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Question Number 181875 by mr W last updated on 01/Dec/22
if a+b+c=0, find the maximum of ((∣a+2b+3c∣)/( (√(a^2 +b^2 +c^2 )))).
$${if}\:{a}+{b}+{c}=\mathrm{0},\:{find}\:{the}\:{maximum}\:{of} \\ $$$$\frac{\mid{a}+\mathrm{2}{b}+\mathrm{3}{c}\mid}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }}. \\ $$
Commented by mr W last updated on 01/Dec/22
Q181318 reposted, asking for more solutions.
$${Q}\mathrm{181318}\:{reposted},\:{asking}\:{for}\:{more} \\ $$$${solutions}. \\ $$
Commented by SEKRET last updated on 01/Dec/22
what happened to previos solutions
$$\:\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{happened}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{previos}}\:\boldsymbol{\mathrm{solutions}} \\ $$
Commented by mr W last updated on 01/Dec/22
they are there, nothing happened.
$${they}\:{are}\:{there},\:{nothing}\:{happened}. \\ $$
Answered by SEKRET last updated on 01/Dec/22
max ((√(2 )) )
$$\boldsymbol{\mathrm{max}}\:\left(\sqrt{\mathrm{2}\:}\:\right) \\ $$
Commented by mr W last updated on 01/Dec/22
please show your solution.
$${please}\:{show}\:{your}\:{solution}. \\ $$
Answered by SEKRET last updated on 01/Dec/22
c = −(b+a) a+2b−3(b+a)= −(b+2a) a^2 +b^2 +(a+b)^2 =2a^2 +2b^2 +2ab ((∣b+2a∣)/( (√2) ∙(√(a^2 +b^2 +ab)))) b= ax ((∣a∣∙∣x+2∣)/( (√2) ∙∣a∣∙(√(x^2 +x+1))))= ((∣x+2∣)/( (√2)∙(√(x^2 +x+1))))=f(x) f ′ (x)=0 ((−3x)/(2(√(2 ))∙(√((x^2 +x+1)^3 )))) =0 x=0 f(0)=(2/( (√2))) = (√2)
$$\:\:\:\boldsymbol{\mathrm{c}}\:=\:−\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}\right)\:\:\:\:\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{b}}−\mathrm{3}\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}\right)=\:−\left(\boldsymbol{\mathrm{b}}+\mathrm{2}\boldsymbol{\mathrm{a}}\right) \\ $$$$\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)^{\mathrm{2}} =\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{ab}} \\ $$$$\:\frac{\mid\boldsymbol{\mathrm{b}}+\mathrm{2}\boldsymbol{\mathrm{a}}\mid}{\:\sqrt{\mathrm{2}}\:\centerdot\sqrt{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{ab}}}}\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{b}}=\:\boldsymbol{\mathrm{ax}} \\ $$$$\:\:\:\:\:\frac{\mid\boldsymbol{\mathrm{a}}\mid\centerdot\mid\boldsymbol{\mathrm{x}}+\mathrm{2}\mid}{\:\sqrt{\mathrm{2}}\:\centerdot\mid\boldsymbol{\mathrm{a}}\mid\centerdot\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}}}=\:\frac{\mid\boldsymbol{\mathrm{x}}+\mathrm{2}\mid}{\:\sqrt{\mathrm{2}}\centerdot\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}}}=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\:\boldsymbol{\mathrm{f}}\:'\:\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{0}\:\:\:\:\:\:\frac{−\mathrm{3}\boldsymbol{\mathrm{x}}}{\mathrm{2}\sqrt{\mathrm{2}\:}\centerdot\sqrt{\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{3}} }}\:=\mathrm{0} \\ $$$$\:\:\boldsymbol{\mathrm{x}}=\mathrm{0}\:\:\:\:\boldsymbol{\mathrm{f}}\left(\mathrm{0}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}\:=\:\sqrt{\mathrm{2}} \\ $$$$ \\ $$
Commented by mr W last updated on 01/Dec/22
thanks!
