathrm{sin}\:\mathrm{2}\alpha\right)\left[\sqrt{\mathrm{sin}^{\mathrm{4}} \:\alpha−\lambda\left(\lambda−\mathrm{sin}\:\mathrm{2}\alpha\right)}−\mathrm{sin}^{\mathrm{2}} \:\alpha\right]}{\mathrm{sin}^{\mathrm{2}} \:\alpha} \\ $$$${the}\:{green}\:{curves}\:{represent}\:{the}\:{equation} \\ $$$${F}\left(\lambda,\alpha\right)=\lambda\left(\mathrm{1}+\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }−\frac{\lambda}{\mathrm{tan}\:\alpha}\right)−{P}=\mathrm{0} \\ $$$${with}\:{different}\:{values}\:{of}\:{P}: \\ $$$${curve}\:\mathrm{1}:\:{with}\:{P}\:{which}\:{fulfills}\:{the}\:{condition} \\ $$$${curve}\:\mathrm{3}:\:{with}\:{P}\:{which}\:{doesn}'{t}\:{fulfill}\:{the}\:{condition} \\ $$$${curve}\:\mathrm{2}:\:{with}\:{P}_{{max}} \:{which}\:\:{fulfills}\:{the}\:{condition} \\ $$
Commented by ajfour last updated on 10/Feb/20
$${Thanks}\:{Sir},\:{for}\:{everything}, \\ $$$${but}\:{i}\:{still}\:{shall}\:{try}\:{at}\:{an} \\ $$$${analytical}\:{way}\:{to}\:{fetch}\:{the} \\ $$$${answer},\:{wish}\:{me}\:{good}\:{luck}.. \\ $$
Commented by mr W last updated on 10/Feb/20
$${yes},\:{i}\:{do}\:{sir}!\:{i}\:{wished}\:{also}\:{a}\:{better}\:{and} \\ $$$${analytical}\:{way},\:{but}\:{had}\:{no}\:{idea}. \\ $$