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Math Help - Related Rates #2

  1. #1
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    Related Rates #2

    The illumination at a point is inversely proportional to the square of the distance of the point from the light source and directly proportional to the intensity of the light source. If two light sources are 20 feet apart and their intensities are 40 and 30 respectively, at what point between them will the sum of their illuminations be a minimum?

    Let x be the distance from the brighter source at which the sum of the illuminations is a minimum. Then x = how many feet?
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  2. #2
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    If we define I to be the total illumination, then

    I=i_1+i_2,

    where

    \begin{aligned}<br />
i_1&=\frac{40k}{x^2}\\<br />
i_2&=\frac{30k}{(20-x)^2}.<br />
\end{aligned}

    Because I approaches \infty at x=0,20 and is differentiable everywhere else, the minimum will occur at a point at which I'=0.
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  3. #3
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    So, I tried to take the derivative...

    But I got stuck...

    it looks like:

    The Derivative of: 40k/x^2 = (0 * x^2 - 2x * 40) / x^4

    The Derivative of: 30k/(20-x)^2 = (0 - 30*2(20-x)*-1/(20-x)^4

    But these are giving the wrong answer. Where am I going wrong?
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  4. #4
    jsl
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    Simplify your derivatives, add them together, and set them equal to 0. (-80/x^3)+(60/(20-x)^3)=0.

    60/(20-x))^3=80/x^3

    Then invert the equation
    [(20-x)^3]/60=[x^3]/80

    Then solve for x.
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