Quadratic Equations 2

Discussion in 'Community Discussion' started by waloshin, Apr 2, 2009.

  1. waloshin macrumors 68040

    Oct 9, 2008
    It seems my recent post has been moved to the wasteland already: http://forums.macrumors.com/showthread.php?t=679777&highlight=quadratic+equation

    So in a follow up am i doing this right?

    2) A farmer wishes to fence off a rectangular plot of land along the side of a river. Since the river is very deep, he does not need to build a fence along the river. If he has 600 m of fencing, what dimensions should he use in order to maximize the area of the field What is the maximum area.

    x + y = 600
    y = 600 - x

    a= x(600-x)
    a= 600x-x^2

    a= -1
    b= 600
    c= 0






    vertex ( 300, 90000)
  2. swiftaw macrumors 603


    Jan 31, 2005
    Omaha, NE, USA
    Since the river is one side of the rectangle, the fence has to cover the other 3 sides.

    So x+2y = 600
  3. darklyt macrumors regular

    Jun 5, 2007
    You're trying to maximize area? A regular polygon maximizes the amount of area bounded by a given perimeter for any number of sides (and a circle maximizes the bounded area for a given perimeter). Since you're looking for a rectangle, the solution is going to be a square. You only have to give three sides of this square with the fourth side the river, so you need three equal lengths of fence. Therefore, we have 600/3 = 200. Thus largest area is 40,000 square feet.
  4. swiftaw macrumors 603


    Jan 31, 2005
    Omaha, NE, USA
    Not true.

    Let the rectangle be x by y. Then, since the fence has to cover 3 sides, we have that x+y+y = 600, or x = 600-2y.

    Thus the area of the rectangle = x*y = (600-2y)*y = 600y - 2y^2

    Thus dA/dy = 600 - 4y

    Solving for zero gives us y = 150, thus x = 600-2*150 = 300

    Thus the max area is 300*150 = 45000
  5. darklyt macrumors regular

    Jun 5, 2007
    Nice catch. I should have spent more than a few seconds looking at it lol.

    If it were four sides involved, then it would have been necessary to have a square since if P=total perimeter, P/2 * P/2 = (P^2)/4, while (P/2 - x)(P/2 + x) = (P^2)/4 - x^2 < (P^2)/4 for x>0. It is interesting that the solution you found is half a square with sides of length 300. Also interesting is that if it were at a plot of land that lied next to a river that (magically) formed a right angle next to this land, then we would have the solution is a square.

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