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Optimisation Problem Using Derivatives

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data
    a cylindrical tin can with volume 0.3l is being made, with the top and bottom sufaces twice the thickness as the sides.

    Show that a height to radius ration of [itex]h=4r[/itex] will minimise the amount of aluminium required.

    2. Relevant equations
    [itex]V=\pi r^2 h \\
    A = 2 \pi r^2 + 2 \pi r h
    [/itex]

    3. The attempt at a solution
    I have kind of done it, but I was not sure how to factor in the fact that the top and bottom is twice the thickness of the sides. But my answer is a half of what it should be, or rather I am getting a ratio of [itex]h=2r[/itex] , so I know why my answer is wrong, but would appreciate a little help on how to do it correctly.

    First what I did was, using the first equation in the relevent equations section, solved for height and then subsituted it into the equation for surface area and got...
    [itex]
    A=2 \pi r^2+ \frac{2 \pi r V}{\pi r^2} = 2 \pi r^2+ \frac{2 V}{\pi r}
    [/itex]

    Then took the derivative with respect to the radius and set to zero...
    [itex]
    \frac{dA}{dr}=4 \pi r - \frac{2V}{r^2}=0
    [/itex]
    Then I multiplied through by r^2
    [itex]
    0=4 \pi r^3 - 2V \\
    r=\sqrt[3]{\frac{2V}{4 \pi}} = \sqrt[3]{\frac{V}{2 \pi}}
    [/itex]

    Then found the radius
    [itex]
    r= \sqrt[3]{\frac{0.3}{2 \pi}}=0.3628m
    [/itex]

    Then with that the height
    [itex]
    h=\frac{V}{\pi r^2}=\frac{0.3}{\pi \times 0.3628^2} = 0.7255m
    [/itex]

    And one can see that my height is twice the radius rather than four times. And I know that the fact the top and bottom are twice as thick is why, so could I just leave it as it is and say that, or there is probably a much easier way to factor in the twice thickness thing before hand to get the correct result so would appreciate a little help/advice. Thanks :)
     
  2. jcsd
  3. Nov 12, 2014 #2

    DrClaude

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    Staff: Mentor

    You don't take into account thickness in any way, even for the sides. You have to rethink your approach.
     
  4. Nov 12, 2014 #3
    The area of the cylinder is not directly useful - you should instead find (say) M, the volume of material required for the can. This will be some multiple of t, the thickness of the walls of the can.
     
  5. Nov 12, 2014 #4

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    You can factor in the extra thickness of the top and bottom of the can by increasing the area of those two surfaces proportionately, as if the can was being made of a single sheet of metal having the same thickness. You wind up using the same amount of aluminum to make the can.
     
  6. Nov 12, 2014 #5
    Thanks.

    So that would mean multiply the area of those by 2 and then the total area would then be
    [itex]
    4 \pi r^2+ \frac{2 V}{\pi r}
    [/itex]

    Then finding derivative
    [itex]
    \frac{dA}{dr}=8 \pi r - \frac{2V}{r^2}=0 \\
    8 \pi r^3 - 2V=0 \\
    r=\sqrt[3]{\frac{2V}{8 \pi }} = \sqrt[3]{\frac{V}{4 \pi }} = 0.2879m
    [/itex]

    And the height then
    [itex]
    h=\frac{V}{\pi r^2}=\frac{0.3}{\pi \times 0.2879^2} = 1.152m
    [/itex]

    Which then satisfies the ratio
    [itex]h=4r \\
    \frac{h}{r}=4 \\
    \frac{1.152}{0.2879}=4
    [/itex]

    Thanks guys. I knew it would be something so simple :)
     
  7. Nov 12, 2014 #6

    SteamKing

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    Homework Helper

    Now, the volume of the can, 0.3 L, has to handled using appropriate units in order to find the correct values of r and h for the can. Merely plugging in 0.3 L for the volume is not correct, because a liter is a unit of capacity and not length. If you consider the r and h values you calculated, it's a can with pretty awkward dimensions.
     
  8. Nov 12, 2014 #7
    The question actually gave it as 300ml but I changed it to 0.3L, as I am so used to change milimeters/centimenters to metres to use in equations I just did it off the bat with the volume too. Isn't 300ml the same as 300cm^3 so 0.3L the same as 0.3m^3 ? So would I need to use 0.3^3 ?
     
    Last edited: Nov 12, 2014
  9. Nov 12, 2014 #8
    I think we're beyond a keg there - I would call that a barrel. :)

    A litre is a volume 1dm^3. 1dm, one decimetre, is a unit that has almost no usage outside the definition of a litre, but is 0.1m which is also 10 cm.

    Yes, 300ml = 300 cm^3 = 0.3L

    However, 100cm = 1m, so, basic maths, how many cm^3 in a m^3 ?
     
    Last edited: Nov 12, 2014
  10. Nov 12, 2014 #9
    Sorry yes, in my head I was still thinking mili- rather than centi- . So if 300ml=300cm^3, it should equal 3m^3 . .
     
  11. Nov 12, 2014 #10

    SteamKing

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    Is that your final answer? Think how many cc are in 1 cubic meter. The answer isn't 100 cc = 1 cubic meter.
     
  12. Nov 12, 2014 #11
    lol. I didnt even try to think about it, I just assumed but just worked it out I think.

    One centimetre is a 1/100 m so...
    [itex]
    \frac{1}{100}\times \frac{1}{100}\times \frac{1}{100} \\
    \frac{1}{100 \times 100 \times 100} \\
    \frac{1}{1 \times 10^6}
    [/itex]

    So therefore there is 10^6 cc's in one cubic metre
     
  13. Nov 12, 2014 #12

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    So, if you have a can with a volume of 300 cc, how many cubic meters is that equivalent to?
     
    Last edited: Nov 12, 2014
  14. Nov 12, 2014 #13
    [itex]300 \times 10^{-6}=3 \times 10^{-4}[/itex]

    Using that value for the volume I then get like 28.79cm for the radius and 115.21cm for the height.
     
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