Calculus 3: Relative Minimums and Maximums

I'm given this curve 3x² + 3xy + 3y² =1 that's located on the xy plane and I have to find the the point that's closest to the origin. Can this be solved by setting z = 3x² + 3xy + 3y² -1 and plugging z into D² = x² + y² + z² as like example 3 here: http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx

Question

Calculus 3: Relative Minimums and Maximums

I'm given this curve 3x² + 3xy + 3y² =1 that's located on the xy plane and I have to find the the point that's closest to the origin. Can this be solved by setting z = 3x² + 3xy + 3y² -1 and plugging z into D² = x² + y² + z² as like example 3 here: http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx

anonymous · Answer

So is this a curve in the x-y plane or a surface in the x-y-z space? I'm not sure what you intend to use the $z$ for.

anonymous · Answer

it is a curve on the x-y plane

anonymous · Answer

No, that example features a plane in three dimensions whereas you're dealing with a two dimensional curve in the x-y plane. Hence, example 3 is not relevant.

anonymous · Answer

By the way, you're dealing with an ellipse in R^2, so forget about the "z" variable...

anonymous · Answer

Oh I see, but do I still use the distance formula D² = x² + y², in xy plane, to find the point that is the closest to the origin?

anonymous · Answer

Actually I don't think I can because I can't solve for x or y, so how would I proceed to find the point?

amistre64 · Answer

|dw:1384348094970:dw|

there are other ways then this, but in my head im thinking:

the closest distance between 2 points relates to perpendicular property.  If we know the normal line equation to a given point on the curve, then we can find when the point (0,0) isa solution to the equation.  This will would give us a collection of points to test out.