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Math Help - Function of a cylinder

  1. #1
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    Function of a cylinder

    A can producer must make a cylindrical can that contains a volume of 16 cubic centimeters. Find the function this producer must analyze to make sure the amount of material is the least possible.
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  2. #2
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    Quote Originally Posted by bret80
    A can producer must make a cylindrical can that contains a volume of 16 cubic centimeters. Find the function this producer must analyze to make sure the amount of material is the least possible.
    V=\frac{1}{3}\pi r^2 h
    The amount of matherial used is,
    A=2\pi r h+2\pi r^2=
    You want to minimize A.

    Note in first equation we have,
    \frac{3V}{h\pi}=r^2
    Thus,
    A=2\pi \sqrt{\frac{3V}{h\pi }} h+2\pi \frac{3V}{h\pi}
    Simplify (steps omitted),
    2\sqrt{3V\pi}\cdot \sqrt{h}+\frac{6V}{h}
    Derivative time,
    A'=\sqrt{3V\pi}h^{-1/2}-6Vh^{-2}
    Make equal to zero,
    \sqrt{3V\pi}h^{-1/2}=6Vh^{-2}
    Divide both sides by h^{-2},
    \sqrt{3V\pi}h^{3/2}=6V
    Thus,
    h^{3/2}=\frac{1}{\sqrt{\pi}}2\sqrt{3}\sqrt{V}
    Note we substitute the value for "V":
    h^{3/2}=\frac{2\sqrt{3}\cdot 4}{\sqrt{\pi}}
    Thus,
    h^{3/2}=\frac{8\sqrt{3}}{\sqrt{\pi}}
    Square both sides,
    h^3=\frac{2^6 \cdot 3}{\pi}
    Now take cube root,
    h=\frac{4 \cdot \sqrt[3]{3}}{\sqrt[3]{\pi}}
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  3. #3
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    The correct answer I have is

    32/r + 2r^2
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  4. #4
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    Quote Originally Posted by bret80
    The correct answer I have is

    32/r + 2r^2
    \

    This time, TPH, you've gone too far!!

    (All he wanted was the equation to minimize. )

    bret80: Follow the first two lines of TPH's thread, but instead of solving the Volume equation for r, solve it for h. Sub that value for h into the Area equation.

    -Dan
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  5. #5
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    Quote Originally Posted by ThePerfectHacker
    V=\frac{1}{3}\pi r^2 h
    Isn't that the volume of a cone?

    RonL
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  6. #6
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    Hello, Bret!

    The correct answer I have is: 32/r + 2πr . . . . not quite

    The volume of a cylinder is: . V\:=\:\pi r^2h
    We are told that the volume is 16\pi cm.

    So we have: . \pi r^2h \:=\:16\pi\qiad\Rightarrow\quad h \:=\:\frac{16}{r^2} [1]


    The surface area of a cylinder is: .top + bottom + side.

    The top and bottom are circles of radius r. .Their area is: 2 \times \pi r^2

    The side is a rectangle of length 2\pi r and height h: area = 2\pi rh

    Hence, the surface area of a cylinder is: . S \;=\;2\pi r^2 + 2\pi rh

    Substitute [1]: . S\;=\;2\pi r^2 + 2\pi r\left(\frac{16}{r^2}\right)

    Therefore: . \boxed{S \;= \;2\pi r^2 + \frac{32\pi}{r}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    By the way, you can use &# 960; (without the space) to make π.


    More codes (until you learn LaTeX):

    &# 178; . . squared:

    &# 179; . . cubed:

    &# 8776; . approx: ≈

    &# 8800; . not equal: ≠

    &# 952; . . theta: θ

    &# 177; . . plus/minus:

    &# 176; . . degree:

    &# 8730; . radical: √

    &# 183; . . dot:

    &# 189; . . one-half:

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