Area of a Half-Mowed YardDate: 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/ |
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