# Thread: closest point mmn14 4a

1. ## closest point mmn14 4a

on the surface $\displaystyle x^2y^2z^2=1$
find a point on this plane where it is the closest to (0,0,0)

i know we should find a function f(x,y) which is the distance between the point on the suface oand (0,0,0)

how to find this f(x,y)

?

2. Originally Posted by transgalactic
on the surface $\displaystyle x^2y^2z^2=1$
find a point on this plane where it is the closest to (0,0,0)

i know we should find a function f(x,y) which is the distance between the point on the suface oand (0,0,0)

how to find this f(x,y)

?
Minimise

$\displaystyle d^2=x^2+y^2+z^2$

subject to

$\displaystyle x^2y^2z^2=1$

Lagrange multiplier problem, now show us what you have done or explain where you are having difficulties.

CB

3. Originally Posted by CaptainBlack
Minimise

$\displaystyle d^2=x^2+y^2+z^2$

subject to

$\displaystyle x^2y^2z^2=1$

Lagrange multiplier problem, now show us what you have done or explain where you are having difficulties.

CB
ok
$\displaystyle z=g(x,y)=\sqrt{\frac{1}{x^2y^2}}$

$\displaystyle d^2=x^2+y^2+z^2$
$\displaystyle f(x,y)=d^2=x^2+y^2+(\sqrt{\frac{1}{x^2y^2}})^2$

can i use the d^2 formula
and find its minimal points instead of the "d" formula
?

4. Originally Posted by transgalactic
ok
$\displaystyle z=g(x,y)=\sqrt{\frac{1}{x^2y^2}}$

$\displaystyle d^2=x^2+y^2+z^2$
$\displaystyle f(x,y)=d^2=x^2+y^2+(\sqrt{\frac{1}{x^2y^2}})^2$

can i use the d^2 formula
and find its minimal points instead of the "d" formula
?
d is a minimum if and only if d^2 is a minimum when d is a distance.

CB