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March 26, 2017

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Donald wants to build an aquarium with internal volume of 1 m^3 that made of flat glass sheet. The bottom and side walls of the aquarium have to be made of 7 and 5 mm-thick glass sheet, respectively. Current price of 5 mm and 7 mm galss sheet are $3.5 and $5.6 per m^2, respectively. Determine the dimensions (wide x length x depth) of the aquarium such that the price of required glass sheets is minimum.

  • math - ,

    with width x, length y, height z, in m,

    volume v = xyz = 1
    bottom area is xy
    side areas are xz and yz

    cost
    c = 5.6xy + 2*3.5(xz+yz)
    c = 5.6xy + 7xz + 7yz

    Using Lagrange multipliers, let
    f(x,y,z) = 5.6xy + 7xz + 7yz
    g(x,y,z) = xyz
    we have to solve

    ∇f = λ∇g
    g(x,y,z) = 1

    5.6y+7z = λyz
    5.6x+7z = λxz
    7x+7y = λxy
    xyz = 1

    5.6xy + 7xz = λxyz
    5.6xy + 7yz = λxyz
    7xz + 7yz = λxyz
    xyz=1
    dividing out the xyz, we get
    5.6xy + 7xz = λ
    5.6xy + 7yz = λ
    7xz + 7yz = λ

    7xz + 7yz = 5.6xy + 7xz
    now, since z = 1/xy,
    7x/y + 7y/x = 5.6xy + 7x/y

    7y/x = 5.6xy
    7/x = 5.6x
    x^2 = 7/5.6
    x = 1.25

    Now, going back to our original equations,

    c = 5.6(1.25)y + 7z(1.25+y)
    z = 1/(1.25y)

    c = 7y + 5.6 + 7/y
    dc/dy = 7 - 7/y^2
    dc/dy = 0 when y = 1

    so, the aquarium is

    1.25 x 1 x 0.8

    cost of materials is
    c(1.25,1,0.8) = 5.6(1.25) + 7(0.8)(1.25+1)
    = 7 + 5.6(2.25)
    = 19.60

    Hmmm. According to wolframalpha, minimum c occurs at
    (x,y,z) = (∛(5/4),∛(5/4),∛(16/25))
    c = 21∛(4/5) = 19.4947

    That is, in fact, less than my value using Lagrange multipliers. Better check my math (as usual)

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