A rectangular play area is to be fenced off in a person's yard and is to contain 108 yd^2. The next-door neighbor agrees to pay half the cost of the fence on the side of the play area that lies along the property line. What dimensions will minimize the cost of the fence?
Calculus - Steve, Tuesday, October 16, 2012 at 3:26pm
cost is proportional to perimeter, so we want to minimize the perimeter enclosing 108 yd^2. If x is the length of the fence on the property line, 108/x is the other dimension.
p(x) = 2(x + 108/x)
dp/dx = 2(1 - 108/x^2)
dp/dx=0 when x^2 = 108
x = 6√3 = 10.39
the pool is thus
6√3 by 6√3
as we know, a square has maximum area for a given perimeter; conversely a square has minimum perimeter for a given area.
Now, if we want to minimize the cost to the pool owner, that would involve subtracting half the cost of the side on the property line:
p(x) = 2(x + 108/x) - x/2
dp/dx = 2(1 - 108/x^2) - 1/2
dp/dx=0 when x = 12
now the fence would be 12 by 9
The longer border length makes the neighbor pay more.