Suppose Sarah wants to build and enclose a rectangular fence in her backyard for planting fruits with only 3,250 feet of wire fencing. (Hint: all sides of the fence must be taken into consideration)

Question

a)     Find the area of the rectangular fence as a quadratic function in terms of length.

b)     Find the length of the rectangular fence. 

Note: do not solve this statement by using any methods from Calculus.

c)     Find the maximum area.

Note: do not solve this statement by using any methods from Calculus.

Answer 1

a) we have the perimeter of our rectangle as P = 2 x length + 2 x width

We can write Width as 2W(L) = P - 2L

Area is width x length = 1/2[2W] x L = 1/2[P-2L] x L this is the quadratic in L

We can write this in Vertex form as A(L) = -(L-P/4)² + (P/4)²

We know by inspection that if the long length of our fence is the perimeter divided in two that the area our fence encloses is zero square feet. The Vertex form of this quadratic in L agrees with this prediction.

The area of our enclosure can be zero or a positive value so the range of our length has a maximum value of P/2, it must be less than P/2 to yield an area greater than zero. The length also has a minimum value of zero, again the area is zero for this length. 0<L<P/2

c) the maximum area is the maximum of the vertex equation which is given when L=P/4 since the constant fro the quadratic is negative. The maximum area that can be enclosed is (P/4)² this is the same as the area of the square enclosed by this same perimeter.

one can use foil on the Vertex form to see the match to the standard form

Calculus has not been used to realize that A(L) as expressed in vertex form has a maximum and that this parabola in L turns down with vertex (L/4,( L/4)²)