Question Number 51774 by ajfour last updated on 30/Dec/18
Commented by ajfour last updated on 30/Dec/18
$${Find}\:{maximum}\:{area}\:{A}\:{in}\:{light}\:{blue}. \\ $$$$\left({source}\::\:{ajfour}\right) \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18
Commented by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18
![area of OABCDA is=2[(1/2)×h×b+(1/2)×k×a] =hb+ka area if pinnk circle sector=(b^2 /2)×(π−2α) area of yelolow circle sector=(a^2 /2)×(π−2β) 2α+2β=(π/2) α+β=(π/4) area of light blue area hb+ka−[(π/2)(a^2 +b^2 )−(a^2 β+b^2 α)] to my view...tangent from point o to circle radius b...tangent from o to circle radis a should be equal ldngth... so h=k tanα=(b/h) α=tan^(−1) ((b/h)) tanβ=(a/k)=(a/h) β=tan^(−1) ((a/h)) tan(α+β)=((tanα+tanβ)/(1−tanαtanβ))=(((b/h)+(a/h))/(1−((ab)/h^2 )))=(((a+b)/h)/(1−((ab)/h^2 )))=1 ((a+b)/h)=1−((ab)/h^2 ) (a+b)h=h^2 −ab h^2 −h(a+b)−ab=0 h=(((a+b)±(√((a+b)^2 +4ab)))/2) h=(((a+b)+(√((a+b)^2 +4ab)))/2)=L(say) so light blue area hb+ka−[(π/2)(a^2 +b^2 )−(a^2 β+b^2 α)] =L(a+b)−[(π/2)(a^2 +b^2 )−{a^2 tan^(−1) ((a/L))+b^2 tan^(−1) ((b/L))] where L=f(a,b) pls chek upto this step...](https://www.tinkutara.com/question/Q51780.png)
$${area}\:{of}\:{OABCDA}\:{is}=\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}×{h}×{b}+\frac{\mathrm{1}}{\mathrm{2}}×{k}×{a}\right] \\ $$$$={hb}+{ka} \\ $$$${area}\:{if}\:{pinnk}\:{circle}\:{sector}=\frac{{b}^{\mathrm{2}} }{\mathrm{2}}×\left(\pi−\mathrm{2}\alpha\right) \\ $$$${area}\:{of}\:{yelolow}\:{circle}\:{sector}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}×\left(\pi−\mathrm{2}\beta\right) \\ $$$$\mathrm{2}\alpha+\mathrm{2}\beta=\frac{\pi}{\mathrm{2}}\:\:\:\:\:\alpha+\beta=\frac{\pi}{\mathrm{4}} \\ $$$${area}\:{of}\:{light}\:{blue}\:{area} \\ $$$${hb}+{ka}−\left[\frac{\pi}{\mathrm{2}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} \beta+{b}^{\mathrm{2}} \alpha\right)\right] \\ $$$${to}\:{my}\:{view}…{tangent}\:{from}\:{point}\:{o}\:{to}\:{circle} \\ $$$${radius}\:{b}…{tangent}\:{from}\:{o}\:{to}\:{circle}\:{radis}\:{a} \\ $$$$\:{should}\:{be}\:{equal}\:{ldngth}… \\ $$$${so}\:{h}={k} \\ $$$${tan}\alpha=\frac{{b}}{{h}}\:\:\:\:\alpha={tan}^{−\mathrm{1}} \left(\frac{{b}}{{h}}\right) \\ $$$${tan}\beta=\frac{{a}}{{k}}=\frac{{a}}{{h}}\:\:\:\beta={tan}^{−\mathrm{1}} \left(\frac{{a}}{{h}}\right) \\ $$$${tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}=\frac{\frac{{b}}{{h}}+\frac{{a}}{{h}}}{\mathrm{1}−\frac{{ab}}{{h}^{\mathrm{2}} }}=\frac{\frac{{a}+{b}}{{h}}}{\mathrm{1}−\frac{{ab}}{{h}^{\mathrm{2}} }}=\mathrm{1} \\ $$$$\frac{{a}+{b}}{{h}}=\mathrm{1}−\frac{{ab}}{{h}^{\mathrm{2}} } \\ $$$$\left({a}+{b}\right){h}={h}^{\mathrm{2}} −{ab} \\ $$$${h}^{\mathrm{2}} −{h}\left({a}+{b}\right)−{ab}=\mathrm{0} \\ $$$${h}=\frac{\left({a}+{b}\right)\pm\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{4}{ab}}}{\mathrm{2}} \\ $$$${h}=\frac{\left({a}+{b}\right)+\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{4}{ab}}}{\mathrm{2}}={L}\left({say}\right) \\ $$$${so}\:{light}\:{blue}\:{area} \\ $$$${hb}+{ka}−\left[\frac{\pi}{\mathrm{2}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} \beta+{b}^{\mathrm{2}} \alpha\right)\right] \\ $$$$={L}\left({a}+{b}\right)−\left[\frac{\pi}{\mathrm{2}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left\{{a}^{\mathrm{2}} {tan}^{−\mathrm{1}} \left(\frac{{a}}{{L}}\right)+{b}^{\mathrm{2}} {tan}^{−\mathrm{1}} \left(\frac{{b}}{{L}}\right)\right]\right. \\ $$$${where}\:{L}={f}\left({a},{b}\right) \\ $$$${pls}\:{chek}\:{upto}\:{this}\:{step}… \\ $$$$ \\ $$
Commented by ajfour last updated on 30/Dec/18
$${tough}\:{to}\:{check}\:{Sir},\:{i}\:{have}\:{myself} \\ $$$${tried}\:{along}\:{mrW}\:{Sir}'{s}\:{line}..! \\ $$
Answered by mr W last updated on 30/Dec/18
Commented by mr W last updated on 30/Dec/18
