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Find-the-minimum-value-of-x-2-y-2-z-2-with-the-condition-ax-by-cz-p-

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Question Number 15741 by ajfour last updated on 13/Jun/17
Find the minimum value of x^2 +y^2 +z^2 , with the condition ax+by+cz=p .
Findtheminimumvalueofx2+y2+z2,withtheconditionax+by+cz=p.
Answered by mrW1 last updated on 13/Jun/17
(√(x^2 +y^2 +z^2 )) is the distance of a point (x,y,z) on the plane ax+by+cz−p=0 to the origin (0,0,0). since the distance from (0,0,0) to the plane ax+by+cz−p is d=((∣p∣)/( (√(a^2 +b^2 +c^2 )))) (√(x^2 +y^2 +z^2 )) ≥d x^2 +y^2 +z^2 ≥d^2 =(p^2 /(a^2 +b^2 +c^2 ))
x2+y2+z2isthedistanceofapoint(x,y,z)ontheplaneax+by+cz−p=0totheorigin(0,0,0).sincethedistancefrom(0,0,0)totheplaneax+by+cz−pisd=∣p∣a2+b2+c2x2+y2+z2⩾dx2+y2+z2⩾d2=p2a2+b2+c2
Commented by ajfour last updated on 13/Jun/17
good viewpoint sir, i thought it would again involve partial derivatives.
goodviewpointsir,ithoughtitwouldagaininvolvepartialderivatives.
Commented by mrW1 last updated on 13/Jun/17
Solution in conventional way: z=((p−(ax+by))/c)=p′−(a′x+b′y) with p′=(p/c), a′=(a/c), b′=(b/c) S=x^2 +y^2 +z^2 =x^2 +y^2 +p′^2 −2p′(a′x+b′y)+(a′x+b′y)^2 =x^2 +y^2 +p′^2 −2p′(a′x+b′y)+a′^2 x^2 +b′^2 y^2 +2a′b′xy =(1+a′^2 )x^2 +(1+b′^2 )y^2 +2a′b′xy−2p′a′x−2p′b′y+p′^2 (∂S/∂x)=2(1+a′^2 )x+2a′b′y−2p′a′=0 ⇒(1+a′^2 )x+a′b′y=p′a′ (∂S/∂y)=2(1+b′^2 )y+2a′b′x−2p′b′=0 ⇒(1+b′^2 )y+a′b′x=p′b′ ⇒y=((p′b′)/((1+b′^2 )))−((a′b′)/((1+b′^2 )))x ⇒(1+a′^2 )x−(((a′b′)^2 )/((1+b′^2 )))x=p′a′−((p′a′b′^2 )/((1+b′^2 ))) ⇒(((1+a′^2 )(1+b′^2 )−(a′b′)^2 )/((1+b′^2 )))x=((p′a′)/((1+b′^2 ))) ⇒x=((p′a′)/((1+a′^2 )(1+b′^2 )−(a′b′)^2 ))=((p′a′)/(1+a′^2 +b′^2 )) ⇒y=((p′b′)/((1+a′^2 )(1+b′^2 )−(a′b′)^2 ))=((p′b′)/(1+a′^2 +b′^2 )) S_(min) =(1+a′^2 )(((p′a′)/(1+a′^2 +b′^2 )))^2 +(1+b′^2 )(((p′b′)/(1+a′^2 +b′^2 )))^2 +2a′b′(((p′a′)/(1+a′^2 +b′^2 )))(((p′b′)/(1+a′^2 +b′^2 )))−2p′a′(((p′a′)/(1+a′^2 +b′^2 )))−2p′b′(((p′b′)/(1+a′^2 +b′^2 )))+p′^2 =(1/((1+a′^2 +b′^2 )^2 ))[(1+a′^2 )p′^2 a′^2 +(1+b′^2 )p′^2 b′^2 +2p′^2 a′^2 b′^2 −2p′^2 (a′^2 +b′^2 )(1+a′^2 +b′^2 )+p′^2 (1+a′^2 +b′^2 )^2 ] =((p′^2 )/((1+a′^2 +b′^2 )^2 ))[(a′^2 +b′^2 )(1+a′^2 +b′^2 )+(1−a′^2 −b′^2 )(1+a′^2 +b′^2 )] =((p′^2 )/((1+a′^2 +b′^2 )^2 ))[a′^2 +b′^2 +1−a′^2 −b′^2 ](1+a′^2 +b′^2 ) =((p′^2 )/(1+a′^2 +b′^2 )) =((p^2 /c^2 )/(1+(a^2 /c^2 )+(b^2 /c^2 )))=(p^2 /(a^2 +b^2 +c^2 ))
Solutioninconventionalway:z=p−(ax+by)c=p′−(a′x+b′y)withp′=pc,a′=ac,b′=bcS=x2+y2+z2=x2+y2+p′2−2p′(a′x+b′y)+(a′x+b′y)2=x2+y2+p′2−2p′(a′x+b′y)+a′2x2+b′2y2+2a′b′xy=(1+a′2)x2+(1+b′2)y2+2a′b′xy−2p′a′x−2p′b′y+p′2∂S∂x=2(1+a′2)x+2a′b′y−2p′a′=0⇒(1+a′2)x+a′b′y=p′a′∂S∂y=2(1+b′2)y+2a′b′x−2p′b′=0⇒(1+b′2)y+a′b′x=p′b′⇒y=p′b′(1+b′2)−a′b′(1+b′2)x⇒(1+a′2)x−(a′b′)2(1+b′2)x=p′a′−p′a′b′2(1+b′2)⇒(1+a′2)(1+b′2)−(a′b′)2(1+b′2)x=p′a′(1+b′2)⇒x=p′a′(1+a′2)(1+b′2)−(a′b′)2=p′a′1+a′2+b′2⇒y=p′b′(1+a′2)(1+b′2)−(a′b′)2=p′b′1+a′2+b′2Smin=(1+a′2)(p′a′1+a′2+b′2)2+(1+b′2)(p′b′1+a′2+b′2)2+2a′b′(p′a′1+a′2+b′2)(p′b′1+a′2+b′2)−2p′a′(p′a′1+a′2+b′2)−2p′b′(p′b′1+a′2+b′2)+p′2=1(1+a′2+b′2)2[(1+a′2)p′2a′2+(1+b′2)p′2b′2+2p′2a′2b′2−2p′2(a′2+b′2)(1+a′2+b′2)+p′2(1+a′2+b′2)2]=p′2(1+a′2+b′2)2[(a′2+b′2)(1+a′2+b′2)+(1−a′2−b′2)(1+a′2+b′2)]=p′2(1+a′2+b′2)2[a′2+b′2+1−a′2−b′2](1+a′2+b′2)=p′21+a′2+b′2=p2c21+a2c2+b2c2=p2a2+b2+c2
Commented by mrW1 last updated on 13/Jun/17
this way is lengthy, but the result is the same.
thiswayislengthy,buttheresultisthesame.

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