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Quadratic equations

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Building a rectangular room with fixed perimeter of 280ft
what dimentions would yield the maximum area?
What dimentions would yeild the minimum area?

  • Quadratic equations - ,

    A rectangle of largest area is a square.
    So divide your perimeter by 4.

    To build a room of minimum area is a rather silly question.

  • Quadratic equations - ,

    I apologize for my silly question. I followed your advice, revised my question and found my answer. Thank you.

  • Quadratic equations - ,

    x = length
    y = width
    A = x y
    x+y=140
    A = x(140-x)
    A = 140 x - x^2
    look at shape of parabola by completing the square
    x^2 -140 x = -A
    x^2 - 140 x + 4900 = -A + 4900
    (x-70)^2 = -A + 4900
    the vertex will be at x = 70 and A = 4900 and the -A means that the parabola opens down (sheds water) so that vertex is the maximum where x = y =70 as Reiny told you.
    since a negative x or y does not make much sense, the area may not get any smaller than zero when x --> 0 or y -->0 0

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