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If-A-gt-0-B-gt-0-and-A-B-pi-3-then-maximum-value-of-tanAtanB-is-

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

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