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burhan101

  • one year ago

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?

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  1. burhan101
    • one year ago
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  2. burhan101
    • one year ago
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    Am I doing it right?

  3. e.mccormick
    • one year ago
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    I see an issue with your derivative.

  4. burhan101
    • one year ago
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    please point it out

  5. e.mccormick
    • one year ago
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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.

  6. burhan101
    • one year ago
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    i get |dw:1371523952498:dw|

  7. e.mccormick
    • one year ago
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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.

  8. e.mccormick
    • one year ago
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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'.

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

  10. e.mccormick
    • one year ago
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    Yes.

  11. e.mccormick
    • one year ago
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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.

  12. burhan101
    • one year ago
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    |dw:1371526270946:dw|

  13. burhan101
    • one year ago
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    the r^2 cancel out. what else

  14. e.mccormick
    • one year ago
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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.

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

  16. burhan101
    • one year ago
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    and then I just solve for 'r' ?

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

  18. burhan101
    • one year ago
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    cant I do |dw:1371526627893:dw|

  19. e.mccormick
    • one year ago
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    Forgot your cube.

  20. burhan101
    • one year ago
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    what?

  21. e.mccormick
    • one year ago
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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.

  22. burhan101
    • one year ago
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    ohhh okay let me fix that

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

  24. burhan101
    • one year ago
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    theres a denominator too right ^

  25. e.mccormick
    • one year ago
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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....

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

  27. burhan101
    • one year ago
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    oh okay, so i only focus on the top

  28. burhan101
    • one year ago
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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 ?

  29. e.mccormick
    • one year ago
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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.

  30. e.mccormick
    • one year ago
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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.

  31. e.mccormick
    • one year ago
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    |dw:1371527305050:dw|

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

  33. e.mccormick
    • one year ago
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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.

  34. burhan101
    • one year ago
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    would I leave the calculator in radians?

  35. e.mccormick
    • one year ago
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    Does not matter. No degrees involved.

  36. burhan101
    • one year ago
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    the pi ?

  37. burhan101
    • one year ago
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    |dw:1371528718316:dw|

  38. e.mccormick
    • one year ago
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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

  39. burhan101
    • one year ago
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    Oh my, thank you SO much !

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

  41. e.mccormick
    • one year ago
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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!

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

  43. burhan101
    • one year ago
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    ohhh, i can pick any two numbers?

  44. e.mccormick
    • one year ago
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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.

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

  46. e.mccormick
    • one year ago
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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!

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

  48. burhan101
    • one year ago
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    thanks for the heads up !

  49. e.mccormick
    • one year ago
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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.

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

  51. burhan101
    • one year ago
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    \[\large C=7998.49m^2\]

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

  53. burhan101
    • one year ago
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    15.50 x 1000= 15,500

  54. e.mccormick
    • one year ago
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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...

  55. burhan101
    • one year ago
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    back of the book says $0.80 each

  56. e.mccormick
    • one year ago
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    Hmmm... I must have converted wrong somewhere.

  57. e.mccormick
    • one year ago
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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.

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

  59. e.mccormick
    • one year ago
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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!

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

  61. burhan101
    • one year ago
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    thanks !!

  62. e.mccormick
    • one year ago
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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.

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