algebra
posted by Anonymous on .
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

if w is the length of the pen, then let the side parallel to the barn be l
l + 2w = 450
l = 4502w
area = lw = (4502w)w
= 2w^2 + 450w 
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.