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Thread: Right circular cylinder: find radius, height, given volume

  1. #1
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    Right circular cylinder: find radius, height, given volume

    I am trying to find the radius and height of a container with a capacity of 50 in ^3.
    I know that the total voume of the cylinder must be 50 in^3, so volume = 50, or pie times radius ^2 = 50 and height = 50/pie times radius ^2.

    I need to substitute the expression for height into the equation for area. I have to find the minimum of area (radius) =

    . . .(2 times pie times radius [50/(pie times radius times ^2] )

    . . . . . . .+

    . . .2 times pie times radius ^2. . .=. . .???

    Please help.
    K
    Kaye

  2. #2
    Elite Member stapel's Avatar
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    Note: The volume of a right circular cylinder with radius "r" and height "h" is given by:

    . . . . .V = (pi)(r<sup>2</sup>)(h)

    ...not ("pie")(r<sup>2</sup>). With only the volume given, the height and radius cannot be fixed; they can only be related.

    Eliz.

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    I am trying to find the radius and height of a container with a capacity of 50 in ^3. I know that the total voume of the cylinder must be 50 in^3, so volume = 50, or pie times radius ^2 = 50 and height = 50/pie times radius ^2.

    I need to substitute the expression for height into the equation for area. I have to find the minimum of area (radius) =

    . . .(2 times pie times radius [50/(pie times radius times ^2] )

    . . .. . .+

    . . .2 times pie times radius ^2. . .=. . .???

    Please help.
    Having read your problem statement over and over, I cannot help but come to the conclusion that you are seeking the container having the minimum surface area for the 50in^3 volume. On that basis:

    V = 50 = PiR^2h making h = 50 - PiR^2
    A = PiR^2 + =2PiRh
    Substituting for h yields A = PiR^2 + 100PiR - 2Pi^2R^3
    The first derivitive yields dA/dR = 2PiR + 100Pi - 6Pi^2R^2 = 0
    Simplifying, 6PiR^2 - 2R - 100 = 0
    Using the quadratic formula,

    . . .R = [2+/-sqrt(4 + =2400Pi)]/12Pi which yields two answers.
    . . .R = 2.3569 and h = 2.8650 and
    . . .R = 2.3233 and h = 2.9739

    Taking the average of the two, R = 2.3396 and h = 2.9076
    The surface area becomes A = 59.938...in^2
    The volume calculates out to 49.99995...in^3

    If this is not what you were looking for, please clarify yourneeds in the problem statement.
    TchrWill

    No matter how insignificant it might appear, learn something new every day.

  4. #4
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    I got something a little different.

    Assuming your container has a top and bottom, the surface area is given by:

    [tex]\L\\2{\pi}r^{2}+2{\pi}rh[/tex].....[1]

    The volume is [tex]\L\\{\pi}r^{2}h=50[/tex]....[2]

    Solve for h in [2] and sub into [1].

    [tex]\L\\\frac{50}{{\pi}r^{2}}=h[/tex]

    [tex]\L\\2{\pi}r^{2}+2{\pi}r(\frac{50}{{\pi}r^{2}})=2{\ pi}r^{2}+\frac{100}{r}[/tex]

    Differentiate:

    [tex]\L\\\frac{d}{dr}[2{\pi}r^{2}+\frac{100}{r}]=4{\pi}r-\frac{100}{r^{2}}[/tex]

    Set to 0 and solve for r:

    [tex]\L\\4{\pi}r+\frac{100}{r^{2}}=0[/tex]

    Multiply by [tex]r^{2}[/tex]

    [tex]\L\\4{\pi}r^{3}-100=0[/tex]

    [tex]\L\\r=(\frac{25}{\pi})^{\frac{1}{3}}\approx{1.996}[/tex]

    You could round to 2

    This gives [tex]\L\\h=\approx{3.993}[/tex]

    You could round to 4.

    This gives a surface area of 75.13 square units as the minimum.

    And please...."pie"?. Give me a break!

  5. #5
    Senior Member skeeter's Avatar
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    I like this "pie" ...






    ... better than I like this "pi"



  6. #6
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    CUTE ~ ~ ~

    That was mean...I do not know how to show the symbol for pie.......sorry to get your taste buds all in an uproar.

    K
    Kaye

  7. #7
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    Everyone is just teasing because you spelled 'pi' as 'pie'.

    Did you actually think a popular dessert was a part of the Greek alphabet?.

  8. #8
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    Quote Originally Posted by galactus
    I got something a little different...
    Yes, galactus, your answer for two end covers is right. While kiones did not specify whether the cylinder had a cover or not, I chose to assume no top just to convey the method. It is well known that the cylinder of minimum surface area (having both a top and bottom) for a specified volume is one where the height is equal to the diameter. I just assumed that if kiones wanted two covers, she at least had the method to derive it if so desired.

    I like your sequence of steps. Clearly, mine could have been arranged differently..
    TchrWill

    No matter how insignificant it might appear, learn something new every day.

  9. #9
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    No problem, TchrWill. I assumed kjones wanted ends because of the way (gender unknown, insert appropriate pronoun here) had it formatted.

    Namely,

    .(2 times pie times radius [50/(pie times radius times ^2] )

    . . . . . . .+

    . . .2 times pie times radius ^2. . .=. . .???

    Good luck kjones. Don't mind the light-hearted ribbing regarding your 'pie' faux pas.

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