1. ## 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

2. 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 }
$

....

3. 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?

4. 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

5. 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

6. 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...