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Question-153901

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Question Number 153901 by mr W last updated on 11/Sep/21
Commented by mr W last updated on 12/Sep/21
1) find (ab)_(max) =? 2) prove that S_1 =S_2 .
$$\left.\mathrm{1}\right)\:{find}\:\left({ab}\right)_{{max}} =? \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:{S}_{\mathrm{1}} ={S}_{\mathrm{2}} . \\ $$
Answered by EDWIN88 last updated on 12/Sep/21
Commented by mr W last updated on 12/Sep/21
thanks alot sir!
$${thanks}\:{alot}\:{sir}! \\ $$
Commented by Tawa11 last updated on 12/Sep/21
Weldone sir.
$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 12/Sep/21
Commented by mr W last updated on 12/Sep/21
(1) α+β=45 ((tan α+tan α)/(1−tan α tan β))=1 (1/(tan α))+(1/(tan β))=(1/(tan α))×(1/(tan β))−1 ⇒(1/(tan β))=(((1/(tan α))+1)/((1/(tan α))−1)) ((sin (α+45))/(sin α))=(l/a) (1/(tan α))+1=(((√2)l)/a) ...(i) similarly (1/(tan β))+1=(((√2)l)/b) ...(ii) ((2l^2 )/(ab))=((1/(tan α))+1)((1/(tan β))+1) =2((1/(tan α))+1)(((1/(tan α))/((1/(tan α))−1))) let t=(1/(tan α)) ⇒P=(l^2 /(ab))=((t(t+1))/(t−1)) (dP/dt)=((2t+1)/(t−1))−((t(t+1))/((t−1)^2 ))=0 t^2 −2t−1=0 ⇒t=(1/(tan α))=1+(√2) ⇒tan α=(√2)−1 ⇒α=22.5°=β ⇒tan β=(√2)−1 ((2l^2 )/((ab)_(max) ))=(2+(√2))^2 ⇒(ab)_(max) =((√2)−1)^2 l^2 ≈0.171l^2
$$\left(\mathrm{1}\right) \\ $$$$\alpha+\beta=\mathrm{45} \\ $$$$\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\alpha}{\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\frac{\mathrm{1}}{\mathrm{tan}\:\beta}=\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}×\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\beta}=\frac{\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}−\mathrm{1}} \\ $$$$ \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\mathrm{45}\right)}{\mathrm{sin}\:\alpha}=\frac{{l}}{{a}} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\mathrm{1}=\frac{\sqrt{\mathrm{2}}{l}}{{a}}\:\:\:…\left({i}\right) \\ $$$${similarly} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\beta}+\mathrm{1}=\frac{\sqrt{\mathrm{2}}{l}}{{b}}\:\:\:…\left({ii}\right) \\ $$$$\frac{\mathrm{2}{l}^{\mathrm{2}} }{{ab}}=\left(\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{tan}\:\beta}+\mathrm{1}\right) \\ $$$$=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\mathrm{1}\right)\left(\frac{\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}}{\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}−\mathrm{1}}\right) \\ $$$${let}\:{t}=\frac{\mathrm{1}}{\mathrm{tan}\:\alpha} \\ $$$$\Rightarrow{P}=\frac{{l}^{\mathrm{2}} }{{ab}}=\frac{{t}\left({t}+\mathrm{1}\right)}{{t}−\mathrm{1}} \\ $$$$\frac{{dP}}{{dt}}=\frac{\mathrm{2}{t}+\mathrm{1}}{{t}−\mathrm{1}}−\frac{{t}\left({t}+\mathrm{1}\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\sqrt{\mathrm{2}}−\mathrm{1}\:\Rightarrow\alpha=\mathrm{22}.\mathrm{5}°=\beta \\ $$$$\Rightarrow\mathrm{tan}\:\beta=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\frac{\mathrm{2}{l}^{\mathrm{2}} }{\left({ab}\right)_{{max}} }=\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left({ab}\right)_{{max}} =\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} {l}^{\mathrm{2}} \approx\mathrm{0}.\mathrm{171}{l}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 12/Sep/21
(2) (L/c)=((sin (45+α))/(sin 45))=((sin (45+45−β))/(sin 45))=(√2)cos β (L/d)=((sin (45+β))/(sin 45))=((sin (45+45−α))/(sin 45))=(√2)cos α (L^2 /(cd))=2 cos α cos β ⇒cd=(L^2 /(2cos α cos β)) e=(L/(cos α)) f=(L/(cos β)) ⇒ef=(L^2 /(cos α cos β))=2cd S_1 =((cd sin 45°)/2) S_1 +S_2 =((ef sin 45°)/2)=2S_1 ⇒S_1 =S_2
$$\left(\mathrm{2}\right) \\ $$$$\frac{{L}}{{c}}=\frac{\mathrm{sin}\:\left(\mathrm{45}+\alpha\right)}{\mathrm{sin}\:\mathrm{45}}=\frac{\mathrm{sin}\:\left(\mathrm{45}+\mathrm{45}−\beta\right)}{\mathrm{sin}\:\mathrm{45}}=\sqrt{\mathrm{2}}\mathrm{cos}\:\beta \\ $$$$\frac{{L}}{{d}}=\frac{\mathrm{sin}\:\left(\mathrm{45}+\beta\right)}{\mathrm{sin}\:\mathrm{45}}=\frac{\mathrm{sin}\:\left(\mathrm{45}+\mathrm{45}−\alpha\right)}{\mathrm{sin}\:\mathrm{45}}=\sqrt{\mathrm{2}}\mathrm{cos}\:\alpha \\ $$$$\frac{{L}^{\mathrm{2}} }{{cd}}=\mathrm{2}\:\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta \\ $$$$\Rightarrow{cd}=\frac{{L}^{\mathrm{2}} }{\mathrm{2cos}\:\alpha\:\mathrm{cos}\:\beta} \\ $$$${e}=\frac{{L}}{\mathrm{cos}\:\alpha} \\ $$$${f}=\frac{{L}}{\mathrm{cos}\:\beta} \\ $$$$\Rightarrow{ef}=\frac{{L}^{\mathrm{2}} }{\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta}=\mathrm{2}{cd} \\ $$$${S}_{\mathrm{1}} =\frac{{cd}\:\mathrm{sin}\:\mathrm{45}°}{\mathrm{2}} \\ $$$${S}_{\mathrm{1}} +{S}_{\mathrm{2}} =\frac{{ef}\:\mathrm{sin}\:\mathrm{45}°}{\mathrm{2}}=\mathrm{2}{S}_{\mathrm{1}} \\ $$$$\Rightarrow{S}_{\mathrm{1}} ={S}_{\mathrm{2}} \\ $$
Commented by liberty last updated on 12/Sep/21

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