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I tried sketching it and possibly couldnt come up with the ewuation. Need help .... The question is

A rancher wants to fence in an area of 2,400,000 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?

2007-11-17 07:03:30 · 4 answers · asked by Amer 1 in Science & Mathematics Mathematics

4 answers

Let L represent the length of the field.
Let W represent the width of the field.
Let F represent the length of fence needed.
L*W=2400000
2L+3W=F Both length sides+both width sides+ fence across the width.
Let's look at the first equation again:
L*W=2400000
W=(2400000/L)
Now we can substitute (2400000/L) for W in the second equation.
2L+3(2400000/L)=F
2L+7200000/L=F
Now we have a function. We can input any length, and we'll know length needed. But how can we find the length needed so that F is as small as it can be? If you know derivatives, there's an exact way. The next easiest way to do it is to graph.

Manner 1: Derivative
Set the derivative to zero...when the length stops going down and starts coming back up, it'll be at it's lowest point.

2L+7200000L^-1=F
F'=2-7200000L^-2
0=2-7200000L^-2
7200000/(L^2)=2
7200000=2L^2
3600000=L^2
L=1897.366
Now, we know that L=2400000/W
so 1897.366=2400000/W
W=2400000/1897.366
W=1264.91106 feet.
Now,
2L+3W=2(1897.366)+3(1264.91106)
Minimum fencing needed: 7589.465

Method 2: Graphing. Graph y=2x+7200000/x (y=fence size, x=length of side)
to to the "bottom" or the graph. Zoom in really far. As we zoom in closer and closer, we find that x=1897.366 and y=7589.465.
Thus, the minimum amount of fencing needed is 7589.465 feet.

2007-11-17 07:24:05 · answer #1 · answered by SaintPretz59 4 · 3 0

square root of 2,400,000 square feet = 1,549..19 feet.

the rectangular fiel has 4 sides plus the division = 5 sides.
then : 1,549.19 * 5 = 7,745 .95 feet

answer = 7,745.95 feet aprox.

2007-11-17 07:39:55 · answer #2 · answered by Eduardo (lalo) Leal 2 · 0 2

area = L*W = 2,400,000
fence = 2L + 3W
minimize fence
L = 2,400,000/W
2(2,400,000/W) + 3W = fence
4,800,000W^(-1) +3W = fence
-4,800,000W^-2 + 3 = f'(fence) = 0
-4,800,000 = -3W^2
1,600,000 = W^2
1265 = W (rounded)
1897 = L (rounded)
3*W + 2*L = 7589' fence

2007-11-17 07:35:44 · answer #3 · answered by Steve A 7 · 0 0

square root of 2,400,000 times five

2007-11-17 07:11:23 · answer #4 · answered by Anonymous · 0 0

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