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anonymous

  • 5 years ago

A fence is to be built to enclose a rectangular area of 280 square feet. The fence along three sides is to be made of material that costs 5 dollars per foot, and the material for the fourth side costs 15 dollars per foot. Find the dimensions of the enclosure that is most economical to construct.

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  1. amistre64
    • 5 years ago
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    Area = LW Perimeter = L1 + L2 + 2W 180 = LW 2L(5) + W(5) + L(15) = cost

  2. amistre64
    • 5 years ago
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    since 180 = LW, we can say that W = 180/L

  3. amistre64
    • 5 years ago
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    that L(15) up there should be W(15) 2L(5) + (180/L)(5) + (180/L)(15) = Cost

  4. amistre64
    • 5 years ago
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    10L +900/L + 2700/L = cost

  5. amistre64
    • 5 years ago
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    maybe we need to minimise the perimeter :) P = 2L + 2W P = 2L + 2(180/L) P = 2L + 360/L P = 2L^2 + 360 ---------- take the derivative L

  6. amistre64
    • 5 years ago
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    dP = L(2L) - (2L^2+360)(1) ------------------- L(L)

  7. amistre64
    • 5 years ago
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    4L^2 - 2L^2 - 360 2L^2-360 ----------------- = --------- L^2 L^2 I adjusted for my stupidity :)

  8. amistre64
    • 5 years ago
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    when the top is 0 we got a min or a max

  9. anonymous
    • 5 years ago
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    sorry just a little confused because it 280 but it still works i can just go back and plug in my numbers

  10. amistre64
    • 5 years ago
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    2(L^2 -180) = 0 L^2 - 180 = 0 L^2 = 180 L = +sqrt(180)

  11. amistre64
    • 5 years ago
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    ...... lol ..........

  12. amistre64
    • 5 years ago
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    i forgot to adjust for all my stupidity lol

  13. anonymous
    • 5 years ago
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    it ok you have it i see just a mix up of numbers is all...

  14. amistre64
    • 5 years ago
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    L = sqrt(280) :)

  15. amistre64
    • 5 years ago
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    W = 280/sqrt(280) W = sqrt(280) as well, so its a square I guess...

  16. amistre64
    • 5 years ago
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    Well, that would maximise the area ..... but it doesnt help with costs :) lets see what costs gets us

  17. amistre64
    • 5 years ago
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    2L(5) + (280/L)(5) + (280/L)(15) = Cost 10L + 900/L + 4200/L = cost

  18. amistre64
    • 5 years ago
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    10L^2 + 900 + 4200 ------------------ = cost L

  19. amistre64
    • 5 years ago
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    stop me when I get lost :)

  20. amistre64
    • 5 years ago
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    10L^2 + 5100 ------------ = cost L L(20L) - 1(10L^2 + 5100) ----------------------- = cost' L^2

  21. anonymous
    • 5 years ago
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    wait i am confused....it wants to know the dimensions...so why are we looking for cost

  22. amistre64
    • 5 years ago
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    20L^2 -10L^2 -5100 = 0 10L^2 - 5100 = 0 10(L^2 -510) = 0 L^2 - 510 = 0 L=sqrt(510) because..... the diminsions dont need to be "minimized for area, they need to be minimized for cost...makes sense?

  23. anonymous
    • 5 years ago
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    yea...

  24. amistre64
    • 5 years ago
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    lets plug sqrt(510) into the calulator to give us a roundabout: 22.58 is what I get for one side :) 280 = LW 280 = 22.58W 280/22.58 = W 12.40 = W

  25. amistre64
    • 5 years ago
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    Now figure out how much we need for the most expensive side :)

  26. amistre64
    • 5 years ago
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    2(22.58)(5) + 12.40(5) + 12.40(15) =? 2(12.40)(5) + 22.58(5) + 22.58(15)? which is cheaper?

  27. anonymous
    • 5 years ago
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    my answer are not correct....thats what my website i have to submit my answer into is telling me...so i don't know if there was a mix up somewhere

  28. amistre64
    • 5 years ago
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    maybe..... I could of thought it wrong as well. but to me, to minimize cost, we would derive the equation for Price and not Area, figure out what a side would be for that price then figure the rest out...

  29. amistre64
    • 5 years ago
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    280/L or 280/W doesnt matter, since its just a name for a side..right?

  30. amistre64
    • 5 years ago
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    polpak is here.... hell know what I did wrong ;)

  31. amistre64
    • 5 years ago
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    hit "post" lol

  32. anonymous
    • 5 years ago
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    Ok, we have \[Area = LW = 280 \rightarrow W = 280/L\] \[Cost = 5(L + 2W) + 15L \] \[\rightarrow Cost = 20L + (5*2)(280)/L \]

  33. anonymous
    • 5 years ago
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    Thanks figured it out...

  34. anonymous
    • 5 years ago
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    \[\frac{d}{dL}Cost = 20 - \frac{2800}{L^2}\] The cost is a min when the derivative is 0 \[\rightarrow 0 = 20-\frac{2800}{L^2}\]

  35. anonymous
    • 5 years ago
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    Ok cool.

  36. amistre64
    • 5 years ago
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    so where did I miss it?

  37. anonymous
    • 5 years ago
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    I can't tell, but you have a lot of cost functions and none of them look the same as the one I have.

  38. amistre64
    • 5 years ago
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    :) they are similar.... I was just going over that to see if I made a mistake

  39. amistre64
    • 5 years ago
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    I did 2L(5) + 180(5)/L + 280(15)/L for cost... but that should be correct either way....

  40. amistre64
    • 5 years ago
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    make that 280 up there

  41. anonymous
    • 5 years ago
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    If those are both 280 instead of 180 it should be ok.

  42. anonymous
    • 5 years ago
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    It just means that your L is my W.

  43. amistre64
    • 5 years ago
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    it is...they are :)

  44. amistre64
    • 5 years ago
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    10L +900/L + 4200/L = cost

  45. anonymous
    • 5 years ago
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    Ok then you would have \[Cost' = 10 -900/L^2 - 4200/L^2\]

  46. amistre64
    • 5 years ago
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    right... you derived them seperately which is fine :)

  47. amistre64
    • 5 years ago
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    10L^2 - 5100 = 0 L^2 - 510 = 0

  48. amistre64
    • 5 years ago
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    L=sqrt(510).... is it right? the same as yours?

  49. amistre64
    • 5 years ago
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    you get L = sqrt(140)

  50. anonymous
    • 5 years ago
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    Yeah.

  51. anonymous
    • 5 years ago
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    But my L is your W remember.

  52. amistre64
    • 5 years ago
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    hmmm....... maybe thats right, havent checked that out yet :)

  53. amistre64
    • 5 years ago
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    11.83 = you 12.40 = me.....

  54. anonymous
    • 5 years ago
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    Hrm. Lemme check that cost again.

  55. amistre64
    • 5 years ago
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    280*5 = 1400 ... not 900

  56. anonymous
    • 5 years ago
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    Ah. That'll do it.

  57. amistre64
    • 5 years ago
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    sqrt(560).... might be better :)

  58. amistre64
    • 5 years ago
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    Yep... thats it.... I just forgot how to multiply integers LOL

  59. anonymous
    • 5 years ago
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    Yep, much better.

  60. anonymous
    • 5 years ago
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    Thats why I always save the plugging the numbers part for last.

  61. anonymous
    • 5 years ago
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    Or as late as I can without making it look horrible.

  62. amistre64
    • 5 years ago
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    yeah.... itda helped if I started out with the 280 and worked it, but..... you know :)

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