+1107 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?
+223 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
+129899 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 }
