Question Number 57269 by rahul 19 last updated on 01/Apr/19 | ||
$${If}\:{A}>\mathrm{0},{B}>\mathrm{0},\:{and}\:{A}+{B}=\frac{\pi}{\mathrm{3}}\:,\:{then} \\ $$ $${maximum}\:{value}\:{of}\:{tanAtanB}\:{is}\:? \\ $$ | ||
Commented byrahul 19 last updated on 01/Apr/19 | ||
$${Any}\:{short}\:{method}\:? \\ $$ | ||
Commented bytanmay.chaudhury50@gmail.com last updated on 01/Apr/19 | ||
Commented bytanmay.chaudhury50@gmail.com last updated on 01/Apr/19 | ||
Commented bytanmay.chaudhury50@gmail.com last updated on 01/Apr/19 | ||
$${tanAtanB}\:{max}\:{when}\:{A}={B}=\frac{\pi}{\mathrm{6}} \\ $$ | ||
Commented bymr W last updated on 01/Apr/19 | ||
$${due}\:{to}\:{symmetry}\:\mathrm{tan}\:{A}×\mathrm{tan}\:{B}\:{is} \\ $$ $${maximum}\:{or}\:{minimum}\:{when}\:{A}={B}=\frac{\pi}{\mathrm{6}}, \\ $$ $${max}.=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$ | ||
Commented byrahul 19 last updated on 01/Apr/19 | ||
thank you sir! | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19 | ||
$${tan}\left({A}+{B}\right)=\sqrt{\mathrm{3}}\: \\ $$ $${p}={tanAtanB} \\ $$ $${p}={tanAtan}\left(\frac{\pi}{\mathrm{3}}−{A}\right) \\ $$ $${p}={tanA}×\frac{\sqrt{\mathrm{3}}\:−{tanA}}{\mathrm{1}+\sqrt{\mathrm{3}}\:{tanA}} \\ $$ $${p}=\frac{\sqrt{\mathrm{3}}\:{a}−{a}^{\mathrm{2}} }{\mathrm{1}+\sqrt{\mathrm{3}}\:{a}} \\ $$ $$\frac{{dp}}{{da}}=\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\:{a}\right)\left(\sqrt{\mathrm{3}}\:−\mathrm{2}{a}\right)−\left(\sqrt{\mathrm{3}}\:{a}−{a}^{\mathrm{2}} \right)\left(\mathrm{0}+\sqrt{\mathrm{3}}\:\right)}{\left(\mathrm{1}+\sqrt{\mathrm{3}}\:{a}\right)^{\mathrm{2}} } \\ $$ $${for}\:{max}/{min}\:\frac{{dp}}{{da}}=\mathrm{0} \\ $$ $$\sqrt{\mathrm{3}}\:−\mathrm{2}{a}+\mathrm{3}{a}−\mathrm{2}\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} −\mathrm{3}{a}+\sqrt{\mathrm{3}}{a}^{\mathrm{2}} =\mathrm{0} \\ $$ $$−\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} −\mathrm{2}{a}+\sqrt{\mathrm{3}}\:=\mathrm{0} \\ $$ $$\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} +\mathrm{2}{a}−\sqrt{\mathrm{3}}\:=\mathrm{0} \\ $$ $$\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} +\mathrm{3}{a}−{a}−\sqrt{\mathrm{3}}\:=\mathrm{0} \\ $$ $$\sqrt{\mathrm{3}}\:{a}\left({a}+\sqrt{\mathrm{3}}\:\right)−\mathrm{1}\left({a}+\sqrt{\mathrm{3}}\:\right)=\mathrm{0} \\ $$ $$\left({a}+\sqrt{\mathrm{3}}\:\right)\left(\sqrt{\mathrm{3}}\:{a}−\mathrm{1}\right)=\mathrm{0} \\ $$ $${a}\neq−\sqrt{\mathrm{3}}\:\:\:\:\:{so}\:{a}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$ $$\frac{{dp}}{{da}}=\frac{−\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} −\mathrm{2}{a}+\sqrt{\mathrm{3}\:}}{\left(\mathrm{1}+\sqrt{\mathrm{3}}\:{a}\right)^{\mathrm{2}} } \\ $$ $$\frac{{dp}}{{da}}>\mathrm{0}\:{when}\:{a}<\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$ $$\frac{{dp}}{{da}}=\mathrm{0}\:{at}\:{a}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$ $$\frac{{dp}}{{da}}<\mathrm{0}\:{when}\:{a}>\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$ $${so}\:{sign}\:{of}\:\frac{{dp}}{{da}}\:{changes}\:{from}\:\left(+{ve}\right){to}\:\left(−{ve}\right) \\ $$ $${so}\:{at}\:{a}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\:\:{p}\:{is}\:{maximum} \\ $$ $$ \\ $$ $${tanA}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}={tan}\frac{\pi}{\mathrm{6}} \\ $$ $${A}=\frac{\pi}{\mathrm{6}}\:\:{hence}\:{B}=\frac{\pi}{\mathrm{6}} \\ $$ $${max}\:{value}\:\left({tan}\frac{\pi}{\mathrm{6}}\right)\left({tan}\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ | ||
Commented byrahul 19 last updated on 01/Apr/19 | ||
thank you sir! | ||
Answered by ajfour last updated on 01/Apr/19 | ||
$$\mathrm{tan}\:\left(\mathrm{A}+\mathrm{B}\right)=\frac{\mathrm{tan}\:\mathrm{A}+\mathrm{tan}\:\mathrm{B}}{\mathrm{1}−\mathrm{tan}\:\mathrm{Atan}\:\mathrm{B}}=\sqrt{\mathrm{3}} \\ $$ $$\Rightarrow\:\mathrm{tan}\:\mathrm{Atan}\:\mathrm{B}=\mathrm{1}−\frac{\mathrm{tan}\:\mathrm{A}+\mathrm{tan}\:\mathrm{B}}{\sqrt{\mathrm{3}}} \\ $$ $$\mathrm{let}\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{tan}\:\mathrm{x}+\mathrm{tan}\:\left(\pi/\mathrm{3}−\mathrm{x}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{sec}\:^{\mathrm{2}} \left(\pi/\mathrm{3}−\mathrm{x}\right) \\ $$ $$\:\:\:\:\:\:\:\:=\mathrm{0}\:\:\Rightarrow\:\:\:\mathrm{x}=\frac{\pi}{\mathrm{3}}−\mathrm{x} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{x}=\frac{\pi}{\mathrm{6}} \\ $$ $$\:\:\:\:\:\:\:\:\mathrm{f}\:''\left(\frac{\pi}{\mathrm{6}}\right)=\mathrm{2sec}\:^{\mathrm{2}} \frac{\pi}{\mathrm{6}}\mathrm{tan}\:\frac{\pi}{\mathrm{6}} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2sec}\:^{\mathrm{2}} \frac{\pi}{\mathrm{6}}\mathrm{tan}\:\frac{\pi}{\mathrm{6}}\:>\mathrm{0} \\ $$ $$\:\:\:\:\:\:\:\:\:\mathrm{hence}\:\left(\mathrm{tan}\:\mathrm{Atan}\:\mathrm{B}\right)_{\mathrm{max}} =\frac{\mathrm{1}}{\mathrm{3}}\:. \\ $$ | ||
Commented byrahul 19 last updated on 01/Apr/19 | ||
thank you sir! | ||