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Math Help - optimization problem

  1. #1
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    optimization problem

    I need help as I'm having trouble with this and can't seem to get anywhere

    Problem:
    The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed 10 ft apart, where should an object be placed on the line between the sources so as to receive the least illumination?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by forkball42 View Post
    I need help as I'm having trouble with this and can't seem to get anywhere

    Problem:
    The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed 10 ft apart, where should an object be placed on the line between the sources so as to receive the least illumination?
    Call the power of the weaker source P and the stronger 3P. I'm putting the weaker source on the origin and the stronger one at x = 10 ft. Put the object at x = d. The total intensity at this point is:
    I = \frac{P}{d^2} + \frac{3P}{(10 - d)^2}

    Now find the minimum of I with respect to the position d. (ie. Take the derivative of I with respect to d, set it equal to 0, and solve for d. Then you need to show that this value of d is, in fact, a minimum point rather than a maximum.)

    Warning: You are going to have to solve an ugly cubic equation. Don't even bother trying to factor it: I'd recommend a numerical solution. For reference, I'm getting d around 4.09 ft.

    -Dan
    Last edited by topsquark; December 2nd 2007 at 11:50 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    Call the power of the weaker source P and the stronger 3P. I'm putting the weaker source on the origin and the stronger one at x = 10 ft. Put the object at x = d. The total intensity at this point is:
    I = \frac{P}{d^2} + \frac{P}{(10 - d)^2}

    I = \frac{P}{d^2} + \frac{3P}{(10 - d)^2}

    RonL
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    I = \frac{P}{d^2} + \frac{3P}{(10 - d)^2}

    RonL
    Humbug!

    Yeah, it was a typo and thanks for pointing it out.

    @ forkball42: For the record, my approximate solution was tabulated using the correct equation.

    -Dan
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