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Math Help - Another Optimization problem

  1. #1
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    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
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  2. #2
    Senior Member Peritus's Avatar
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    any point on the line is of the form:

    <br />
\left( {x,mx + b} \right)<br />

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

    <br />
d(x) = \sqrt {x^2  + \left( {mx + b} \right)^2 } <br />


    ....
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  3. #3
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    Quote Originally Posted by Peritus View Post
    any point on the line is of the form:

    <br />
\left( {x,mx + b} \right)<br />

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

    <br />
d(x) = \sqrt {x^2  + \left( {mx + b} \right)^2 } <br />


    ....

    right i understand that i just don't know what the critical points of the distance formula would be?
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  4. #4
    Newbie Black Kawairothlite's Avatar
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    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
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  5. #5
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    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
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  6. #6
    Newbie Black Kawairothlite's Avatar
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    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...
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