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anonymous

  • 5 years ago

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?

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  1. anonymous
    • 5 years ago
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    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...

  2. anonymous
    • 5 years ago
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    You have to take into account the fence down the middle.

  3. anonymous
    • 5 years ago
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    yeah, that's what confuses me...

  4. anonymous
    • 5 years ago
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    is the fence down the middle just the same as one of the width of the rectangular fence

  5. anonymous
    • 5 years ago
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    Yes.

  6. anonymous
    • 5 years ago
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    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.

  7. anonymous
    • 5 years ago
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    ok thanks i will do that if i get stuck i will ask for more help :) but that seems to clarify things

  8. anonymous
    • 5 years ago
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    If you get stuck, post here what equations you're using for area and perimeter.

  9. anonymous
    • 5 years ago
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    A=LW; P=2L+2W I get P=2L+2(2693400/L)

  10. anonymous
    • 5 years ago
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    That perimeter is not quite right because you need to take into account the fence down the middle. So P = 2L + 3W

  11. anonymous
    • 5 years ago
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    ok yea i don't know why i am leaving it out lol

  12. anonymous
    • 5 years ago
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    so my P'=2+8050200/L^2

  13. anonymous
    • 5 years ago
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    Close, but \[ \frac{d}{dx} (\frac{1}{x}) = \frac{-1}{x^2}\] So you've got a sign problem in P'

  14. anonymous
    • 5 years ago
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    yea i saw it so i get 2/8050200=L^2 after i fix sign error then i get sqrt(2/8050200)=L

  15. anonymous
    • 5 years ago
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    uh.. I think you're playing a bit fast/loose here.. \[P' = 2-\frac{8050200}{L^2}\] \[P'=0 \rightarrow \frac{8050200}{L^2} = 2\] \[\rightarrow \frac{8050200}{2} = L^2 \]

  16. anonymous
    • 5 years ago
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    You have it the other way around.

  17. anonymous
    • 5 years ago
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    yea cause its like saying 8050200(1/L^2)

  18. anonymous
    • 5 years ago
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    ok so will L be my answer..

  19. anonymous
    • 5 years ago
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    I am getting a wrong answer...

  20. anonymous
    • 5 years ago
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    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.

  21. anonymous
    • 5 years ago
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    my math is wrong it was 8080200 not 8050200

  22. anonymous
    • 5 years ago
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    I got the right answer...thanks for your help sorry i was just making minor errors but you helped me see them

  23. anonymous
    • 5 years ago
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    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.

  24. anonymous
    • 5 years ago
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    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

  25. anonymous
    • 5 years ago
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    err 2 equations, one for surface area, one for volume.

  26. anonymous
    • 5 years ago
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    is it surface area because we know the amount of material that we use

  27. anonymous
    • 5 years ago
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    it's surface area because they give us the material yes (in square centimeters)

  28. anonymous
    • 5 years ago
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    V=w^2*H and my surface area would be of a cube?

  29. anonymous
    • 5 years ago
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    Yes with the top missing.

  30. anonymous
    • 5 years ago
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    so its just five sides SA=5x^2 or is it SA=2w^2+4wh

  31. anonymous
    • 5 years ago
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    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.

  32. anonymous
    • 5 years ago
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    ok

  33. anonymous
    • 5 years ago
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    i am kinda lost with the derivative...

  34. anonymous
    • 5 years ago
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    What's the equation you have for Volume?

  35. anonymous
    • 5 years ago
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    V=w^2(3072-w^2/4w)

  36. anonymous
    • 5 years ago
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    Shouldn't that be \[ V = \frac{w^2(3072-w^2)}{4w}\]

  37. anonymous
    • 5 years ago
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    could you eliminate the w from the bottom and one w from w^2

  38. anonymous
    • 5 years ago
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    yes.

  39. anonymous
    • 5 years ago
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    \[ V=\frac{1}{4} * [3072w - w^3] \]

  40. anonymous
    • 5 years ago
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    V'=1/4[3072-3w^2]

  41. anonymous
    • 5 years ago
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    yep.

  42. anonymous
    • 5 years ago
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    Thanks again i got it again... i am starting to understand optimization

  43. anonymous
    • 5 years ago
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    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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