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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
avatar+544 
+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

 

 

winkwinkwink

lynx7  Apr 13, 2018
edited by lynx7  Apr 13, 2018
 #2
avatar+86931 
+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   }

 

 

cool cool cool

CPhill  Apr 13, 2018

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