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Math Help - [SOLVED] Completing the Square Problem?

  1. #1
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    [SOLVED] Completing the Square Problem?

    "In an electrical circuit, the voltage, V volts, as a function of time, t minutes, is given by V = 12 - 9t + 2t^2. Determine the greatest and least values of voltage during the first 5 min. When do these values occur?"

    The first thing I did was substitute t = 5.

    <br />
V = 12 - 9t + 2t^2
    V = 12 - 9(5) + 2(5)^2
    V = 12 - 45 + 50
    V = 17

    Then I go check the answer page to see if I was right. The answer is "Least value: 1.875 V at 2.25 min; greatest value: 17 V at 5 min". How was the first part of the answer obtained? Is completing the square involved? If it is, I already have that done.

    <br />
V = 12 - 9t + 2t^2
    V - 12 - 2 = 2(t^2 - \frac{9}{2}t)
    V - 12 - 2(\frac{9}{4}) = 2(t^2 - \frac{9}{2}t + \frac{9}{4})
    V - 12 - \frac{18}{4} = 2(t - \frac{3}{2})^2
    V - \frac{48}{4} - \frac{18}{4} = 2(t - \frac{3}{2})^2
    V - \frac{66}{4} = 2(t - \frac{3}{2})^2
    V = 2(t - \frac{3}{2})^2 + \frac{33}{2}

    I'm guessing that this wasn't needed, as this seems to have no relevance to the aforementioned answer. What needs to be done here? Help would be appreciated.
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  2. #2
    MHF Contributor
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    You made a couple of errors when completing the square, and this is actually a perfectly valid method of solving the problem.

    V - 12 = 2(t^2 - \frac{9}{2}t)

    V - 12 + 2 \cdot \frac{81}{16} = 2(t^2 - \frac{9}{2}t + \frac{81}{16})

    V - 12 + \frac{81}{8} = 2(t - \frac{9}{4})^2

    V - \frac{96}{8} + \frac{81}{8} = 2(t - \frac{9}{4})^2

    V - \frac{15}{8} = 2(t - \frac{9}{4})^2

    V = 2(t - \frac{9}{4})^2 + \frac{15}{8}

    and the minimum value is obtained when t = \frac{9}{4},
    from which the desired conclusion follows.
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  3. #3
    Member
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    Ah, got it now. I forgot to square 9/4. Thanks very much for the help.
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