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Math Help - Quadratic optimisation

  1. #1
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    Quadratic optimisation

    A manufacturer finds that the profit P from assembling x bicycles per day is given by

    P(x) = -x^2 + 50x - 200

    What is the maximum profit?

    I figured to find the max profit i should find the turning point of of the parabola which illustrates this equation.

    What I got was (25, 1675)

    So I figured that it would be $1675

    HOWEVER, the answer says 425 dollars!

    Could you tell me how to do this?
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  2. #2
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    Quote Originally Posted by juliak View Post
    A manufacturer finds that the profit P from assembling x bicycles per day is given by

    P(x) = -x^2 + 50x - 200

    What is the maximum profit?

    I figured to find the max profit i should find the turning point of of the parabola which illustrates this equation.

    What I got was (25, 1675)

    So I figured that it would be $1675

    HOWEVER, the answer says 425 dollars!

    Could you tell me how to do this?

    i think
    Maximum profit P(x)= -x^2+50x-200 at maxima x=25 will be
    P(25) = -(25)^2+50*25-200
    = -625+1250-200
    = 425
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  3. #3
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    Why is the negative sign not squared to make term a positive?
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  4. #4
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    Quote Originally Posted by juliak View Post
    Why is the negative sign not squared to make term a positive?
    -x^2 does not mean (-x)^2. It means -(x^2).
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  5. #5
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    Quote Originally Posted by ramiee2010 View Post
    i think
    Maximum profit P(x)= -x^2+50x-200 at maxima x=25 will be
    P(25) = -(25)^2+50*25-200
    = -625+1250-200
    = 425
    Do you think you might tell us why you think that.

    CB
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  6. #6
    Senior Member pacman's Avatar
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    Mr Fantastic just want to be helpful, things that might be clear to us, but not to newbies.
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  7. #7
    Senior Member pacman's Avatar
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    P(x) = y = -x^2 + 50x - 200,

    x^2 - 50x +200 = -y

    completing the square,

    x^2 - 50x +(-50/2)^2 = -y +(-50/2)^2 - 200

    (x - 25)^2 = -(y - 425)

    (x,y) --> (25, 425), the value of x and y are the maximum that can be obtained, its the apex of the inverted parabola

    x = units,

    y = P(x) = $425.00, profit in dollars

    the graph below confirms the answer.
    Attached Thumbnails Attached Thumbnails Quadratic optimisation-yt.gif  
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