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anonymous

  • 5 years ago

A cylindrical can is to hold 1/2 cup of tomato sauce. What are the dimensions of its diameter and height in centimeters, to minimize the amount of metal to make the can? V=(pi*d^2*h)/4 A=(pi*d^2)+pi*d*h 1qt=946cc

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  1. anonymous
    • 5 years ago
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    this is a tricky one....

  2. amistre64
    • 5 years ago
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    are thos answers? or formulas to apply?

  3. anonymous
    • 5 years ago
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    formulas it's a related rates question

  4. amistre64
    • 5 years ago
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    yes, optimizations and such

  5. anonymous
    • 5 years ago
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    yes they want D, H, and Amin

  6. amistre64
    • 5 years ago
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    V = 1/2 cup, soo; .5 = (pi) (r^2) (h) right?

  7. anonymous
    • 5 years ago
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    Yes

  8. amistre64
    • 5 years ago
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    A = 2 circles and a rectangle :)

  9. anonymous
    • 5 years ago
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    haha :)

  10. amistre64
    • 5 years ago
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    A = 2pi (r^2) + h(2pi r)

  11. amistre64
    • 5 years ago
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    use the volume to calibrate either h or r h = .5/pi r^2 seems easiest right?

  12. amistre64
    • 5 years ago
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    substitute that "value" into the Area formula and then derive

  13. anonymous
    • 5 years ago
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    which value the 946cc?

  14. amistre64
    • 5 years ago
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    the .5 cups...worry about conversions afterwards

  15. anonymous
    • 5 years ago
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    oh okay gotcha

  16. amistre64
    • 5 years ago
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    so...\[h = \frac{1}{2\pi r^2}\] right?

  17. amistre64
    • 5 years ago
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    A = 2pi (r^2) + h(2pi r) sooo \[A = 2\pi r^2 + \frac{2\pi r}{2\pi r^2}\]

  18. anonymous
    • 5 years ago
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    I did get that for A

  19. amistre64
    • 5 years ago
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    \[A = 2\pi r^2 + {1 \over r}\] correct?

  20. anonymous
    • 5 years ago
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    yes

  21. amistre64
    • 5 years ago
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    now derive :) you know how to derive?

  22. anonymous
    • 5 years ago
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    yes i do

  23. amistre64
    • 5 years ago
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    A = 4pi r - 1/r^2 then correct?

  24. amistre64
    • 5 years ago
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    A' that is lol

  25. anonymous
    • 5 years ago
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    Yes that is what I just got

  26. amistre64
    • 5 years ago
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    if we combine these with like denoms we get: \[A'=\frac{4\pi r^3 - 1}{r^2}\] correct?

  27. anonymous
    • 5 years ago
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    Yes that looks good

  28. amistre64
    • 5 years ago
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    \[A'=0=4pir^3 -1\] so what are the zeros?

  29. amistre64
    • 5 years ago
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    r = ? the easiest way is to plug this into wolframs site and itll tell us right away ;)

  30. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/

  31. anonymous
    • 5 years ago
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    okay and the equation that I would type in would be \[\sqrt[3]{1/4\pi}\] right

  32. amistre64
    • 5 years ago
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    4pi r^3 - 1 :) and i got abt .4 as the only root

  33. anonymous
    • 5 years ago
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    oh okay you don't have to get r by itself

  34. amistre64
    • 5 years ago
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    not with wolframs site; itll do it for you

  35. amistre64
    • 5 years ago
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    r = aprrox. .430127

  36. anonymous
    • 5 years ago
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    okay and r^2 is the diameter and I can use that to solve for Amin

  37. amistre64
    • 5 years ago
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    Yes; or simply use it to solve 'h'

  38. anonymous
    • 5 years ago
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    Yeah I guess that would be easier lol

  39. amistre64
    • 5 years ago
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    how many cups to a quart? 8 or 4?

  40. anonymous
    • 5 years ago
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    4 cups

  41. amistre64
    • 5 years ago
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    4 cups to 1 quart and we need half a cup so 1/8th of a quart right?

  42. anonymous
    • 5 years ago
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    yes

  43. amistre64
    • 5 years ago
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    1/2 cup = 946/8 = 118.25 centimeters squared (cc) then

  44. amistre64
    • 5 years ago
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    2 = abt 2.1 centimeters then

  45. amistre64
    • 5 years ago
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    r = abt. 2.1 centimeters

  46. amistre64
    • 5 years ago
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    A = 2pi(2.1)^2 + 1/(2.1)

  47. amistre64
    • 5 years ago
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    Total min area = 28.18 cc

  48. anonymous
    • 5 years ago
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    wouldn't it be 2.1^4 since it was d^2

  49. anonymous
    • 5 years ago
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    oh never mind I see it now

  50. amistre64
    • 5 years ago
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    diameter = 2r = 2(2.1) = 4.2 V = 128.25 = 2pi(2.1)^2 * h 128.25 ---------- = h = abt. 4.63 2pi (2.1)^2

  51. anonymous
    • 5 years ago
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    Thank you very much I've been working on this problem for about 3 hours and you figured it out in 10 minutes lol! You're a lifesaver!

  52. amistre64
    • 5 years ago
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    is it right? cuase i might see an error in it; i allowed the area of the material to be 2 circles, top and bottom of a can; and a height...

  53. anonymous
    • 5 years ago
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    Yes it is it matches up to the parts that I had figured out

  54. amistre64
    • 5 years ago
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    cool :)

  55. anonymous
    • 5 years ago
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    Thanks again!

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