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Ian wants to build a rectangular pen for his animals. One side of the pen will be against the barn; the other three sides will be enclosed with wire fencing. If Ian has 450 feet of fencing, what dimensions would maximize the area of the pen?

a) Let w be the length of the pen perpendicular to the barn. Write an equation to model the area of the pen in terms of w

  • algebra -

    if w is the length of the pen, then let the side parallel to the barn be l
    l + 2w = 450
    l = 450-2w

    area = lw = (450-2w)w
    = -2w^2 + 450w

  • algebra -

    Considering all rectangles with a given perimeter, one side being provided by a straight given boundry, which one encloses the largest area?

    Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P - 2x) = Px - 2x^2.

    Taking the first derivitive and setting equal to zero, dA/dx = P - 4x = 0, x becomes P/4.

    With x = P/4, we end up with a rectangle with side ratio of 2:1.
    .....The short side is P/4.The traditional calculus approach would be as follows.

    .....The long side is (P - 2(P/4)) = P/2.

    Therefore, it can be unequivicably stated that of all possible rectangles with a given perimeter, one side being a given external boundry, the rectangle with side ratio of 2:1 encloses the maximum area.

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