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Math Help - Optimization Segment Minimum

  1. #1
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    Optimization Segment Minimum

    A line through the point (2,2) cuts the x and y axes at A and B. Find the minimum length of the segment AB.
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  2. #2
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    OA=2+x x-distance from x=2 to the right
    tg a = 2/x
    OB=(2+x)tg a=2(2+x)/x
    |AB|^2=|OA|^2+|OB|^2
    f(x)= |AB|^2=(2+x)^2+\frac{4(2+x)^2}{x^2}
    f'(x)=0
    x^3-8=0
    x=2
    OA=4 (x)
    OB=4 (y).
    |AB| min= 4\sqrt{2}
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  3. #3
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    Quote Originally Posted by robertosavin View Post
    A line through the point (2,2) cuts the x and y axes at A and B. Find the minimum length of the segment AB.
    The line has a negative slope.

    If it had a positive slope, it could go through the origin making the distance
    between A and B zero.

    If you draw the line going through (2,2) with a negative slope,
    we have similar triangles by drawing lines from (2,2) to the x and y axes.

    hence tan\theta=\frac{2}{x-2}=\frac{y-2}{2}

    y-2=\frac{4}{x-2}

    let x-2=c, y-2=k

    the length of the line segment is, using Pythagoras' theorem

    \sqrt{(k)^2+2^2}+\sqrt{(c)^2+2^2}=\sqrt{\frac{4^2}  {c^2}+4}+\sqrt{c^2+4}

    You could differentiate this with respect to c and set the result = 0
    to find the value of x causing the segment length to be a minimum.

    Alternatively,

    the segment length is

    \frac{2}{cos\theta}+\frac{2}{sin\theta}

    we can differentiate wrt the angle to find the minimum length

    \frac{2sin\theta}{cos^2\theta}-\frac{2cos\theta}{sin^2\theta}=0

    \frac{2sin^3\theta-2cos^3\theta}{sin^2\theta\ cos^2\theta}=0

    sin^3\theta-cos^3\theta=0

    this occurs when the angle is 45 degrees

    hence the minimum segment length is 2sin45^o+2cos45^o

    or \sqrt{4^2+4^2}
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