$${thanks}! \\ $$
Answered by mr W last updated on 02/Dec/22
I) regular method b=−(a+c) f=((∣a+2b+3c∣)/( (√(a^2 +b^2 +c^2 )))) =((∣a−2a−2c+3c∣)/( (√(a^2 +(a+c)^2 +c^2 )))) =((∣c−a∣)/( (√(2(a^2 +ac+c^2 ))))) =((∣k−1∣)/( (√(2(1+k+k^2 ))))) with k=(c/a) =((∣k−1∣)/( (√(2((k−1)^2 +3(k−1)+3))))) =((∣u∣)/( (√(2(u^2 +3u+3))))) with u=k−1 =(1/( (√(2((3/u^2 )+(3/u)+1))))) =(1/( (√(6[((1/u)+(1/2))^2 +(1/(12))])))) ≤(1/( (√(6×(1/(12))))))=(√2) ⇒maximum =(√2) at u=−2=k−1=(c/a)−1 ⇒c+a=0, b=0
$$\left.\underline{\boldsymbol{{I}}\right)\:\boldsymbol{{regular}}\:\boldsymbol{{method}}} \\ $$$${b}=−\left({a}+{c}\right) \\ $$$${f}=\frac{\mid{a}+\mathrm{2}{b}+\mathrm{3}{c}\mid}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }} \\ $$$$=\frac{\mid{a}−\mathrm{2}{a}−\mathrm{2}{c}+\mathrm{3}{c}\mid}{\:\sqrt{{a}^{\mathrm{2}} +\left({a}+{c}\right)^{\mathrm{2}} +{c}^{\mathrm{2}} }} \\ $$$$=\frac{\mid{c}−{a}\mid}{\:\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{ac}+{c}^{\mathrm{2}} \right)}} \\ $$$$=\frac{\mid{k}−\mathrm{1}\mid}{\:\sqrt{\mathrm{2}\left(\mathrm{1}+{k}+{k}^{\mathrm{2}} \right)}}\:\:\:\:{with}\:{k}=\frac{{c}}{{a}} \\ $$$$=\frac{\mid{k}−\mathrm{1}\mid}{\:\sqrt{\mathrm{2}\left(\left({k}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left({k}−\mathrm{1}\right)+\mathrm{3}\right)}}\: \\ $$$$=\frac{\mid{u}\mid}{\:\sqrt{\mathrm{2}\left({u}^{\mathrm{2}} +\mathrm{3}{u}+\mathrm{3}\right)}}\:\:\:\:{with}\:{u}={k}−\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\left(\frac{\mathrm{3}}{{u}^{\mathrm{2}} }+\frac{\mathrm{3}}{{u}}+\mathrm{1}\right)}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}\left[\left(\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{12}}\right]}} \\ $$$$\leqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}×\frac{\mathrm{1}}{\mathrm{12}}}}=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{maximum}\:=\sqrt{\mathrm{2}}\: \\ $$$${at}\:{u}=−\mathrm{2}={k}−\mathrm{1}=\frac{{c}}{{a}}−\mathrm{1}\:\Rightarrow{c}+{a}=\mathrm{0},\:{b}=\mathrm{0} \\ $$
Commented by SEKRET last updated on 01/Dec/22
super
$$\:\:\:\boldsymbol{\mathrm{super}} \\ $$
Answered by mr W last updated on 02/Dec/22
II) geometric method image we have a plane with following equation: ax+by+cz=0 this plane passes through the origin A(0,0,0). since a+b+c=0, it means the plane passes also through the pointB(1,1,1). we know ((∣a+2b+3c∣)/( (√(a^2 +b^2 +c^2 )))) represents the distance from the point P(1,2,3) to the plane ax+by+cz=0. now we only need to find the largest distance from the point P(1,2,3) to the plane ax+by+cz=0. since the plane passes through the points A(0,0,0) and B(1,1,1), the largest distance from point P to the plane is the distance from point P to the line AB. AB=(1,1,1) AP=(1,2,3) AP×AB=(1,2,3)×(1,1,1)=(1,2,1) d=((∣AP×AB∣)/(∣AB∣))=((√(1^2 +2^2 +1^2 ))/( (√(1^2 +1^2 +1^2 ))))=((√6)/( (√3)))=(√2) that means maximum of ((∣a+2b+3c∣)/( (√(a^2 +b^2 +c^2 )))) is (√2).
