## 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

1. anonymous

this is a tricky one....

2. amistre64

are thos answers? or formulas to apply?

3. anonymous

formulas it's a related rates question

4. amistre64

yes, optimizations and such

5. anonymous

yes they want D, H, and Amin

6. amistre64

V = 1/2 cup, soo; .5 = (pi) (r^2) (h) right?

7. anonymous

Yes

8. amistre64

A = 2 circles and a rectangle :)

9. anonymous

haha :)

10. amistre64

A = 2pi (r^2) + h(2pi r)

11. amistre64

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

12. amistre64

substitute that "value" into the Area formula and then derive

13. anonymous

which value the 946cc?

14. amistre64

the .5 cups...worry about conversions afterwards

15. anonymous

oh okay gotcha

16. amistre64

so...$h = \frac{1}{2\pi r^2}$ right?

17. amistre64

A = 2pi (r^2) + h(2pi r) sooo $A = 2\pi r^2 + \frac{2\pi r}{2\pi r^2}$

18. anonymous

I did get that for A

19. amistre64

$A = 2\pi r^2 + {1 \over r}$ correct?

20. anonymous

yes

21. amistre64

now derive :) you know how to derive?

22. anonymous

yes i do

23. amistre64

A = 4pi r - 1/r^2 then correct?

24. amistre64

A' that is lol

25. anonymous

Yes that is what I just got

26. amistre64

if we combine these with like denoms we get: $A'=\frac{4\pi r^3 - 1}{r^2}$ correct?

27. anonymous

Yes that looks good

28. amistre64

$A'=0=4pir^3 -1$ so what are the zeros?

29. amistre64

r = ? the easiest way is to plug this into wolframs site and itll tell us right away ;)

30. amistre64
31. anonymous

okay and the equation that I would type in would be $\sqrt[3]{1/4\pi}$ right

32. amistre64

4pi r^3 - 1 :) and i got abt .4 as the only root

33. anonymous

oh okay you don't have to get r by itself

34. amistre64

not with wolframs site; itll do it for you

35. amistre64

r = aprrox. .430127

36. anonymous

okay and r^2 is the diameter and I can use that to solve for Amin

37. amistre64

Yes; or simply use it to solve 'h'

38. anonymous

Yeah I guess that would be easier lol

39. amistre64

how many cups to a quart? 8 or 4?

40. anonymous

4 cups

41. amistre64

4 cups to 1 quart and we need half a cup so 1/8th of a quart right?

42. anonymous

yes

43. amistre64

1/2 cup = 946/8 = 118.25 centimeters squared (cc) then

44. amistre64

2 = abt 2.1 centimeters then

45. amistre64

r = abt. 2.1 centimeters

46. amistre64

A = 2pi(2.1)^2 + 1/(2.1)

47. amistre64

Total min area = 28.18 cc

48. anonymous

wouldn't it be 2.1^4 since it was d^2

49. anonymous

oh never mind I see it now

50. amistre64

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

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

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

Yes it is it matches up to the parts that I had figured out

54. amistre64

cool :)

55. anonymous

Thanks again!