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Question Number 155495 by mathdanisur last updated on 01/Oct/21

if a;b;c;d∈R verify a+2b+3c+4d=6 then find min(a^2 +b^2 +c^2 +d^2 )

$$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c};\mathrm{d}\in\mathbb{R}\:\:\mathrm{verify}\:\:\mathrm{a}+\mathrm{2b}+\mathrm{3c}+\mathrm{4d}=\mathrm{6} \\ $$$$\mathrm{then}\:\mathrm{find}\:\:\boldsymbol{\mathrm{min}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} \right) \\ $$

Answered by mr W last updated on 01/Oct/21

(a^2 +b^2 +c^2 +d^2 )_(min) =((6/( (√(1^2 +2^2 +3^2 +4^2 )))))^2 =(6/5)

$$\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} \right)_{{min}} =\left(\frac{\mathrm{6}}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }}\right)^{\mathrm{2}} =\frac{\mathrm{6}}{\mathrm{5}} \\ $$

Commented by mathdanisur last updated on 01/Oct/21

Thanlyou Ser, can you tell me how this part was obtained, if possible

$$\mathrm{Thanlyou}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{can}\:\mathrm{you}\:\mathrm{tell}\:\mathrm{me}\:\mathrm{how}\:\mathrm{this} \\ $$$$\mathrm{part}\:\mathrm{was}\:\mathrm{obtained},\:\mathrm{if}\:\mathrm{possible} \\ $$

Commented by mr W last updated on 01/Oct/21

(√(x^2 +y^2 +z^2 +w^2 )) is the distance from point (x,y,z,w) to the origin (0,0,0,0) in the 4D space. x+2y+3z+4w=6 is an plane in the 4D space. the shortest distance from a point (x,y,z,w) on this plane to the origin is ((√(x^2 +y^2 +z^2 +w^2 )))_(min) =((∣0+2×0+3×0+4×0−6∣)/( (√(1^2 +2^2 +3^2 +4^4 ))))=(6/( (√(30)))) therefore (a^2 +b^2 +c^2 +d^2 )_(min) =((√(a^2 +b^2 +c^2 +d^2 )))_(min) ^2 =((6/( (√(30)))))^2 =(6/5)

$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{w}^{\mathrm{2}} }\:{is}\:{the}\:{distance}\:{from} \\ $$$${point}\:\left({x},{y},{z},{w}\right)\:{to}\:{the}\:{origin}\:\left(\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0}\right) \\ $$$${in}\:{the}\:\mathrm{4}{D}\:{space}.\: \\ $$$${x}+\mathrm{2}{y}+\mathrm{3}{z}+\mathrm{4}{w}=\mathrm{6}\:{is}\:{an}\:{plane}\:{in}\:{the} \\ $$$$\mathrm{4}{D}\:{space}.\: \\ $$$${the}\:{shortest}\:{distance}\:{from}\:{a}\:{point} \\ $$$$\left({x},{y},{z},{w}\right)\:{on}\:{this}\:{plane}\:{to}\:{the}\:{origin}\:{is} \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{w}^{\mathrm{2}} }\right)_{{min}} =\frac{\mid\mathrm{0}+\mathrm{2}×\mathrm{0}+\mathrm{3}×\mathrm{0}+\mathrm{4}×\mathrm{0}−\mathrm{6}\mid}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{4}} }}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{30}}} \\ $$$${therefore} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)_{{min}} =\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }\right)_{{min}} ^{\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{6}}{\:\sqrt{\mathrm{30}}}\right)^{\mathrm{2}} =\frac{\mathrm{6}}{\mathrm{5}} \\ $$

Commented by mathdanisur last updated on 01/Oct/21

Creativ solution thank you Ser

$$\mathrm{Creativ}\:\mathrm{solution}\:\mathrm{thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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