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anonymous

  • 5 years ago

please help me out with this one- a right circular cone has a surface area of 718 square inches. what dimensions (radius and height) will result in a maximum volume?

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  1. anonymous
    • 5 years ago
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    I began the problem, and now it after 30 mins of work it is a nightmare.

  2. Mertsj
    • 5 years ago
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    What class is this for--calculus?

  3. anonymous
    • 5 years ago
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    yup

  4. anonymous
    • 5 years ago
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    so far, i've taken the SA eqn and tried to optimize it with substitution, but then after finding a derivative it just is bad.

  5. Mertsj
    • 5 years ago
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    If you could find satellite of myininaya they could help you

  6. amistre64
    • 5 years ago
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    this one, jwg lol

  7. Mertsj
    • 5 years ago
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    yep

  8. amistre64
    • 5 years ago
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    is there a formula for the surface area of a rt circular cone?

  9. Mertsj
    • 5 years ago
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    SA=pir^2+pirl

  10. amistre64
    • 5 years ago
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    and perferably one for volume as well to relate the 2?

  11. amistre64
    • 5 years ago
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    1/3 something is all i can remember

  12. amistre64
    • 5 years ago
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    1/3 volume of a cylinder perhaps?

  13. Mertsj
    • 5 years ago
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    V=-1/3pir^2h

  14. Mertsj
    • 5 years ago
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    Exactly

  15. amistre64
    • 5 years ago
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    ok; looks like we are given an value for SA so that we can find an r perhaps?

  16. Mertsj
    • 5 years ago
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    The problem is the slant height adds a variable.

  17. amistre64
    • 5 years ago
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    i take it thats the "l" part in the SA

  18. Mertsj
    • 5 years ago
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    yes

  19. amistre64
    • 5 years ago
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    |dw:1327287244737:dw|

  20. Mertsj
    • 5 years ago
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    precisely

  21. amistre64
    • 5 years ago
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    l = sqrt(h^2 + r^2) then 718 = pi r^2 + pi r(sqrt(h^2+r^2)) 718/pir = r + (sqrt(h^2+r^2)) (718/pir) - r = sqrt(h^2+r^2) [(718/pir) - r]^2 = h^2+r^2 [(718/pir) - r]^2 - r^2 = h^2 i wonder of thats going according to plan :)

  22. amistre64
    • 5 years ago
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    at any rate we can define h in terms of r; aint pretty but doable

  23. Mertsj
    • 5 years ago
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    That should be progress. Wonder where the asker is.

  24. amistre64
    • 5 years ago
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    fetal position, sobbing maybe lol

  25. amistre64
    • 5 years ago
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    V = 1/3 pi r^2 h \[V=\frac{pi\ r^2\ \sqrt{(\frac{718}{pi\ r}-r)^2-r^2}}{3}\] looks about right?

  26. Mertsj
    • 5 years ago
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    yes

  27. Mertsj
    • 5 years ago
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    Now I suppose we have to differentiate that and set to 0?

  28. amistre64
    • 5 years ago
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    the 1/3 pi is a constant that can be put to the side and we are left with deriving:\[V=r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}\]

  29. amistre64
    • 5 years ago
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    yep, find the zeros and the undefineds to test

  30. amistre64
    • 5 years ago
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    \[V'=r'^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)'^{1/2}\] \[V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*((\frac{718}{pi\ r}-r)^2-r^2)'\] \[V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*(\frac{718}{pi\ r}-r)'^2-r'^2\] \[V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*2(\frac{718}{pi\ r}-r)(\frac{718}{pi\ r}-r)'-2r\] yep, its a doozie; the wolf might be quicker

  31. Mertsj
    • 5 years ago
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    That's what I did.

  32. amistre64
    • 5 years ago
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    ..... slacker!! lol

  33. Mertsj
    • 5 years ago
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    proudly so

  34. Mertsj
    • 5 years ago
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    r=7.56

  35. amistre64
    • 5 years ago
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    yeah, i was gonna say around 8

  36. amistre64
    • 5 years ago
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    plug that value in to get the V max

  37. Mertsj
    • 5 years ago
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    I just ignored the miserable denominator and set the numerator to 0.

  38. Mertsj
    • 5 years ago
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    Must first find the height

  39. amistre64
    • 5 years ago
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    [(718/pir) - r]^2 - r^2 = h^2 plug in yer r

  40. Mertsj
    • 5 years ago
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    yep

  41. amistre64
    • 5 years ago
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    i dunno if solving in terms of h would have been easier

  42. Mertsj
    • 5 years ago
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    tooooooooooooooooooooooooooooo late

  43. amistre64
    • 5 years ago
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    lol, since h was buried i assume not

  44. Mertsj
    • 5 years ago
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    h= 21.37

  45. amistre64
    • 5 years ago
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    that should do it then

  46. Mertsj
    • 5 years ago
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    Except for checking to make sure those values result in the correct surface area.

  47. amistre64
    • 5 years ago
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    it has to, we did everything correctly

  48. Mertsj
    • 5 years ago
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    Of course...no mistakes...ever!!

  49. amistre64
    • 5 years ago
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    they wanted max volume

  50. amistre64
    • 5 years ago
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    the wolf helped so we can always pass the blame ;)

  51. Mertsj
    • 5 years ago
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    working on that.

  52. Mertsj
    • 5 years ago
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    1279.02= max volume

  53. Mertsj
    • 5 years ago
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    Good to have a scapegoat.

  54. Mertsj
    • 5 years ago
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    Hey!! It's right. SA = 717.9758

  55. Mertsj
    • 5 years ago
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    Thank you very much.

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