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Math Help - - Optimization Problems - Proving Question

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    - Optimization Problems - Proving Question

    Prove that any cylindrical can of volume k cubic units that is to be made using a minimum of material must have the height equal to the diameter.


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    Quote Originally Posted by darkknight100 View Post
    Prove that any cylindrical can of volume k cubic units that is to be made using a minimum of material must have the height equal to the diameter.
    volume ...

    k = \pi r^2 h

    material is equivalent to the surface area ...

    A = 2\pi r^2 + 2\pi r h

    solve the volume equation for h in terms of k and r, then substitute the derived expression for h into the surface area equation ...

    find \frac{dA}{dr} and determine the value of r that minimizes A ... finally determine h from your derived expression.

    also remember that k is a constant.
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    So, the given condition is

    = k (Eq. 1)

    And we need to minimize the surface area given by

    (Eq. 2)

    Step 1: Solve equation 1 for h in terms of r and substitue into equation 2.

    Step 2: Differentiate equation 2 wrt r. Set derivative equal to zero and solve for r.

    Step 3: Substitute r back into equation 1 to get h.

    You should then see that r and h will be equal.

    Good Luck!



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    Quote Originally Posted by dats13 View Post
    So, the given condition is

    = k (Eq. 1)

    And we need to minimize the surface area given by

    (Eq. 2)

    Step 1: Solve equation 1 for h in terms of r and substitue into equation 2.

    Step 2: Differentiate equation 2 wrt r. Set derivative equal to zero and solve for r.

    Step 3: Substitute r back into equation 1 to get h.

    You should then see that h = 2r.
    ...
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    can you guys actually show the work?
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    Quote Originally Posted by darkknight100 View Post
    can you guys actually show the work?
    you've been given the setup and directions ... follow through with it, please.
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    volume equation for in terms of and , then substitute the derived expression for

    what do you mean by that?
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    Quote Originally Posted by darkknight100 View Post
    volume equation for in terms of and , then substitute the derived expression for

    what do you mean by that?
    this ...

    k = \pi r^2 h

    \frac{k}{\pi r^2} = h
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  9. #9
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    Quote Originally Posted by skeeter View Post
    ...
    Yes, h=2r. Sorry, my mistake.
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    how do you solve for r for the following

    4pi r^3 + 2r - 2k
    ______________ = 0
    r^2
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  11. #11
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    Quote Originally Posted by darkknight100 View Post
    how do you solve for r for the following

    4pi r^3 + 2r - 2k
    ______________ = 0
    r^2
    how did you get that equation?
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    derivative of the surface area equation
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  13. #13
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    h = \frac{k}{\pi r^2}

    A = 2\pi r^2 + 2\pi r h

    substitute \frac{k}{\pi r^2} for h ...

    A = 2\pi r^2 + 2\pi r \cdot \frac{k}{\pi r^2}

    clean up the algebra ...

    A = 2\pi r^2 + \frac{2k}{r}

    A = 2\left(\pi r^2 + \frac{k}{r}\right)

    try finding \frac{dA}{dr} again.
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  14. #14
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    what's the derivative of
    k ?
    r
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  15. #15
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    Quote Originally Posted by darkknight100 View Post
    what's the derivative of
    k ?
    r
    what's the derivative of kr^{-1} ?

    (remember that k is a constant)
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