A rancher wants to fence in an area of 2500000 square feet in a rectangular field...?

Question

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?

Answer 1

I'm to tired to do the math behind this, but I can tell you the method of solving it. The fence will sourround the enclosed rectangular field, meaning you are looking for the smallest possible perimeter that can hold in 2,500,000 sq feet. Smallest perimeter will always be the two factors (two numbers that multiply into 2,500,000), that are as close to each other as possible (I think its 1250 and 2000, but that may be wrong). Then add two times each of the factors, as a rectangle has two lenghts and two widths, and if the rancher wants to half his property he would add another layer of the fence down along the shortest length, in this case the 1250. If my two numbers, 1250 and 2000, are the correct factors, then the total shortest possible lenght of fence would be (2x1250)+(2x2000)+1250=7750 feet. I hope this helped. (Again, my numbers may be wrong, you may want to make sure there are no factors closer to each other then mine)

Answer 2

A is section P is Perimeter W is Width L is length we are attempting to shrink the fringe, that's the size of fence. P = 3W + 2L A = W*L = 2500000 resolve for L to get L = 2500000/W Plug L into P. P = 3W + 5000000/W Take the by-made of P. P' = 3 - 5000000/(W^2) Set P' = 0 0 = 3 - 5000000/(W^2) 3*(W^2) = 5000000 W^2 = 5000000/3 W = 1290.ninety 9 Plug W back into W*L = 2500000 L = 2500000/1290.ninety 9 = 1936.5 Plug W & L into P P = 3W + 2L = 7745.ninety seven

Answer 3

Related rate problem, you have area but want to minimize perimeter+extra length.
Area = length x width (L x W)
Fence length (F) = 2 x L + 2 x W + L or 3 L + 2 W
So substitute 2.5x10^6/L for W, take dF/dW, set to zero, solve for W (if quadratic, you may have to select the correct root).

Answer 4

i guess it would be

7905.69 feet

length = 5 * area^1/2

smallest fencing will always be a perfect square area as I remember. hmmm