burhan101 2 years 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? • This Question is Closed 1. burhan101 2. burhan101 Am I doing it right? 3. e.mccormick I see an issue with your derivative. 4. burhan101 please point it out 5. e.mccormick $$\cfrac{27900}{r}=27900r^{-1}$$ Use this second form of it and the exponent rules you know and see what you get. 6. burhan101 i get |dw:1371523952498:dw| 7. e.mccormick Yes. Your other part, the $$+62\pi r$$ is fine, so now add those fractions up and see where you go. 8. e.mccormick 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 Was the c' was wrong due to my wrong derivative? 10. e.mccormick Yes. 11. e.mccormick 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 |dw:1371526270946:dw| 13. burhan101 the r^2 cancel out. what else 14. e.mccormick Well, it does not really cancel... it is because there is a 0 on the other side that it does not matter. 15. burhan101 $\huge 0=62\pi r-27900$ 16. burhan101 and then I just solve for 'r' ? 17. e.mccormick Now, what if you multiply through by 1/62? 18. burhan101 cant I do |dw:1371526627893:dw| 19. e.mccormick Forgot your cube. 20. burhan101 what? 21. e.mccormick 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 ohhh okay let me fix that 23. e.mccormick $$62\pi r^3 - 27900=0$$ 24. burhan101 theres a denominator too right ^ 25. e.mccormick 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 cant i just multiply the equation by r^2 to get rid of the denominator 27. burhan101 oh okay, so i only focus on the top 28. burhan101 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 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 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 |dw:1371527305050:dw| 32. burhan101 Okay thanks for that, ill check it as soon as i finish !! :D 33. e.mccormick 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 would I leave the calculator in radians? 35. e.mccormick Does not matter. No degrees involved. 36. burhan101 the pi ? 37. burhan101 |dw:1371528718316:dw| 38. e.mccormick 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 Oh my, thank you SO much ! 40. burhan101 now can I plug this r value intto my cost equation ? 41. e.mccormick 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 Yes, you can put that r into the original. 43. burhan101 ohhh, i can pick any two numbers? 44. e.mccormick 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 I do see one big thing to be careful of in all of this. 46. e.mccormick "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

$$1m^2=10000cm^3$$

48. burhan101

thanks for the heads up !

49. e.mccormick

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

I know, i still understood what you were saying :P

51. burhan101

$\large C=7998.49m^2$

52. e.mccormick

$$\cfrac{15.50}{10000cm^2}\times 5.23cm^2$$ The $$cm^2$$ cancels....

53. burhan101

15.50 x 1000= 15,500

54. e.mccormick

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 back of the book says$0.80 each

56. e.mccormick

Hmmm... I must have converted wrong somewhere.

57. e.mccormick

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

Ah! Remembered. R needs to go back into the formula!

59. e.mccormick

$$\cfrac{1800}{r}+2\pi r^2\implies \cfrac{1800}{5.23}+2\pi (5.23)^2=????$$ That makes much more sense!

60. e.mccormick

Yes, that got me something that will round to 80 cents.

61. burhan101

thanks !!

62. e.mccormick

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.