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if-a-b-c-d-R-verify-a-2b-3c-4d-6-then-find-min-a-2-b-2-c-2-d-2-




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 )
ifa;b;c;dRverifya+2b+3c+4d=6thenfindmin(a2+b2+c2+d2)
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)
(a2+b2+c2+d2)min=(612+22+32+42)2=65
Commented by mathdanisur last updated on 01/Oct/21
Thanlyou Ser, can you tell me how this  part was obtained, if possible
ThanlyouSer,canyoutellmehowthispartwasobtained,ifpossible
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)
x2+y2+z2+w2isthedistancefrompoint(x,y,z,w)totheorigin(0,0,0,0)inthe4Dspace.x+2y+3z+4w=6isanplaneinthe4Dspace.theshortestdistancefromapoint(x,y,z,w)onthisplanetotheoriginis(x2+y2+z2+w2)min=0+2×0+3×0+4×0612+22+32+44=630therefore(a2+b2+c2+d2)min=(a2+b2+c2+d2)min2=(630)2=65
Commented by mathdanisur last updated on 01/Oct/21
Creativ solution thank you Ser
CreativsolutionthankyouSer

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