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Area of a Half-Mowed Yard

Date: 08/21/2003 at 15:33:14
From: Diane Evans
Subject: Area of a half mowed yard

There is a square/rectangular yard, size unknown. You have mowed a 10
foot swath around the perimeter. You are now half done. How big is
the yard?


Date: 08/21/2003 at 18:55:09
From: Doctor Douglas
Subject: Re: Area of a half mowed yard

Hi Diane,

Thanks for writing to the Math Forum.

Here's a diagram of the situation:

   +---+-------------------+---+
   |   |                   |   | 10     The unmowed area is
   +---A-------------------B---+        rectangle ABCD, which
   |   |                   |   |        has area equal to xy.
   |   |                   |   | y
   |   |                   |   |  
   +---D-------------------C---+
   |   |                   |   | 10
   +---+-------------------+---+
     10          x          10

The total area of the entire yard is (x + 20)*(y + 20). The area of 
the unmowed portion is xy.  For this to be half of the total, we must 
have

  2xy = (x + 20)*(y + 20) = xy + 20x + 20y + 400

   xy = 20x + 20y + 400

This allows us to express y in terms of x:

  (x-20)*y = 20x + 400

         y = 20*(x + 20)/(x - 20)

So the area of the unmowed yard is

  Area = xy = (20x)(x + 20)/(x - 20),

where we are free to pick whatever values of x we like, consistent 
with the fact that x and y must both be positive. In the above formula 
for y, the factor of x-20 in the denominator must be positive (because 
the numerator is guaranteed to be positive), so this forces x > 20.  
Similarly, y > 20.  As one of these two values becomes close to 20, 
the other one must become very large. For example, if x = 21, then 
y = 20(21+20)/(21-20) = 20*42 = 840 ft.

Because we have the freedom to choose x and y (subject to these
restrictions), the area A/2 of the unmowed portion can take on many 
possible values.  For example,

  x=30, y=(20)(30+20)/(30-20)=100,   A/2 = 30*100   = 3000 sq ft
  x=40, y=(20)(40+20)/(40-20)=60,    A/2 = 40*60    = 2400 sq ft
  x=50, y=(20)(50+20)/(50-20)=46.67, A/2 = 50*46.67 = 2333 sq ft
  x=60, y=40,                        A/2 = 60*40    = 2400 sq ft
  x=70, y=(20)(20+70)/(70-20)=36,    A/2 = 70*36    = 2520 sq ft

And of course the area A of the whole yard is twice the last column.
Now, the minimum area of the yard occurs when x is exactly equal to y.  
In this special case, the above equations lead to

   x = 20*(x + 20)/(x - 20), or x^2 - 20x = 20x + 400, 
   x^2 - 40x - 400 = 0

This quadratic equation can be solved for x using the quadratic
formula. One of the roots is

   x = {-(-40) - sqrt[40^2 - 4*(-400)(1)]}/2
     = {40 - sqrt[1600 + 1600]}/2
     = 20 - sqrt[800]
     = -8.2843

but this root violates the requirement that x > 20, and is therefore
rejected.  The positive root is

   x = {-(-40) + sqrt[40^2 - 4*(-400)(1)]}/2
     = {40 + sqrt[1600 + 1600]}/2
     = 20 + sqrt[800]
     = 48.2843

For this special case, the area of the mowed portion is

   x^2  = (20 + sqrt[800])^2 = 400 + 800 + 2*20*sqrt(800)
        = 1200 + 40*(28.2843)
        = 2331.4 sq ft

The area of the whole yard is of course twice this.

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Triangles and Other Polygons
Middle School Triangles and Other Polygons
Middle School Word Problems

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