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
$${area}\:{of}\:{OABCDA}\:{is}=\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}×{h}×{b}+\frac{\mathrm{1}}{\mathrm{2}}×{k}×{a}\right] \\ $$$$={hb}+{ka} \\ $$$${area}\:{if}\:{pinnk}\:{circle}\:{sector}=\frac{{b}^{\mathrm{2}} }{\mathrm{2}}×\left(\pi−\mathrm{2}\alpha\right) \\ $$$${area}\:{of}\:{yelolow}\:{circle}\:{sector}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}×\left(\pi−\mathrm{2}\beta\right) \\ $$$$\mathrm{2}\alpha+\mathrm{2}\beta=\frac{\pi}{\mathrm{2}}\:\:\:\:\:\alpha+\beta=\frac{\pi}{\mathrm{4}} \\ $$$${area}\:{of}\:{light}\:{blue}\:{area} \\ $$$${hb}+{ka}−\left[\frac{\pi}{\mathrm{2}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} \beta+{b}^{\mathrm{2}} \alpha\right)\right] \\ $$$${to}\:{my}\:{view}…{tangent}\:{from}\:{point}\:{o}\:{to}\:{circle} \\ $$$${radius}\:{b}…{tangent}\:{from}\:{o}\:{to}\:{circle}\:{radis}\:{a} \\ $$$$\:{should}\:{be}\:{equal}\:{ldngth}… \\ $$$${so}\:{h}={k} \\ $$$${tan}\alpha=\frac{{b}}{{h}}\:\:\:\:\alpha={tan}^{−\mathrm{1}} \left(\frac{{b}}{{h}}\right) \\ $$$${tan}\beta=\frac{{a}}{{k}}=\frac{{a}}{{h}}\:\:\:\beta={tan}^{−\mathrm{1}} \left(\frac{{a}}{{h}}\right) \\ $$$${tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}=\frac{\frac{{b}}{{h}}+\frac{{a}}{{h}}}{\mathrm{1}−\frac{{ab}}{{h}^{\mathrm{2}} }}=\frac{\frac{{a}+{b}}{{h}}}{\mathrm{1}−\frac{{ab}}{{h}^{\mathrm{2}} }}=\mathrm{1} \\ $$$$\frac{{a}+{b}}{{h}}=\mathrm{1}−\frac{{ab}}{{h}^{\mathrm{2}} } \\ $$$$\left({a}+{b}\right){h}={h}^{\mathrm{2}} −{ab} \\ $$$${h}^{\mathrm{2}} −{h}\left({a}+{b}\right)−{ab}=\mathrm{0} \\ $$$${h}=\frac{\left({a}+{b}\right)\pm\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{4}{ab}}}{\mathrm{2}} \\ $$$${h}=\frac{\left({a}+{b}\right)+\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{4}{ab}}}{\mathrm{2}}={L}\left({say}\right) \\ $$$${so}\:{light}\:{blue}\:{area} \\ $$$${hb}+{ka}−\left[\frac{\pi}{\mathrm{2}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} \beta+{b}^{\mathrm{2}} \alpha\right)\right] \\ $$$$={L}\left({a}+{b}\right)−\left[\frac{\pi}{\mathrm{2}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left\{{a}^{\mathrm{2}} {tan}^{−\mathrm{1}} \left(\frac{{a}}{{L}}\right)+{b}^{\mathrm{2}} {tan}^{−\mathrm{1}} \left(\frac{{b}}{{L}}\right)\right]\right. \\ $$$${where}\:{L}={f}\left({a},{b}\right) \\ $$$${pls}\:{chek}\:{upto}\:{this}\:{step}… \\ $$$$ \\ $$
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
$${A}_{{blue}} =\left[{a}+\left({a}+{b}\right)\mathrm{cos}\:\alpha\right]\left[{b}+\left({a}+{b}\right)\mathrm{sin}\:\alpha\right]−\frac{\left({a}+{b}\right)^{\mathrm{2}} \mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\mathrm{2}}+\frac{\alpha{a}^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\frac{\pi}{\mathrm{2}}−\alpha\right){b}^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi{a}^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi{b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${A}_{{blue}} =\left({a}+{b}\right)\left[\left({a}\:\mathrm{sin}\:\alpha+{b}\:\mathrm{cos}\:\alpha\right)+\frac{\left({a}+{b}\right)\mathrm{sin}\:\mathrm{2}\alpha}{\mathrm{4}}+\frac{\alpha\left({a}−{b}\right)}{\mathrm{2}}\right]+{ab}−\frac{\pi{a}^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi{b}^{\mathrm{2}} }{\mathrm{4}}\:\:\:…\left({ii}\right) \\ $$$$\frac{{dA}_{{blue}} }{{d}\alpha}=\mathrm{0} \\ $$$$\Rightarrow{a}\:\mathrm{cos}\:\alpha−{b}\:\mathrm{sin}\:\alpha+\frac{\left({a}+{b}\right)\mathrm{cos}\:\mathrm{2}\alpha}{\mathrm{2}}+\frac{{a}−{b}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left({a}\:\mathrm{cos}\:\alpha−{b}\:\mathrm{sin}\:\alpha\right)+\left({a}+{b}\right)\mathrm{cos}\:\mathrm{2}\alpha={b}−{a} \\ $$$${let}\:\frac{{b}}{{a}}=\mu,\:{t}=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha\left(\mathrm{1}+\mathrm{cos}\:\:\alpha\right)=\mu\:\mathrm{sin}\:\alpha\left(\mathrm{1}+\mathrm{sin}\:\alpha\right) \\ $$$$\Rightarrow\mu=\frac{\mathrm{cos}\:\alpha\left(\mathrm{1}+\mathrm{cos}\:\:\alpha\right)}{\mathrm{sin}\:\alpha\left(\mathrm{1}+\mathrm{sin}\:\alpha\right)}=\frac{\mathrm{1}−{t}}{{t}\left(\mathrm{1}+{t}\right)} \\ $$$$\mu{t}^{\mathrm{2}} +\left(\mu+\mathrm{1}\right){t}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{\sqrt{\mu^{\mathrm{2}} +\mathrm{6}\mu+\mathrm{1}}−\mu−\mathrm{1}}{\mathrm{2}\mu} \\ $$$$\Rightarrow\alpha=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mu^{\mathrm{2}} +\mathrm{6}\mu+\mathrm{1}}−\mu−\mathrm{1}}{\mathrm{2}\mu} \\ $$$$\Rightarrow{max}.\:{A}_{{blue}} \:{from}\:\left({ii}\right) \\ $$
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
$${thank}\:{you}\:{sir}!\:{that}'{s}\:{correct}. \\ $$