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Thread: Parallelepiped

  1. #1
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    Parallelepiped

    I'm stuck with my homework in the following task:

    a rectangular parallelepiped ( b*b*h ) holds K dm^3.
    Show that the total surface area of the side faces and top and bottom surface is minimal if h/b = 1

    see picture in the link: https://myalbum.com/album/k1KaEgC1WieE



    I must first describe a function / formula with one variable , who can help me thus on the way?
    I fail to derive it by myself.
    Last edited by westerwolde; Apr 3rd 2017 at 12:07 AM.
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  2. #2
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    Re: Parallelepiped

    Hey westerwolde.

    Can you show us the total surface area? [Hint - you have six sides to consider].
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    Re: Parallelepiped

    I think it is: 2*b + 4*h
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  4. #4
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    Re: Parallelepiped

    Hint - The surface area of a b x b block is b*b and for b x h it is b*h.

    Take a closer look at the box and write out the terms for all sides you need to include.
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  5. #5
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    Re: Parallelepiped

    Quote Originally Posted by westerwolde View Post
    I think it is: 2*b + 4*h
    This is a length, not an area.
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    Re: Parallelepiped

    I looked at the picture again, i think it should be : 4*bh + 2*bb
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  7. #7
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    Re: Parallelepiped

    Quote Originally Posted by westerwolde View Post
    I looked at the picture again, i think it should be : 4*bh + 2*bb
    correct ...

    $S= 2(2bh+b^2)$

    what method will you use to determine the minimum surface area, $S$ ?
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  8. #8
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    Re: Parallelepiped

    So you want to minimize S= 4bh+ 2b^2 subject to the constraint V= b^2h= K.

    The simpler in concept is to solve the constraint for h= K/b^2 and replace h in the formula for S so that you have a function in the single variable b to minimize. Do you know how to do that? Set the derivative of S, with respect to b, equal to 0 and solve for b.
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    Re: Parallelepiped

    I was going to do it as described by HallsofIvy, in another task has already succeeded but ther was given the formula.
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    Re: Parallelepiped

    I think there is something wrong with my development, can you see what goes wrong?

    h= k/b^2
    s= 4bh+ 2b^2

    I replace h in s

    => s= 4k/b^2 + 2b^2

    ds/db=0 => -4k/b^2 + 4b=0

    and when I now solve for b,I get specific outcomes ..
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  11. #11
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    Re: Parallelepiped

    $S = 2(2bh + b^2)$

    $b^2h = k \implies \color{red}{h = \dfrac{k}{b^2}} \implies S = 2\left(2b \cdot \dfrac{k}{b^2} + b^2\right) = 2\left(\dfrac{2k}{b} + b^2\right)$

    $\dfrac{dS}{db} = 2\left(-\dfrac{2k}{b^2} + 2b\right)$

    $2\left(-\dfrac{2k}{b^2} + 2b\right) = 0 \implies b = \dfrac{k}{b^2}$

    what else equals $\dfrac{k}{b^2}$ ?
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    Re: Parallelepiped

    h is equal to k/b^2 ?

    How do you get b= k/b^2 ?
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    Re: Parallelepiped

    How do you get b= k/b^2 ?
    Quote Originally Posted by skeeter View Post
    $\dfrac{dS}{db} = 2\left(-\dfrac{2k}{b^2} + 2b\right)$

    $\color{red}{2\left(-\dfrac{2k}{b^2} + 2b\right) = 0}$
    I solved the above equation ...
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  14. #14
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    Re: Parallelepiped

    okay, I had solved the equation as b=^3√ k , but that's wrong.

    what do you mean with ; what else equals k/b^2 ? How do I go from here?
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  15. #15
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    Re: Parallelepiped

    I had solved the equation as b=^3√ k , but that's wrong.
    $2\left(-\dfrac{2k}{b^2}+2b\right) = 0 \implies 4\left(-\dfrac{k}{b^2}+b\right) = 0 \implies \color{red}{b = \dfrac{k}{b^2}}$

    ... this is the value of $b$ that minimizes the volume.


    what do you mean with ; what else equals k/b^2 ? How do I go from here?
    from the original equation for volume ...

    $k = b^2 h \implies \color{red}{h = \dfrac{k}{b^2}}$


    take a look again at your first post ... what you were supposed to show?
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