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Given-a-gt-0-b-gt-0-c-gt-2-a-b-1-find-the-minimum-value-of-3ac-b-c-ab-6-c-2-

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Question Number 179090 by CrispyXYZ last updated on 24/Oct/22
Given a>0, b>0, c>2, a+b=1. find the minimum value of ((3ac)/b)+(c/(ab))+(6/(c−2)).
$$\mathrm{Given}\:{a}>\mathrm{0},\:{b}>\mathrm{0},\:{c}>\mathrm{2},\:{a}+{b}=\mathrm{1}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\:\frac{\mathrm{3}{ac}}{{b}}+\frac{{c}}{{ab}}+\frac{\mathrm{6}}{{c}−\mathrm{2}}. \\ $$
Commented by Frix last updated on 24/Oct/22
24
$$\mathrm{24} \\ $$
Answered by MJS_new last updated on 24/Oct/22
a>0∧b>0∧a+b=1 ⇒ 0<a<1∧0<b<1 ⇒ let a=sin^2 x ∧ b=cos^2 x I prefer x=arctan t ⇒ a=(t^2 /(t^2 +1))∧b=(1/(t^2 +1)) c>2 ⇒ c=2+u^2 ⇒ ((3ac)/b)+(c/(ab))+(6/(c−2))= =3t^2 (u^2 +2)+(((t^2 +1)^2 (u^2 +2))/t^2 )+(6/u^2 )=f(t, u) (df/dt)=0 6(u^2 +2)t+((2(t^4 −1)(u^2 +2))/t^3 )=0 2(4t^4 −1)(u^2 +2)=0 t=±((√2)/2) f(±((√2)/2), u)=((3(u^2 +2))/2)+((9(u^2 +2))/2)+(6/u^2 )=((6(u^2 +1))/u^2 ) (df/du)=0 ((12(u^4 −1))/u^3 )=0 u=±1 f(±((√2)/2), ±1)=24 with a=(1/3)∧b=(2/3)∧c=3
$${a}>\mathrm{0}\wedge{b}>\mathrm{0}\wedge{a}+{b}=\mathrm{1}\:\Rightarrow\:\mathrm{0}<{a}<\mathrm{1}\wedge\mathrm{0}<{b}<\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{let}\:{a}=\mathrm{sin}^{\mathrm{2}} \:{x}\:\wedge\:{b}=\mathrm{cos}^{\mathrm{2}} \:{x} \\ $$$$\:\:\:\:\:\:\mathrm{I}\:\mathrm{prefer}\:{x}=\mathrm{arctan}\:{t}\:\Rightarrow\:{a}=\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}\wedge{b}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${c}>\mathrm{2}\:\Rightarrow\:{c}=\mathrm{2}+{u}^{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow \\ $$$$ \\ $$$$\frac{\mathrm{3}{ac}}{{b}}+\frac{{c}}{{ab}}+\frac{\mathrm{6}}{{c}−\mathrm{2}}= \\ $$$$=\mathrm{3}{t}^{\mathrm{2}} \left({u}^{\mathrm{2}} +\mathrm{2}\right)+\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({u}^{\mathrm{2}} +\mathrm{2}\right)}{{t}^{\mathrm{2}} }+\frac{\mathrm{6}}{{u}^{\mathrm{2}} }={f}\left({t},\:{u}\right) \\ $$$$\frac{{df}}{{dt}}=\mathrm{0} \\ $$$$\mathrm{6}\left({u}^{\mathrm{2}} +\mathrm{2}\right){t}+\frac{\mathrm{2}\left({t}^{\mathrm{4}} −\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{2}\right)}{{t}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\mathrm{2}\left(\mathrm{4}{t}^{\mathrm{4}} −\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{0} \\ $$$${t}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${f}\left(\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},\:{u}\right)=\frac{\mathrm{3}\left({u}^{\mathrm{2}} +\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{9}\left({u}^{\mathrm{2}} +\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{6}}{{u}^{\mathrm{2}} }=\frac{\mathrm{6}\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{{u}^{\mathrm{2}} } \\ $$$$\frac{{df}}{{du}}=\mathrm{0} \\ $$$$\frac{\mathrm{12}\left({u}^{\mathrm{4}} −\mathrm{1}\right)}{{u}^{\mathrm{3}} }=\mathrm{0} \\ $$$${u}=\pm\mathrm{1} \\ $$$${f}\left(\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},\:\pm\mathrm{1}\right)=\mathrm{24} \\ $$$$\mathrm{with}\:{a}=\frac{\mathrm{1}}{\mathrm{3}}\wedge{b}=\frac{\mathrm{2}}{\mathrm{3}}\wedge{c}=\mathrm{3} \\ $$

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