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Math Help - optimization problems #2

  1. #1
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    optimization problems #2

    I have a question on these two problems where I have worked part of it already. Here is the question...
    1. Investigate the most economical shape of a cylindrical can by finding the height h and radius r that minimize the amount of metal needed to make a can of a given volume V. If we disregard any waste metal in the manufacturing process, then the problem is to minmize the surface area of the cylinder. The surface area is minimized when h=2r or h/r=2, that is, the height is the same as the diameter, resulting in a can with a square profile. But for most cans, the height is greater than the diameter and in fact, the ratio h/r varies from 2 up to about 3.8. Here you want to see if you can explain why this is the case.
    My answer is that the reason for this case is that they wanted to make the can easier to handle with more surface are to hold on to.
    2. The material for cans is cut from sheets of metal. The cylindriacal sides are formed by bending rectangles; these rectangles are cut from the sheet with little or no waste. But if the top and bottom discs are cut from squares of side 2r as in the figure, this leaves considerable waste metal, which may be recycled but has little or no value to the can makers. If this is the case, show that the amount of metal used is minimized when
    h/r=8/pie=about 2.55.
    Part of my answer and my understanding: I need to minimize the surface area
    I apply the variables r=radius and h=height
    my formula is SA=pie r^2
    I have no idea what to do next..Please help me. Thanks a lot.
    Last edited by mr fantastic; October 5th 2009 at 03:29 AM. Reason: Re-titled post
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  2. #2
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    1. another reason ... used to be that the material to make the top and bottom of the can was more expensive than the material to make the side.

    2. fixed volume, V ...

    V = \pi r^2 h

    h = \frac{V}{\pi r^2}

    A = 8r^2 + 2\pi r h

    A = 8r^2 + \frac{2V}{r}

     \frac{dA}{dr} = 16r - \frac{2V}{r^2} = 0

     r^3 = \frac{V}{8}

    \frac{h}{r} = \frac{V}{\pi r^3} = \frac{V}{\pi} \cdot \frac{8}{V} = \frac{8}{\pi}


    btw ... the Greek letter \pi is spelled "pi".



    this is "pie".
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  3. #3
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    opitimization problem

    I still don't understand how u got from

    to . Where did the r from 2pirh went?
    Thanks again for your answer.
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  4. #4
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    V = \pi r^2 h

    solve for h ...

    h = \frac{V}{\pi r^2}

    substitute \frac{V}{\pi r^2} for h in the term 2\pi r h ...

    2\pi r \cdot \frac{V}{\pi r^2}

    ... you simplify the last expression.
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  5. #5
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    Optimization

    In the beginning you got A= 8r^2+2pirh
    Where did you get 8r^2 from, when the given is h/r=8/pi = about 2.55?
    Thanks.
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  6. #6
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    the entire square that the one circular end was cut from had to be included in the total material used ... one square has dimensions 2r by 2r = 4r^2
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