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Given-x-y-real-number-such-that-0-lt-y-x-lt-1-2-Find-minimum-value-of-2y-x-y-3x-x-2y-

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Question Number 150963 by EDWIN88 last updated on 17/Aug/21
Given x ,y real number such that 0<(y/x)<(1/2). Find minimum value of ((2y)/(x−y)) +((3x)/(x+2y)) .
$${Given}\:{x}\:,{y}\:{real}\:{number}\:{such}\:{that} \\ $$$$\:\mathrm{0}<\frac{{y}}{{x}}<\frac{\mathrm{1}}{\mathrm{2}}.\:{Find}\:{minimum}\:{value} \\ $$$${of}\:\frac{\mathrm{2}{y}}{{x}−{y}}\:+\frac{\mathrm{3}{x}}{{x}+\mathrm{2}{y}}\:.\: \\ $$
Answered by john_santu last updated on 17/Aug/21
let (y/x) = t ⇒f(t)=((2t)/(1−t))+(3/(1+2t)) where 0<t<(1/2). taking derivative of f(t) f ′(t)=(2/((1−t)^2 ))−(6/((1+2t)^2 )) = 0 ⇒(1+2t)^2 =3(1−t)^2 ⇒4t^2 +4t+1=3−6t+3t^2 ⇒t^2 +10t−2=0 ⇒(t+5)^2 −27=0 ⇒t=−5±3(√3) f(t)_(min) when t=−5+3(√3) so min f(t)=((2(−5+3(√3)))/(1−(−5+3(√3))))+(3/(1+2(−5+3(√3)))) min f(t)=((6(√3)−10)/(6−3(√3)))+(1/( 2(√3)−3)) min f(t)= (((6(√3)−10)(6+3(√3)))/9)+((2(√3)+3)/3) min f(t)=((6(√3) −6)/9)+((2(√3)+3)/3) min f(t)=((2(√3)−2)/3)+((2(√3)+3)/3) =((4(√3)+1)/3)
$$\:\mathrm{let}\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:\mathrm{t}\:\Rightarrow\mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{2t}}{\mathrm{1}−\mathrm{t}}+\frac{\mathrm{3}}{\mathrm{1}+\mathrm{2t}}\: \\ $$$$\mathrm{where}\:\mathrm{0}<\mathrm{t}<\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$$\mathrm{taking}\:\mathrm{derivative}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{t}\right) \\ $$$$\mathrm{f}\:'\left(\mathrm{t}\right)=\frac{\mathrm{2}}{\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{2}} }−\frac{\mathrm{6}}{\left(\mathrm{1}+\mathrm{2t}\right)^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{2t}\right)^{\mathrm{2}} =\mathrm{3}\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{1}=\mathrm{3}−\mathrm{6t}+\mathrm{3t}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{t}^{\mathrm{2}} +\mathrm{10t}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{t}+\mathrm{5}\right)^{\mathrm{2}} −\mathrm{27}=\mathrm{0}\:\Rightarrow\mathrm{t}=−\mathrm{5}\pm\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\mathrm{f}\left(\mathrm{t}\right)_{\mathrm{min}} \:\mathrm{when}\:\mathrm{t}=−\mathrm{5}+\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\mathrm{so}\:\mathrm{min}\:\mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{2}\left(−\mathrm{5}+\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{1}−\left(−\mathrm{5}+\mathrm{3}\sqrt{\mathrm{3}}\right)}+\frac{\mathrm{3}}{\mathrm{1}+\mathrm{2}\left(−\mathrm{5}+\mathrm{3}\sqrt{\mathrm{3}}\right)} \\ $$$$\mathrm{min}\:\mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{10}}{\mathrm{6}−\mathrm{3}\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}} \\ $$$$\mathrm{min}\:\mathrm{f}\left(\mathrm{t}\right)=\:\frac{\left(\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{10}\right)\left(\mathrm{6}+\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{9}}+\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}}{\mathrm{3}} \\ $$$$\mathrm{min}\:\mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{6}\sqrt{\mathrm{3}}\:−\mathrm{6}}{\mathrm{9}}+\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}}{\mathrm{3}} \\ $$$$\mathrm{min}\:\mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}}{\mathrm{3}}\:=\frac{\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$

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