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Question Number 182000 by mr W last updated on 03/Dec/22

if a−2b+3c−4d+5e−6f=0, find the maximum of ((∣a+b+c+d+e+f∣)/( (√(a^2 +b^2 +c^2 +d^2 +e^2 +f^2 )))).

$${if}\:\boldsymbol{{a}}−\mathrm{2}\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{c}}−\mathrm{4}\boldsymbol{{d}}+\mathrm{5}\boldsymbol{{e}}−\mathrm{6}\boldsymbol{{f}}=\mathrm{0},\:{find} \\ $$$${the}\:{maximum}\:{of} \\ $$$$\frac{\mid\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}+\boldsymbol{{d}}+\boldsymbol{{e}}+\boldsymbol{{f}}\mid}{\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} +\boldsymbol{{d}}^{\mathrm{2}} +\boldsymbol{{e}}^{\mathrm{2}} +\boldsymbol{{f}}^{\mathrm{2}} }}. \\ $$

Commented by mr W last updated on 04/Dec/22

A(0,0,0,0,0,0) B(1,−2,3,−4,5,−6) P(1,1,1,1,1,1) ∣AP∣=(√(1^2 +1^2 +1^2 +1^2 +1^2 +1^2 ))=(√6) ∣AB∣=(√(1^2 +(−2)^2 +3^2 +(−4)^2 +5^2 +(−6)^2 ))=(√(91)) cos θ=((1−2+3−4+5−6)/( (√6)×(√(91))))=−(3/( (√6)×(√(91)))) sin θ=((√(6×91−3^2 ))/( (√6)×(√(91))))=((√(537))/( (√6)×(√(91)))) ∣AP∣ sin θ=(√6)×((√(537))/( (√6)×(√(91))))=(√((537)/(91))) ⇒(((∣a+b+c+d+e+f∣)/( (√(a^2 +b^2 +c^2 +d^2 +e^2 +f^2 )))))_(max) = (√((537)/(91)))

$${A}\left(\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0}\right) \\ $$$${B}\left(\mathrm{1},−\mathrm{2},\mathrm{3},−\mathrm{4},\mathrm{5},−\mathrm{6}\right) \\ $$$${P}\left(\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1}\right) \\ $$$$\mid{AP}\mid=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }=\sqrt{\mathrm{6}} \\ $$$$\mid{AB}\mid=\sqrt{\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\left(−\mathrm{6}\right)^{\mathrm{2}} }=\sqrt{\mathrm{91}} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+\mathrm{5}−\mathrm{6}}{\:\sqrt{\mathrm{6}}×\sqrt{\mathrm{91}}}=−\frac{\mathrm{3}}{\:\sqrt{\mathrm{6}}×\sqrt{\mathrm{91}}} \\ $$$$\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{6}×\mathrm{91}−\mathrm{3}^{\mathrm{2}} }}{\:\sqrt{\mathrm{6}}×\sqrt{\mathrm{91}}}=\frac{\sqrt{\mathrm{537}}}{\:\sqrt{\mathrm{6}}×\sqrt{\mathrm{91}}} \\ $$$$\mid{AP}\mid\:\mathrm{sin}\:\theta=\sqrt{\mathrm{6}}×\frac{\sqrt{\mathrm{537}}}{\:\sqrt{\mathrm{6}}×\sqrt{\mathrm{91}}}=\sqrt{\frac{\mathrm{537}}{\mathrm{91}}} \\ $$$$\Rightarrow\left(\frac{\mid\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}+\boldsymbol{{d}}+\boldsymbol{{e}}+\boldsymbol{{f}}\mid}{\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} +\boldsymbol{{d}}^{\mathrm{2}} +\boldsymbol{{e}}^{\mathrm{2}} +\boldsymbol{{f}}^{\mathrm{2}} }}\right)_{{max}} \:=\:\sqrt{\frac{\mathrm{537}}{\mathrm{91}}} \\ $$

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