March 12, 2009
What is the largest possible total area of the four pens?
Can you answer busy_b901's question about pens?:
Consider the following problem: A farmer with 710 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?
Filed under More Pen Answers by Guest Author
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a - lentgh of one side, b - lentgth of other side
fencing used to provide 2a+5b, or 2b+5a (it's not important)
So 2a+5b=710 a=(710-5b)/2
Area of rectangle S=ab=(710b-5b^2)/2
To find largest area calculate S'=0
(710-10b)/2=0
b=71 ft a=177.5 ft
S=12602.5 ft^2
some books consider a square a rectangle…
to maximize an area w/ a given perimeter for a 4 side polygon, it must b a square… proven using differential calculus…
draw a square and divide it into 4 equal parts by drawing a horizontal and vertical line passing at the center of the square…
the length the 6 lines are equal (property of a square) therefore
6(x) = 710
x = 118.333
the L & W of the pens is equal to x/2
L=W= 118.333/2 = 59.17
area = 59.17 * 59.17 = 3500.69 sq.ft