A veterinarian has 500 feet of fencing suitable for making animal runs. He wants to use it all to fence in a rectangular area with three interior fences running from end to end of the rectangular area so that four parallel rectangular runs are made for his larger animals. If he builds so that the overall area is maximized, what will be the longer dimension of each run?
2007-01-28 02:54:16 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Let the two sides be 'x' and 'y' where x > y.
The perimeter: 500 = 2x + 5y [3 inner fence and 1 outer fence].
Area:
A = xy
= x(100 - 0.4x)
= 100x - 0.4x^2
= (2/5)(250x - x^2)
= (2/5)[125^2 - (125^2 - 2*125x + x^2)]
= (2/5)[125^2 - (125 - x)^2]
A is maximum when (125 - x)^2 is minimum i.e. (125 - x)^2 = 0 i.e. x = 125
Thus the longer side is 125 ft and shorter side is 50 ft.
2007-01-28 03:14:06 · answer #1 · answered by psbhowmick 6 · 3⤊ 0⤋
I believe the maximum area of a rectangle is obtained when all of the sides are equal (really a square). If there are 3 interior fences running end to end there will be 7 equal length fences altogether. So the longer dimension of each run will be 500/7 or 71.43 feet.
My assumption is wrong ! Did not consider effect of interior fences on the perimeter.
2007-01-28 11:20:37 · answer #2 · answered by indyacom 3 · 0⤊ 0⤋
5L + 2W =500 --> W = 1/2(500 -5L)
A= LW = L(1/2(500-5L))
dA/dL = (500-5L)/2
Set = 0 to get max
500-5L=0
L = 50 feet
W = (500 -250)/2 = 125 = longer dimension
2007-01-28 11:13:32 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