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Math Help - Rate at maximum

  1. #1
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    Rate at maximum

    A question asks at what time of day is the rate of power consumption the greatest. t>=0 with noon coresponding t=0

    The rate of consumption I derived, which is correct is 450cos(\frac{pi*t}{12})+700

    I know I have to set the Rate equal to zero to get the maximum
    450cos(\frac{pi*t}{12})+700=0
    450cos(\frac{pi*t}{12})=-700
    cos(\frac{pi*t}{12})=\frac{-700}{450}
    arccos(cos(\frac{pi*t}{12}))=arccos(\frac{-700}{450})
    \frac{pi*t}{12}=arccos(\frac{-700}{450})

    I stopped right here because my calculator didnt like the Arccos(700/450). Can I get a point in the right direction?
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  2. #2
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    Re: Rate at maximum

    What was the original function?
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  3. #3
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    Re: Rate at maximum

    E(t) = 700t + \frac{5400}{pi}sin(\frac{pi*t}{12})
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