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Thread: Calculus - Lightbulbs

  1. #1
    Junior Member Fnus's Avatar
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    Calculus - Lightbulbs

    The question is on the picture (:
    Attached Thumbnails Attached Thumbnails Calculus - Lightbulbs-unavngivet.jpg  
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  2. #2
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    Hello, Fnus!

    We are told: .$\displaystyle I \:=\:\frac{kP}{d^2}$

    Let $\displaystyle P$ be a point between the two sources.
    . . Let $\displaystyle x$ = distance from "40".
    . . Then $\displaystyle 6-x$ = distance from "5".

    The total intensity is: .$\displaystyle I \;=\;\frac{40k}{x^2} + \frac{5k}{(6-x)^2} \;=\;40kx^{-2} + 5k(6-x)^{-2}$

    Then we have: .$\displaystyle I'\;=\;\frac{-80k}{x^3} + \frac{10k}{(6-x)^3} \;=\;0$

    Multiply by $\displaystyle \frac{x^3(6-x)^3}{10k}\!:\;\;-8(6-x)^3 + x^3 \;=\;0$

    This simplifies to: .$\displaystyle x^3-16x^2+96x-192 \;=\;0$

    . . which factors: .$\displaystyle (x-4)(x^2-12x+48) \;=\;0$

    . . and has the real root: .$\displaystyle \boxed{x \:=\:4}$


    Since $\displaystyle I'' \;=\;\frac{240k}{x^4} + \frac{30k}{(6-x)^4} $ is always positive,
    . . the function is always concave up.

    Therefore, $\displaystyle x = 4$ provides minimum $\displaystyle I.$

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