Simon has 160 meters of fencing to build a rectangular garden.

The garden's area (in square meters) as a function of the garden's width \(w\) (in meters) is modeled by

\(A(w)=−w(w−80)\)

**What is the maximum area possible?**

MathPeacock Apr 13, 2018

#1**+2 **

The perimeter of the rectangle:

2L + 2 W = 160

A = WL

L + W = 80

A = W ( 80 -w) = -W 2 + 80 W

The quadratic: - W2 +80 W

f( W)= - W2 + 80 W has maximum at Vertex:

W = -80/-2 = 40 yard

L = 80 - 40 = 40

Maximum area obtained by square with side 40 Yard

The area : 40 * 40 = 1600 sq yds^2

lynx7 Apr 13, 2018

#2**+2 **

A(w) = -w (w - 80) = -w^2 + 80w

The width that maximizes the area can be found by :

-80 / [2* -1) = -80/-2 = 40

And the maximum area is

-40 (40 - 80) = -40 * -40 = 1600 m^2

[ This isn't surprising....a square always maximizes the area for a given perimeter....so....really....all we need to do is to divide the perimeter by 4 and square this result....so we would get (160 /4)^2 =

(40)^2 = 1600 m ^2 }

CPhill Apr 13, 2018