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Math Help - optimization!!

  1. #1
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    optimization!!

    ok ive always had trouble with these sort of problems, if someone can help please, i in a hurry as well thanks

    a grapefruit is tossed straight up with an initial velocity of 50 ft/sec. the grapefruit is 5 feet above the ground when it is released. its height at time t is given by

    y = -16t^2+50t+5

    how high does is go before returning to the ground?
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  2. #2
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    Not sure what math class you're in, so I'll do it the "algebra" way as opposed to the calculus way.

    Your formula is a quadratic, so its graph is a parabola that opens downward (since the leading coefficient, -16, is negative). So its vertex will be its highest point - remember that the vertex is the point where, in this case, the graph stops increasing and starts decreasing. So if we can find the vertex, we're done.

    The x-coordinate for the vertex for a quadratic in the form y =ax^2 + bx + c is always given by:

    x = -\frac{b}{2a}.

    In your case, a = -16, b = 50, so the x-coordinate of the vertex, which is represented by t, the time, in your equation, is \frac{25}{16} (after some reducing). We need to know how HIGH the projectile gets - we now know that it reaches its highest point after \frac{25}{16} seconds. To find its height, plug that time into the equation, giving:

    y = -16(\frac{25}{16})^2 + 50(\frac{25}{16}) + 5.

    I got about 44 seconds.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by mathlete View Post
    ok ive always had trouble with these sort of problems, if someone can help please, i in a hurry as well thanks

    a grapefruit is tossed straight up with an initial velocity of 50 ft/sec. the grapefruit is 5 feet above the ground when it is released. its height at time t is given by

    y = -16t^2+50t+5

    how high does is go before returning to the ground?
    take the derivative to find the max point.

    \frac{dy}{dt}=-32t+50

    setting the derivative equal to zero to find the crit points we get

    0=-32t+50 \iff t =\frac{25}{16}

    This gives us the time of the max height so plugging into the original equation we get

    y=-16\left( \frac{25}{16}\right)^2+50 \cdot \frac{25}{16}+5=\frac{705}{16} \approx 44
    Last edited by TheEmptySet; April 1st 2008 at 10:18 PM. Reason: my fraction was wrong
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by mathlete View Post
    ok ive always had trouble with these sort of problems, if someone can help please, i in a hurry as well thanks

    a grapefruit is tossed straight up with an initial velocity of 50 ft/sec. the grapefruit is 5 feet above the ground when it is released. its height at time t is given by

    y = -16t^2+50t+5

    how high does is go before returning to the ground?
    This graph should form a parabola with the open end facing down. You are interested in the point where it's tangent line is equal to zero, the zenith, the pinnacle, whatever you want to call it (apparently it's vertex, thx mathnasium).

    Since at that point the tangent line is horizontal, it's slope will be zero, so we differentiate the problem, and set it equal to zero (the value we are looking for) and find what time will cause the slope to go to zero.

    y = -16t^2 + 50t + 5
    \frac {dy}{dx} = -32t + 50

    Now this is our equation for the slope. We are interested in where the slope equals zero so set it equal to zero and solve:
    \frac {dy}{dx} = 0 = -32t + 50

    -50 = -32t

    \frac {25}{16} =t

    Now we know the time in which this desired event occurs, but we still need to find the height at this time, so we plug it back into the initial equation and simplify to find the height.
    Last edited by angel.white; April 1st 2008 at 10:36 PM.
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    0=-32t+50 \iff t =\frac{16}{25}
    Solving this equation for t gives t = \frac{25}{16}, not t = \frac{16}{25}.
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