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Thread: Optimization

  1. #1
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    Optimization

    Find the points on the graph f(x)=2x that is closest to the point (1,0) .

    How would you go about this?
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  2. #2
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    Quote Originally Posted by av8or91 View Post
    Find the points on the graph f(x)=2x that is closest to the point (1,0) .

    How would you go about this?
    The square distance of an arbitary point $\displaystyle (x,y)$ from $\displaystyle (1,0)$ is:

    $\displaystyle d^2=(x-1)^2+y^2$

    Now substitute from $\displaystyle y=2x$ for either $\displaystyle x$ or $\displaystyle y$ then differentiate and set the derivative to zero, and solve.

    CB
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  3. #3
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    Quote Originally Posted by av8or91 View Post
    Find the points on the graph f(x)=2x that is closest to the point (1,0) .

    How would you go about this?

    A general point on the graph of $\displaystyle f(x)$ is of the form $\displaystyle (x,f(x))$ , so the distance between such a point and (1,0) is;

    $\displaystyle \sqrt{(x-1)^2+(f(x)-0)^2}=\sqrt{x^2-2x+1+4x^2}$ . Well, do some order here and check this is a function of x so to find is extreme points you have to...

    Tonio
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