![A_(blue) =[a+(a+b)cos α][b+(a+b)sin α]−(((a+b)^2 sin α cos α)/2)+((αa^2 )/2)+((((π/2)−α)b^2 )/2)−((πa^2 )/2)−((πb^2 )/2) A_(blue) =(a+b)[(a sin α+b cos α)+(((a+b)sin 2α)/4)+((α(a−b))/2)]+ab−((πa^2 )/2)−((πb^2 )/4) ...(ii) (dA_(blue) /dα)=0 ⇒a cos α−b sin α+(((a+b)cos 2α)/2)+((a−b)/2)=0 ⇒2(a cos α−b sin α)+(a+b)cos 2α=b−a let (b/a)=μ, t=tan (α/2) ⇒cos α(1+cos α)=μ sin α(1+sin α) ⇒μ=((cos α(1+cos α))/(sin α(1+sin α)))=((1−t)/(t(1+t))) μt^2 +(μ+1)t−1=0 t=tan (α/2)=(((√(μ^2 +6μ+1))−μ−1)/(2μ)) ⇒α=2 tan^(−1) (((√(μ^2 +6μ+1))−μ−1)/(2μ)) ⇒max. A_(blue) from (ii)](https://www.tinkutara.com/question/Q51787.png)
$${A}_{{blue}} =\left[{a}+\left({a}+{b}\right)\mathrm{cos}\:\alpha\right]\left[{b}+\left({a}+{b}\right)\mathrm{sin}\:\alpha\right]−\frac{\left({a}+{b}\right)^{\mathrm{2}} \mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\mathrm{2}}+\frac{\alpha{a}^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\frac{\pi}{\mathrm{2}}−\alpha\right){b}^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi{a}^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi{b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${A}_{{blue}} =\left({a}+{b}\right)\left[\left({a}\:\mathrm{sin}\:\alpha+{b}\:\mathrm{cos}\:\alpha\right)+\frac{\left({a}+{b}\right)\mathrm{sin}\:\mathrm{2}\alpha}{\mathrm{4}}+\frac{\alpha\left({a}−{b}\right)}{\mathrm{2}}\right]+{ab}−\frac{\pi{a}^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi{b}^{\mathrm{2}} }{\mathrm{4}}\:\:\:…\left({ii}\right) \\ $$$$\frac{{dA}_{{blue}} }{{d}\alpha}=\mathrm{0} \\ $$$$\Rightarrow{a}\:\mathrm{cos}\:\alpha−{b}\:\mathrm{sin}\:\alpha+\frac{\left({a}+{b}\right)\mathrm{cos}\:\mathrm{2}\alpha}{\mathrm{2}}+\frac{{a}−{b}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left({a}\:\mathrm{cos}\:\alpha−{b}\:\mathrm{sin}\:\alpha\right)+\left({a}+{b}\right)\mathrm{cos}\:\mathrm{2}\alpha={b}−{a} \\ $$$${let}\:\frac{{b}}{{a}}=\mu,\:{t}=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha\left(\mathrm{1}+\mathrm{cos}\:\:\alpha\right)=\mu\:\mathrm{sin}\:\alpha\left(\mathrm{1}+\mathrm{sin}\:\alpha\right) \\ $$$$\Rightarrow\mu=\frac{\mathrm{cos}\:\alpha\left(\mathrm{1}+\mathrm{cos}\:\:\alpha\right)}{\mathrm{sin}\:\alpha\left(\mathrm{1}+\mathrm{sin}\:\alpha\right)}=\frac{\mathrm{1}−{t}}{{t}\left(\mathrm{1}+{t}\right)} \\ $$$$\mu{t}^{\mathrm{2}} +\left(\mu+\mathrm{1}\right){t}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{\sqrt{\mu^{\mathrm{2}} +\mathrm{6}\mu+\mathrm{1}}−\mu−\mathrm{1}}{\mathrm{2}\mu} \\ $$$$\Rightarrow\alpha=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mu^{\mathrm{2}} +\mathrm{6}\mu+\mathrm{1}}−\mu−\mathrm{1}}{\mathrm{2}\mu} \\ $$$$\Rightarrow{max}.\:{A}_{{blue}} \:{from}\:\left({ii}\right) \\ $$
Commented by ajfour last updated on 30/Dec/18
$${let}\:\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:=\:{t} \\ $$$$\Rightarrow\:\:\mu=\:\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{{t}\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\:\Rightarrow\:\:\:\:\boldsymbol{\mu}\:=\:\frac{\mathrm{1}−\boldsymbol{{t}}}{\boldsymbol{{t}}\left(\mathrm{1}+\boldsymbol{{t}}\right)}\: \\ $$$$\mu{t}^{\mathrm{2}} +\left(\mu+\mathrm{1}\right){t}−\mathrm{1}=\mathrm{0} \\ $$$${t}\:=\:\frac{−\left(\mu+\mathrm{1}\right)+\sqrt{\left(\mu+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\mu}}{\mathrm{2}\mu} \\ $$$$\:\alpha\:=\:\mathrm{2tan}^{−\mathrm{1}} \left(\frac{−\left(\mu+\mathrm{1}\right)+\sqrt{\left(\mu+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\mu}}{\mathrm{2}\mu}\right) \\ $$
Commented by ajfour last updated on 30/Dec/18
$${Yes}\:{Sir},\:\left({sorry}\right),\:{its}\:\mathcal{A}{bsolutely}\:{right}! \\ $$
Commented by mr W last updated on 30/Dec/18
$${thank}\:{you}\:{sir}!\:{that}'{s}\:{correct}. \\ $$