anonymous
  • anonymous
A rancher wants to fence in an area of 2693400 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?
Mathematics
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anonymous
  • anonymous
A rancher wants to fence in an area of 2693400 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?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
hmm.. the "fence down the middle" confuses me.... but here's what I think.. The minimum perimeter is a square... so the length = sqrt (2693400) = 1641.158128 There are a total of 5 equal lengths of the fence due to the middle fence... so the total fence = 1641.158128 x 5 = 8205.790638 Just round it to whatever your teacher needs...
anonymous
  • anonymous
You have to take into account the fence down the middle.
anonymous
  • anonymous
yeah, that's what confuses me...

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anonymous
  • anonymous
is the fence down the middle just the same as one of the width of the rectangular fence
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Set up an equation for the area of the field. Set that equal to 2693400. Set up the equation for the perimeter of the fence. Use the equation of the area to make the perimeter equation have only one variable instead of 2. Use the derivative to find the local mins of the perimeter function.
anonymous
  • anonymous
ok thanks i will do that if i get stuck i will ask for more help :) but that seems to clarify things
anonymous
  • anonymous
If you get stuck, post here what equations you're using for area and perimeter.
anonymous
  • anonymous
A=LW; P=2L+2W I get P=2L+2(2693400/L)
anonymous
  • anonymous
That perimeter is not quite right because you need to take into account the fence down the middle. So P = 2L + 3W
anonymous
  • anonymous
ok yea i don't know why i am leaving it out lol
anonymous
  • anonymous
so my P'=2+8050200/L^2
anonymous
  • anonymous
Close, but \[ \frac{d}{dx} (\frac{1}{x}) = \frac{-1}{x^2}\] So you've got a sign problem in P'
anonymous
  • anonymous
yea i saw it so i get 2/8050200=L^2 after i fix sign error then i get sqrt(2/8050200)=L
anonymous
  • anonymous
uh.. I think you're playing a bit fast/loose here.. \[P' = 2-\frac{8050200}{L^2}\] \[P'=0 \rightarrow \frac{8050200}{L^2} = 2\] \[\rightarrow \frac{8050200}{2} = L^2 \]
anonymous
  • anonymous
You have it the other way around.
anonymous
  • anonymous
yea cause its like saying 8050200(1/L^2)
anonymous
  • anonymous
ok so will L be my answer..
anonymous
  • anonymous
I am getting a wrong answer...
anonymous
  • anonymous
The L you find by setting P' to 0 will be the length at which your perimeter function will be at its minimum value. You then need to plug in this L to your normal perimeter function to find how much fence the farmer needs.
anonymous
  • anonymous
my math is wrong it was 8080200 not 8050200
anonymous
  • anonymous
I got the right answer...thanks for your help sorry i was just making minor errors but you helped me see them
anonymous
  • anonymous
could you help me with another If 3072 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
anonymous
  • anonymous
Same type deal. Set up equation for surface area, and volume. Eliminate a variable, take the derivative of the volume function and find a local max (where V' = 0). Plug in that dimension back into V to find the max Volume
anonymous
  • anonymous
err 2 equations, one for surface area, one for volume.
anonymous
  • anonymous
is it surface area because we know the amount of material that we use
anonymous
  • anonymous
it's surface area because they give us the material yes (in square centimeters)
anonymous
  • anonymous
V=w^2*H and my surface area would be of a cube?
anonymous
  • anonymous
Yes with the top missing.
anonymous
  • anonymous
so its just five sides SA=5x^2 or is it SA=2w^2+4wh
anonymous
  • anonymous
That last one. The length and width is the same because the bottom is square, but it doesn't stipulate that the sides are square. Although it should only be w^2 not 2w^2 since the top is open.
anonymous
  • anonymous
ok
anonymous
  • anonymous
i am kinda lost with the derivative...
anonymous
  • anonymous
What's the equation you have for Volume?
anonymous
  • anonymous
V=w^2(3072-w^2/4w)
anonymous
  • anonymous
Shouldn't that be \[ V = \frac{w^2(3072-w^2)}{4w}\]
anonymous
  • anonymous
could you eliminate the w from the bottom and one w from w^2
anonymous
  • anonymous
yes.
anonymous
  • anonymous
\[ V=\frac{1}{4} * [3072w - w^3] \]
anonymous
  • anonymous
V'=1/4[3072-3w^2]
anonymous
  • anonymous
yep.
anonymous
  • anonymous
Thanks again i got it again... i am starting to understand optimization
anonymous
  • anonymous
A rectangle is inscribed with its base on the x -axis and its upper corners on the parabola y=2-x^2 What are the dimensions of such a rectangle with the greatest possible area? width= height= i figured out height was 4/3 cause i thought width was \[\sqrt{2/3}\] and plugged it in y=2-x^2

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