$$\left.\underline{\boldsymbol{{II}}\right)\:\boldsymbol{{geometric}}\:\boldsymbol{{method}}} \\ $$$${image}\:{we}\:{have}\:{a}\:{plane}\:{with}\:{following} \\ $$$${equation}: \\ $$$${ax}+{by}+{cz}=\mathrm{0} \\ $$$${this}\:{plane}\:{passes}\:{through}\:{the}\:{origin} \\ $$$${A}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right). \\ $$$${since}\:{a}+{b}+{c}=\mathrm{0},\:{it}\:{means}\:{the}\:{plane} \\ $$$${passes}\:{also}\:{through}\:{the}\:{pointB}\left(\mathrm{1},\mathrm{1},\mathrm{1}\right). \\ $$$${we}\:{know}\:\frac{\mid{a}+\mathrm{2}{b}+\mathrm{3}{c}\mid}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }}\:{represents}\:{the} \\ $$$${distance}\:{from}\:{the}\:{point}\:{P}\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\:{to}\:{the} \\ $$$${plane}\:{ax}+{by}+{cz}=\mathrm{0}.\: \\ $$$${now}\:{we}\:{only}\:{need}\:{to}\:{find}\:{the}\:{largest} \\ $$$${distance}\:{from}\:{the}\:{point}\:{P}\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\:{to}\:{the} \\ $$$${plane}\:{ax}+{by}+{cz}=\mathrm{0}.\: \\ $$$${since}\:{the}\:{plane}\:{passes}\:{through}\:{the} \\ $$$${points}\:{A}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:{and}\:{B}\left(\mathrm{1},\mathrm{1},\mathrm{1}\right),\:{the} \\ $$$${largest}\:{distance}\:{from}\:{point}\:{P}\:{to}\:{the}\: \\ $$$${plane}\:{is}\:{the}\:{distance}\:{from}\:{point}\:{P}\:{to}\: \\ $$$${the}\:{line}\:{AB}. \\ $$$$\boldsymbol{{AB}}=\left(\mathrm{1},\mathrm{1},\mathrm{1}\right) \\ $$$$\boldsymbol{{AP}}=\left(\mathrm{1},\mathrm{2},\mathrm{3}\right) \\ $$$$\boldsymbol{{AP}}×\boldsymbol{{AB}}=\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)×\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)=\left(\mathrm{1},\mathrm{2},\mathrm{1}\right) \\ $$$${d}=\frac{\mid\boldsymbol{{AP}}×\boldsymbol{{AB}}\mid}{\mid\boldsymbol{{AB}}\mid}=\frac{\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{3}}}=\sqrt{\mathrm{2}} \\ $$$${that}\:{means}\:{maximum}\:{of}\:\frac{\mid{a}+\mathrm{2}{b}+\mathrm{3}{c}\mid}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }} \\ $$$${is}\:\sqrt{\mathrm{2}}. \\ $$
Commented by mr W last updated on 01/Dec/22
Commented by mr W last updated on 02/Dec/22
alternative: cos θ=(((1,2,3)∙(1,1,1))/( (√((1^2 +1^2 +1^2 )(1^2 +2^2 +3^2 )))))=((√6)/( (√7))) sin θ=(1/( (√7))) d=AP sin θ=(√(14))×(1/( (√7)))=(√2)
$${alternative}: \\ $$$$\mathrm{cos}\:\theta=\frac{\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\centerdot\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)}{\:\sqrt{\left(\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \right)\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)}}=\frac{\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{7}}} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}} \\ $$$${d}={AP}\:\mathrm{sin}\:\theta=\sqrt{\mathrm{14}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}=\sqrt{\mathrm{2}} \\ $$

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