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burhan101

A cylinder can is to have a volume of 900 cm cubed. The metal costs $15.50/squared meter. What dimensions produce a can with minimum cost? What is the cost of making the can?

  • 10 months ago
  • 10 months ago

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  1. burhan101
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    • 10 months ago
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  2. burhan101
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    Am I doing it right?

    • 10 months ago
  3. e.mccormick
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    I see an issue with your derivative.

    • 10 months ago
  4. burhan101
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    please point it out

    • 10 months ago
  5. e.mccormick
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    \(\cfrac{27900}{r}=27900r^{-1}\) Use this second form of it and the exponent rules you know and see what you get.

    • 10 months ago
  6. burhan101
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    i get |dw:1371523952498:dw|

    • 10 months ago
  7. e.mccormick
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    Yes. Your other part, the \(+62\pi r\) is fine, so now add those fractions up and see where you go.

    • 10 months ago
  8. e.mccormick
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    Your mistake was in how you took the derivative of that. Other than that, the next steps looked pretty good, but got fowled up by involving the wrong C'.

    • 10 months ago
  9. burhan101
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    Was the c' was wrong due to my wrong derivative?

    • 10 months ago
  10. e.mccormick
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    Yes.

    • 10 months ago
  11. e.mccormick
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    You come up with a new, reduced equation? In your first one, you elimitated the \(r^2\) on the bottom. In this, there is a little more you can cancel out. The 62.

    • 10 months ago
  12. burhan101
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    |dw:1371526270946:dw|

    • 10 months ago
  13. burhan101
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    the r^2 cancel out. what else

    • 10 months ago
  14. e.mccormick
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    Well, it does not really cancel... it is because there is a 0 on the other side that it does not matter.

    • 10 months ago
  15. burhan101
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    \[\huge 0=62\pi r-27900 \]

    • 10 months ago
  16. burhan101
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    and then I just solve for 'r' ?

    • 10 months ago
  17. e.mccormick
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    Now, what if you multiply through by 1/62?

    • 10 months ago
  18. burhan101
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    cant I do |dw:1371526627893:dw|

    • 10 months ago
  19. e.mccormick
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    Forgot your cube.

    • 10 months ago
  20. burhan101
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    what?

    • 10 months ago
  21. e.mccormick
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    Ah, in the earlier one too. You can't cancel the part of the \(r^3\) above. Like I said, it is not that it cancels but that it does not matter.

    • 10 months ago
  22. burhan101
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    ohhh okay let me fix that

    • 10 months ago
  23. e.mccormick
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    \(62\pi r^3 - 27900=0\)

    • 10 months ago
  24. burhan101
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    theres a denominator too right ^

    • 10 months ago
  25. e.mccormick
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    Because if the top of the fraction is 0, it is 0. Only if the bottom woulc cause an asymptote does it matter....

    • 10 months ago
  26. burhan101
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    cant i just multiply the equation by r^2 to get rid of the denominator

    • 10 months ago
  27. burhan101
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    oh okay, so i only focus on the top

    • 10 months ago
  28. burhan101
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    So like mathematically on my paper that my prof would mark i would just ignore it, like cant i be docked marks ?

    • 10 months ago
  29. e.mccormick
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    Yah. Mathematically, it is the same as multiplying through by \(r^2\) because the right hand side is 0 so \(0\times r^2=0\) means it does not change.

    • 10 months ago
  30. e.mccormick
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    I have the entire reducing this fraction in long form with every step. Just finished writing it up. So I'll post it when we get there and you can check what you have against it.

    • 10 months ago
  31. e.mccormick
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    |dw:1371527305050:dw|

    • 10 months ago
  32. burhan101
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    Okay thanks for that, ill check it as soon as i finish !! :D

    • 10 months ago
  33. e.mccormick
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    Well, solve for the root, and tell me what you get. Then I'll post what I got, what I did, and even a graph that shows some things.

    • 10 months ago
  34. burhan101
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    would I leave the calculator in radians?

    • 10 months ago
  35. e.mccormick
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    Does not matter. No degrees involved.

    • 10 months ago
  36. burhan101
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    the pi ?

