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Math Help - Quadratic word problems.

  1. #1
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    Quadratic word problems.

    I'm having a really difficult time trying to solve these problems. Could someone please explain them to me?


    1. The total annual enrollment (in millions) in U.S. elementary schools for the years 1975-1996 is given by the model

    E = -0.058x^2 - 1.162x + 50.604

    Where x=0 corresponds to 1975, x=1 corresponds to 1976, and so on. For this period, when was ennrollment the lowest? To the nearest tenth of a million, what was the enrollment?

    2. A parabolic arch has an equation of x^2 + 20y - 400 = 0, where x is measured in feet. Find the maximum height of the arch.

    Thanks in advance.
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  2. #2
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    1. The total annual enrollment (in millions) in U.S. elementary schools for the years 1975-1996 is given by the model

    E = -0.058x^2 -1.162x +50.604 ----------(i)

    Where x=0 corresponds to 1975, x=1 corresponds to 1976, and so on. For this period, when was ennrollment the lowest? To the nearest tenth of a million, what was the enrollment?


    If equation (i) is correct, then that is a vertical parabola that opens downward---because of the negative coeefficient of the x^2. So its vertex is a maximum point.
    Its vertex is at
    x = -b/2a = -(-1.162)/[2(-0.058)] = -10.017
    That means the vertex is at year 1975 -10 = 1965. Which means the enrollment was highest in 1965.
    The parabola further means after 1965, the enrollment decreases every year "parabolically"---not linearly.

    Therefore, even without solving, the lowest enrollment in the interval [1975,1996] is in 1996, where the enrollment then is:
    1996 -1975 = 21 years
    E = -0.058(21^2) -1.162(21) +50.604
    E = 0.624
    To the nearest tenth,
    E = 0.6 millions
    E = 600,000 students only

    Is that realistic?
    In 1996, there were only about 600,000 students enrolled in all US elementary schools?

    The posted model is correct?

    [In 1975, E = -0.058(0^2) -1.162(0) +50.604 = 50.604 millions elementary students enrolled.]

    ---------------------------------------------

    2. A parabolic arch has an equation of
    x^2 +20y -400 = 0
    where x is measured in feet. Find the maximum height of the arch.


    So, rearranging,
    20y = -x^2 +400
    y = -(1/20)x^2 +20 -----------(ii)

    The parabola is vertical, opening downward, so its vertex is a maximum.

    The vertex is at
    x = -b/(2a) = -0/[2(-1/20)] = 0

    So, the y at x=0 is
    y = -(1/20)(0^2) +20 = 20.

    Therefore, the maximum height of the arch is 20 ft. -------------answer.
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  3. #3
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    Oops! Major error...

    The equation should read:
    E = 0.058x^2 -1.162x +50.604

    Thank you for your help. It is greatly appreciated!
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  4. #4
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    Quote Originally Posted by mi986 View Post
    The equation should read:
    E = 0.058x^2 -1.162x +50.604

    Thank you for your help. It is greatly appreciated!
    Aha, just as I thought it should have been.

    Now it is a vertical parabola that opens upward so its vertex is a minimum.

    The x of the vertex is at
    x = -b/2a = -(-1.162)/[2(0.058)] = 10

    So in the year (1975 +10) = 1985, the enrollment was the lowest. It is only
    E = 0.058(10^2) -1.162(10) +50.604 = 44.8 million enrollees.

    [In 1975, it was 0.058(0^2) -1.162(0) +50.604 = 50.6 millions.
    In 1996, 0.058(21^2) -1.162(21) +50.604 = 51.8 millions.]
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  5. #5
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    Cool, thanks for your help. I'm finally starting to get the hang of these word problems. :-p
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