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

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

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

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

Commented by mathdanisur last updated on 01/Oct/21

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

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

$$\sqrt{x^2+y^2+z^2+w^2}isthedistancefrom$$$$point(x,y,z,w)totheorigin(0,0,0,0)$$$$inthe4Dspace.$$$$x+2y+3z+4w=6isanplaneinthe$$$$4Dspace.$$$$theshortestdistancefromapoint$$$$(x,y,z,w)onthisplanetotheoriginis$$$${\left(\sqrt{x^2+y^2+z^2+w^2}\right)}_{min}=\frac{\mid 0+2\times 0+3\times 0+4\times 0-6\mid }{\sqrt{1^2+2^2+3^2+4^4}}=\frac{6}{\sqrt{30}}$$$$therefore$$$${(a^2+b^2+c^2+d^2)}_{min}={\left(\sqrt{a^2+b^2+c^2+d^2}\right)}_{min}^2$$$$={\left(\frac{6}{\sqrt{30}}\right)}^2=\frac{6}{5}$$

Commented by mathdanisur last updated on 01/Oct/21

$$\mathrm{Creativ}\mathrm{solution}\mathrm{thank}\mathrm{you}\mathbf{S}\mathrm{er}$$

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