I don't understand what I did wrong in my process? The question is: Find the point on the plane 6x − y + 6z = 60 nearest the origin. The answer is: < 360/73, -60/73, 360/73 > I got: < 6(360/73), -1(360/73), 6(360/73) >

Question

$g(x)=6x-y+6z=60$
$gradg(x)=<6,-1,6>$

$D^2=(x-0)^2+(y-0)^2+(z-0)^2$
$6x-6z-60=y$
$D^2(x,z)=x^2+(6x+6z-60)^2+z^2$
$D^2(x,z)=37x^2+72xz-720x-720z+3600+37z^2$

$gradD^2(x,z)=<74x+72z-720,74z+72x-720>$
$gradD^2(x,z)=<37x+36z-360,37z+36x-360>$
$a) 37x+36z-360=0$
$b) 37z+36x-360=0$
$a)-b) = 37x-36x+36z-37z-360+360=0$
$a)-b) = x-z=0 -> x=z$

$x=z, -> 37(x)+36x-360=0$
$x=360/73$

$< x,y,z > * <6,-1,6> = <6(360/73),-1(360/73),6(360/73)>$

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Answer 1 · 2018-06-10T14:04:47

See below

You're using a Lagrange multiplier:

Condition:

$g (x,y,z) = 6x − y + 6z - 60 = 0$

To be optimised:

$f(x,y,z) = D^2=(x-0)^2+(y-0)^2+(z-0)^2$

$(2x, 2y, 2z) = lambda (6,-1,6)$

$implies bb( lambda = x/3 = - 2y = z/3 )$

$g (x,y,z) = 6(3 lambda) − (- lambda/2) + 6(3 lambda) - 60 = 0$

$implies bb(lambda = 120/73 )$

$(x,y,z) = (3 lambda,- lambda/2,3 lambda)$

$= (360/73,-60/73,360/73)$