# Find points nearest/farthest from origin on the intersection of a plane and a parabol

1. Oct 6, 2012

### skyturnred

1. The problem statement, all variables and given/known data

Find the points nearest and furthest from the origin on the intersection of a plane with a paraboloid.

Plane:
x+y+2z=30

Paraboloid:
z=x$^{2}$+y$^{2}$

2. Relevant equations

3. The attempt at a solution

Obviously the first step is to find the equation of the ellipse that is formed

I do that by plugging the equation of the paraboloid into the equation of the plane and get:

x+y+2x$^{2}$+2y$^{2}$=30

But my problem is that the equation above isn't really the equation I need. The equation above gives the shape of the ellipse *on the plane* but I need an equation for that shape on the original 3D Cartesian plane. How do I go about doing that?

Also, once I find said equation, would the following way to go about finding the max/min be correct?

Choose the equation of the ellipse as the constraint, choose f(x,y,z)=$\sqrt{x^{2}+y^{2}+z^{2}}$ and then just treat it like any straight forward max/min question?

Thanks

2. Oct 6, 2012

### Ray Vickson

Re: Find points nearest/farthest from origin on the intersection of a plane and a par

You don't need to find the ellipse of intersection; you just need to minimize (or maximize) x^2 + y^2 + z^2, subject to the two equality constraints x+y+2z = 30 and z = x^2 + y^2. You could, for example, use Lagrange multipliers to solve this problem; you would have two Lagrange multipliers, one for each constraint. Of course, you COULD find the intersection ellipse and proceed that way, but it is not the only way and maybe not even the easiest way.

Note: min/max $x^2+y^2+z^2$ has exactly the same solutions as min/max $\sqrt{x^2 + y^2 + z^2}$ but is much easier to work with.

RGV

3. Oct 6, 2012

### skyturnred

Re: Find points nearest/farthest from origin on the intersection of a plane and a par

Thanks! That simplifies things a lot.