Posted by **Anonymous** on Thursday, December 8, 2011 at 6:36pm.

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 -
**Reiny**, Thursday, December 8, 2011 at 6:43pm
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 -
**tchrwill**, Saturday, December 10, 2011 at 4:01pm
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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