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Math Help - Optimization

  1. #1
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    Optimization

    A house lies on a plain at point (3,4). There is a highway along the graph f(x) = x^3. I want to make the shortest path between the house and the highway. I need to find the point where the lane should meet the highway.
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  2. #2
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    Attachment 17056
    Quote Originally Posted by steezin View Post
    A house lies on a plain at point (3,4). There is a highway along the graph f(x) = x^3. I want to make the shortest path between the house and the highway. I need to find the point where the lane should meet the highway.
    If you draw the graph, you have a cubic function y=x^3

    and a circle centred at (3,4) which touches the graph at a single point.
    The radius of this circle is the minimum distance.

    The point (x,y) where the circle meets the curve must satisfy (x-3)^2+(y-4)^2=d^2\ and\ y=x^3

    Hence (x-3)^2+(x^3-4)^2=d^2

    This is the distance of all points (x,y) on the curve from (3,4)

    hence differentiating (x-3)^2+(x^3-4)^2 and setting the result to zero will find the x corresponding to minimum distance.
    Attached Thumbnails Attached Thumbnails Optimization-optimization.jpg  
    Last edited by Archie Meade; May 30th 2010 at 01:53 PM.
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  3. #3
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    Here's another way to do it: the shortest line from a point to a curve must be perpendicular to the tangent line to the curve at the point where they cross.

    The derivative of x^3 is 3x^2 so we must have a slope of \frac{1}{3x_0^2} for the perpendicular line that crosses the curve at (x_0, x_0^3). Since the line passes through (3, 4), it must be of the form y= \frac{1}{3x_0^2}(x- 3)+ 4. That line must also pass through (x_0, x_0^3) so we must have x_0^3= \frac{1}{3x_0^2}(x_0- 3)+ 4

    Yet another way- minimize the square of the distance (x- 3)^2+ (y- 4)^2 subject to the constraint y- x^3= 0 using Lagrange Multipliers.
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  4. #4
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    We just need to change the polarity of the normal slope as it is negative-going.

    y-4=m(x-3)

    x^3-4=-\frac{1}{3x^2}(x-3)

    x^3=-\frac{1}{3x^2}(x-3)+4

    3x^5=-(x-3)+12x^2

    3x^5-12x^2+x-3=0

    If we differentiate the square of the distance (no need to take the square root)

    d^2=(x-3)^2+(x^3-4)^2

    \frac{d}{dx}d^2=2(x-3)+2(x^3-4)3x^2=2x-6+6x^5-24x^2

    setting the derivative to zero...

    6x^5-24x^2+2x-6=0

    3x^5-12x^2+x-3=0
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