Finding closest and farthest points on a curve using partial derivatives

[AbuYusufEg]

how to find the "most far and most close" ( points on curve ) to another point ?

i'm studying a chapter on how to find maxima and minima values of a function using partial derivatives.

one of the problems is the following:
"if plane $z=x+y+1$ intersects cone $z^2=x^2+y^2$
let the curve that result from intersection, is C
What is the most far and most close points on C, with respect to (0,0,0) ? and also with respect to $(x_1,y_1,z_1)$"

i think that that curve would be something like a circle, and that there would be some function that depends on the length between "(0,0,0) or $(x_1,y_1,z_1)$", and "any point on C".

But what is that function ?
And how to work out that problem ?

* I've exam in that chapter after about 10 hours, so please try to answer me with detailed answer as I've no time for discussions for now, may be i do that later.

 

[tiny-tim]

"if plane $z=x+y+1$ intersects cone $z^2=x^2+y^2$
let the curve that result from intersection, is C
What is the most far and most close points on C, with respect to (0,0,0) ? and also with respect to $(x_1,y_1,z_1)$"

Hi AbuYusufEg!

Hint: the distance² from (0,0,0) to (x,y,z) is x² + y² + z²

and to help find the intersection, I suggest you rearrange z = x + y + 1 and then square it.

 

[AbuYusufEg]

yes i got that, but i want the points that ONLY on the curve C, i think that (x,y,z) is any point.
So, How can i get the points that only on the curve C ?

 

[tiny-tim]

yes i got that, but i want the points that ONLY on the curve C, i think that (x,y,z) is any point.
So, How can i get the points that only on the curve C ?

Rearrange z = x + y + 1 and then square it, and compare with z² = x² + y²