    • 10 months ago
  37. burhan101
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    |dw:1371528718316:dw|

    • 10 months ago
  38. e.mccormick
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    pi is a constant, not an angle. So yah. \[\cfrac{62\pi r^3 - 27900}{r^2}=0\implies \\ \\ r^2\times\cfrac{62\pi r^3 - 27900}{r^2}=r^2\times0\implies \\ \\ 62\pi r^3 - 27900=0\implies \\ \\ \cfrac{1}{62}\times(62\pi r^3 - 27900)=\cfrac{1}{62}\times0\implies \\ \\ \pi r^3 - 450=0\implies \\ \\ \pi r^3 - 450+450=0+450\implies \\ \\ \pi r^3 =450\implies \\ \\ \cfrac{1}{\pi}\pi r^3 =\cfrac{1}{\pi}450\implies \\ \\ r^3 =\cfrac{450}{\pi}\implies \\ \\ \sqrt[3]{ r^3} =\sqrt[3]{ \cfrac{450}{\pi}}\implies \\ \\ r =\sqrt[3]{ \cfrac{450}{\pi}} \]And in the graph you can see how what I start with and end with both overlap, and they are 0 right where you said, about 5.23. https://www.desmos.com/calculator/mz0fk3daat

    • 10 months ago
  39. burhan101
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    Oh my, thank you SO much !

    • 10 months ago
  40. burhan101
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    now can I plug this r value intto my cost equation ?

    • 10 months ago
  41. e.mccormick
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    Here is another interesting point that is very good to know for these types of problems. Let's take the test points of 5 and 6 and put them into \(\pi r^3 - 450\) \(\pi (5)^3 - 450\approx -57\) \(\pi (6)^3 - 450\approx 229\) So as x is increasing, this is moving from negative, to zero, then positive. That means the original equation has a slope there that is negative, bottoms out, then goes positive. This confirms that what you found is a minimum!

    • 10 months ago
  42. e.mccormick
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    Yes, you can put that r into the original.

    • 10 months ago
  43. burhan101
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    ohhh, i can pick any two numbers?

    • 10 months ago
  44. e.mccormick
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    Well, close numbers are best for this sort of test. They just become test points on each side of the critical point to confirm if it is a min or max. If you cross over two critical points, that sort of test is invalid. but we only have one critical point, so it is no big deal.

    • 10 months ago
  45. e.mccormick
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    I do see one big thing to be careful of in all of this.

    • 10 months ago
  46. e.mccormick
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    "A cylinder can is to have a volume of 900 cm cubed." \(\leftarrow\) in cm. "The metal costs $15.50/squared meter." \(\leftarrow\) in m! Watch out for your units!

    • 10 months ago
  47. e.mccormick
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    \(1m^2=10000cm^3\)

    • 10 months ago
  48. burhan101
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    thanks for the heads up !

    • 10 months ago
  49. e.mccormick
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    Oops. made a mistake there... I put cm cubed, but it is squared. Because 1m = 100 cm, so square both and you get: \(1m^2=10000cm^2\) The cubic relationship here is: \(900 cm^3 = .0009 m^3\) So you need to be careful because you found the radius in cm. So your can will be in cm.

    • 10 months ago
  50. burhan101
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    I know, i still understood what you were saying :P

    • 10 months ago
  51. burhan101
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    \[\large C=7998.49m^2\]

    • 10 months ago
  52. e.mccormick
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    \(\cfrac{$15.50}{10000cm^2}\times 5.23cm^2\) The \(cm^2\) cancels....

    • 10 months ago
  53. burhan101
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    15.50 x 1000= 15,500

    • 10 months ago
  54. e.mccormick
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    So like $0.0081 per can.... less than a penny each. Make sure you got the units right in the original, but if so, this is like the bulk manufacture of soda can...

    • 10 months ago
  55. burhan101
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    back of the book says $0.80 each

    • 10 months ago
  56. e.mccormick
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    Hmmm... I must have converted wrong somewhere.

    • 10 months ago
  57. e.mccormick
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    I don't see where though... odd.... because that would be like it was a linear conversion and this is a square conversion.

    • 10 months ago
  58. e.mccormick
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    Ah! Remembered. R needs to go back into the formula!

    • 10 months ago
  59. e.mccormick
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    \(\cfrac{1800}{r}+2\pi r^2\implies \cfrac{1800}{5.23}+2\pi (5.23)^2=????\) That makes much more sense!

    • 10 months ago
  60. e.mccormick
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    Yes, that got me something that will round to 80 cents.

    • 10 months ago
  61. burhan101
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    thanks !!

    • 10 months ago
  62. e.mccormick
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    Yah, it also explains where I made a mistake. I put it back into the wrong equation! I took the linear radius when I needed the square surface area! Be very careful of that.

    • 10 months ago
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