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Math Help - Maths assignment

  1. #1
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    Melaka, Malaysia
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    Maths assignment

    Well this is my end year exam assignment and i would appreciate it if sum1 can help me do it, I'm stuck :P.

    Aluminium Can

    The Muhibbah Company is a manufacturer of cylindrical aluminium tins. The manager plans to reduce the cost of production. The production cost is proportional to the area of the aluminium sheet used. The volume that each tin can hold is 1000cm³(1 Litre)

    1. Determine the value of h, r and hence calculate the ratio of h/r when the total surface area of each tin is minimum. Here, h cm denotes the height and r cm the radius of the tin.

    2. The top and bottom piece of the tin of height h cm are cut from square-shaped aluminium sheets.

    Determine the value for r, h and hence calculate the ratio h/r so that the total area of the aluminium sheets used for making the tin is minimum.

    3. Investigate cases where the top and bottom surfaces are cut from
    i) Equilateral Triangle
    ii) Regular hexagon
    Find the ratio of h/r for each case.

    Further Investigation

    Investigate cases where the top and bottom faces of the tin are being cut from aluminium sheets consisting shapes of regular polygons. From the result of your investigation, what conclusion can you derive from the relationship of the ratio of h/r and the number of sides of a regular polygon?

    Wastage occurs when circles are cut from aluminium sheet, which is not round in shape. Suggest the best possible shape of aluminium sheets to be used so as to reduce the production cost.

    My progress so far:

    I assumed that Q1 can be solved by simultaneous equation so i got these
    A=Area, h=height r= radius

    1000=πr²h
    h=1000/πr² ---(1)

    sub (1) into

    A=2πrh+2πr²
    A=2πr(1000/πr&#178+2πr²
    A=(2000πr/πr&#178+2πr²
    A=(2000r^-1)+2πr²

    dA/dr=-(2000/r&#178+4πr

    when Area is minimum, dA/dr=0

    -(2000/r&#178+4πr=0
    -2000+4πr^3=0

    We have: .4πr³ .= .2000
    . . . . . . . . . r³ .= .500/π
    . . . . . . . . . .r .= .cbrt(500/π) .≈ .5.419

    Then: .h .= .1000/πr² .= .1000/π[cbrt(500/π)]² .≈ .10.838



    . . . . . . . . . . . . . . .h . . . 1000/πr² . . .1000
    We have the ratio: . -- .= .---------- .= .------
    . . . . . . . . . . . . . . .r . . . . . . r . . . . . . πr³


    . . . . . . . . . . . . . . . . . . . . . . . .h . . . . 1000
    Since r = cbrt(500/π), we have: . -- .= .----------- .= .2
    . . . . . . . . . . . . . . . . . . . . . . . .r . . . π(500/π)


    The height is always twice the radius.
    The side view of the tin is a square.

    Q1 Solved, Move on 2 Q2

    Q2 must also use Simultaneous EQ

    1000=πr²h
    h=1000/πr² ----(1)

    A=8r²+2πrh ------(2)

    sub (1) into (2)

    A=8r²+2πr(1000/πr&#178
    A=8r²+(2000/r)

    dA/dr=16r-2000/r²

    When A minimum dA/dr=0

    16r-2000/r²=0
    16r³-2000=0
    16r³=2000
    r³=2000/16
    r³=125
    r=cbrt(125)
    r=5

    sub r= 5 into (1)

    h=1000/π(5)²
    h=1000/25π
    h=40/π
    h=12.7324

    ratio h/r=(1000/πr&#178/r=1000/πr³

    r=5, h/r=28/11

    Therefore, the ratio of h:r=28:11

    (Solved, please check again)Thx

    Moving on 2 Q3(i)

    height of equilateral triangle=ht
    length of each sides of triangle=x
    radius of circle=r
    Ac=Area of circle

    ht=x sin 60
    ht=(1/2)(sqrt[3x])
    r=(1/6)(sqrt[3x])
    3r=(1/2)(sqrt[3x])
    3r=ht

    ht= x sin 60
    3r/0.5=x
    6r=x
    .. ht=3r x=6r

    Ac=3r(3r)
    Ac=9r²

    Total Surface Area= A
    h=1000/πr²
    A=18r²+2πrh
    A=18r²+(2000/r)
    dA/dr=36r-(2000/r&#178

    when area min, dA/dr=0
    36r-(2000/r&#178=0
    36r³-2000=0
    r³=2000/36
    r³=500/9
    r=3.8157

    h=21.8626

    h/r=(1000/πr&#178/r=1000/πr³

    r³=500/9
    h/r=1000/π(500/9)
    h/r=63/11

    (Q3(i) Solved)Rock on to Q3(ii)(As usual, check this please)
    Last edited by bryankek; October 20th 2006 at 09:16 AM. Reason: Q2 done
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
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    Thanks
    542
    Hello, Bryankek!

    Your work on #1 is correct . . . excellent work!
    . . I didn't approximate like you did.


    The Muhibbah Company is a manufacturer of cylindrical aluminium tins.
    The manager plans to reduce the cost of production.
    The production cost is proportional to the area of the aluminium sheet used.
    The volume that each tin can hold is 1000 cm(1 Litre)

    1. Determine the value of h, r and hence calculate the ratio of h/r
    when the total surface area of each tin is minimum.
    Here, h denotes the height and r the radius of the tin.

    My progress so far:

    I assumed that Q1 can be solved by simultaneous equation so i got these
    A = Area, h = height, r = radius

    1000 = πrh . . h = 1000/πr . (1)

    sub (1) into: .A .= .2πrh + 2πr
    . . A .= .2πr(1000/πr) + 2πr
    . . A .= .2000πr/πr + 2πr
    . . A .= .2000r^{-1} + 2πr

    dA/dr .= .-2000/r + 4πr .= .0

    Multiply by r: . -2000 + 4πr .= .0
    . . 4πr .= .2000
    . . . .r .= .1750/11 . . . ok

    We have: .4πr .= .2000
    . . . . . . . . . r .= .500/π
    . . . . . . . . . .r .= .cbrt(500/π) . .5.419

    Then: .h .= .1000/πr .= .1000/π[cbrt(500/π)] . .10.838



    . . . . . . . . . . . . . . .h . . . 1000/πr . . .1000
    We have the ratio: . -- .= .---------- .= .------
    . . . . . . . . . . . . . . .r . . . . . . r . . . . . . πr


    . . . . . . . . . . . . . . . . . . . . . . . .h . . . . 1000
    Since r = cbrt(500/π), we have: . -- .= .----------- .= .2
    . . . . . . . . . . . . . . . . . . . . . . . .r . . . π(500/π)


    The height is always twice the radius.
    The side view of the tin is a square.

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  3. #3
    Newbie
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    Oct 2006
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    Melaka, Malaysia
    Posts
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    wow!thx, am working on 2nd 1 now
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