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 = 450-2w
area = lw = (450-2w)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.
To determine the area of the pen, we need to multiply the length and width of the rectangular pen. Since one side of the pen is against the barn, the length of the pen perpendicular to the barn is represented by w.
Let's assume the width of the pen is x, which means the other two sides will also be x.
So, the equation to model the area of the pen in terms of w will be:
Area = w * x * x
or
Area = w * x^2