# Another Optimization problem

• Oct 25th 2008, 04:08 PM
bdavidson
Another Optimization problem
Hello again this problem is giving me trouble

Find the point on the line y=mx+b that is closest to the orign. A complete answer should use calculus to show that the point is closest. We assume that m not= 0 and m,b are Reals.

i know your supposed to use the distance formula but i'm not sure how to go about it. Any help would be greatly appreciated. Thanks
• Oct 25th 2008, 04:38 PM
Peritus
any point on the line is of the form:

$
\left( {x,mx + b} \right)
$

thus as you already so astutely observed we use the distance formula:

$
d(x) = \sqrt {x^2 + \left( {mx + b} \right)^2 }
$

....
• Oct 26th 2008, 08:55 PM
bdavidson
Quote:

Originally Posted by Peritus
any point on the line is of the form:

$
\left( {x,mx + b} \right)
$

thus as you already so astutely observed we use the distance formula:

$
d(x) = \sqrt {x^2 + \left( {mx + b} \right)^2 }
$

....

right i understand that i just don't know what the critical points of the distance formula would be?
• Oct 26th 2008, 09:16 PM
Black Kawairothlite
i think the critical points are where

$x^2 + \left( {mx + b} \right)^2<0$

so u gotta solve this inequation... correct me if i'm wrong please
• Oct 26th 2008, 09:43 PM
bdavidson
I'm not sure, I'm pretty sure you need to differentiate the distance formula and find the critical point but i can't figure it out. Any help would be awesome thanks
• Oct 26th 2008, 09:58 PM
Black Kawairothlite
the theory says that u find the critical points when $f'(x)=0$

you gotta find the values of x where the derivate is null...