A rancher wants to fence in an area of 1500000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?
2007-06-30 18:20:29 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I think the shortest possible length will be achieved if the field is a square(which is also a rectangle). So, the length of one side would be:
√1500000 = 1224.74
Therefore, the length of fencing required would be:
1224.74*5 [4 sides + the one in the middle]
=30618.62 ft
//
2007-06-30 18:32:24 · answer #1 · answered by Popo B 3 · 0⤊ 2⤋
6000 feet of fence.
Let L be the length of fence, W the width of the area, D is the depth. The width is divided in half by the middle part of the fence.
W = 1,500,000 / D
L = 2 W + 3 D = 3,000,000 / D + 3 D
dL = 3 - 3,000,000 / D^2
The length of fence is at a minimum when dL = 0.
3 - 3,000,000 / D^2 = 0Â Â Â Solve for D
3 D^2 = 3,000,000
D = 1000
W = 1500, L = 2W + 3D = 6000
2007-07-01 01:37:51 · answer #2 · answered by Engineer-Poet 7 · 0⤊ 0⤋
With any given area the shortest perimeter is given by a square, so:
P = Squ root (1 500 000) x4
=1224.74 x4
= 4898.98
Then add on another side length for the fence down the centre:
TP = 4898.98 + 1224.74
=6123.72 feet of fence
2007-07-01 01:27:42 · answer #3 · answered by Anonymous · 0⤊ 1⤋
x = 3L + 2W
A = L*W = 1,500,000
L = 1,500,000/W
x = 3*1,500,000/W + 2W
x = 4,500,000/W + 2W
dx/dW = - 4,500,000/W^2 + 2 = 0
2W^2 = 4,500,000
W^2 = 2,250,000
W = 1,500 ft
L = 1,500,000/1,500 = 1,000 ft
x = 3*1,000 + 2*1,500
x = 3,000 + 3,000
x = 6,000 ft
Check:
. . W . . . . . L . . . . . . . . . X
1499 1000.667111 6000.001334
1500 1000.000000 6000.000000
1501 999.3337775 6000.001332
2007-07-01 01:43:46 · answer #4 · answered by Helmut 7 · 0⤊ 0⤋
YOU GOTTA DO CALCULUS TEEE HEEEE. I could do this but if you look in your book theres an example exactly like this.
2007-07-01 01:23:34 · answer #5 · answered by EMERGENCY 2 · 0⤊ 1⤋