Results 1 to 3 of 3

Math Help - Help with optimization of a box

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    105

    Help with optimization of a box

    Concrete costs $15 per square meter, and glass costs $100 per square meter. If the volume of the pool must be 1000 cubic meters, what should the dimensions be to minimize the cost of the pool? 1 of the side is made of glass and the bottom and other 3 sides are made out of concrete.

    xyz = 1000
    z= 1000/xy
    surface area = xy +2yz + 2xz
    elimate:
    SA = xy + 2000/x + 2000/y
    Then I take the derivative in terms of x and y
    fx(x,y) = y -2000/x^2
    fy(x,y) = x -2000/y^2

    Then I'm lost here.

    Can anyone help? Thanks
    Last edited by khuezy; November 29th 2008 at 10:36 AM. Reason: additional info
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Let C_0 be the concrete cost ($15 per square meter)
    Let G_0 be the glass cost ($100 per square meter)
    Let V_0 be the volume of the pool (1000 cubic meters)
    Let z be the height of the pool

    xyz\;=\;V_0

    We try to minimize the cost C(x,y,z)

    C(x,y,z)\;=\;G_0yz + C_0(yz+2xz+xy)

    C(x,y)\;=\;G_0\frac{V_0}{x} + C_0(\frac{V_0}{x}+2\frac{V_0}{y}+xy)
    Let us differentiate
    x : -G_0\frac{V_0}{x^2} - C_0\frac{V_0}{x^2}+C_0y\;=\;0

    y : -2\frac{C_0V_0}{y^2} + C_0x\;=\;0

    Second equation gives x\;=\;2\frac{V_0}{y^2}

    Replacing in first equation gives at the end
    y\;=\;(4\frac{C_0}{G_0+C_0}V_0)^{\frac{1}{3}}

    Then
    y=8.05m
    x=30.86m
    z=4.03m

    PS : I suggest to name the constant in order to be able to find an eventual calculation mistake using the units

    For instance
    y\;=\;(4\frac{C_0}{G_0+C_0}V_0)^{\frac{1}{3}}

    \frac{C_0}{G_0+C_0} has no unit

    V_0^{\frac{1}{3}} has meter as unit
    Therefore the result is consistent (but it does not mean that it is true !)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,682
    Thanks
    614
    Hello, khuezy!

    Concrete costs $15/m, and glass costs $100/m.
    If the volume of the pool must be 1000 m,
    what should the dimensions be to minimize the cost of the pool?
    One of the sides is glass; the bottom and other 3 sides are concrete.

    xyz \:= \:1000 \quad\Rightarrow\quad z\:= \:\frac{1000}{xy} .[1]

    Surface area: . xy +2yz + 2xz

    This is all correct, but you forgot about the Cost.
    Code:
             *---------*
            /|        /|
           / |       / |z
          *---------*  |
          |         |  *
         z|    G    | /y
          |         |/
          *---------*
               x

    Suppose the front face is glass.
    Its area is xz m and costs $100/m.
    . . Its cost is: . 100xz dollars.

    The rest of the area is: . xy + 2yz + xz m at $15/m.
    . . Its cost is: . 15(xy + 2yz + xz) dollars.


    The total cost is: . C \;=\;15(xy + 2yz + xz) + 100xz \;=\;15xy + 30yz + 115xz

    Substitute [1]: . C \;=\;15xy + 30y\left(\tfrac{1000}{xy}\right) + 115x\left(\tfrac{1000}{xy}\right)

    . . and we have: . C \;=\;15xy + 30,\!000x^{-1} + 115,\!000 y^{-1}


    Set the partial derivatives equal to zero . . .

    . . \begin{array}{cccccc}C_x &=& 15y - 30,\!000x^{-2} &=& 0 & {\color{blue}[2]} \\<br /> <br />
C_y &=& 15x - 115,\!000y^{-2} &=& 0 & {\color{blue}[3]}  \end{array}

    Multiply [2] by x^2\!:\;\;15x^2y - 30,\!000\:=\:0 \quad\Rightarrow\quad y \:=\:\frac{2000}{x^2} .[4]

    Multiply [3] by y^2\!:\;\;15xy^2 -115,\!000\:=\:0  \quad\Rightarrow\quad 15xy^2 \:=\:115,\!000 .[5]


    Substitute [4] into [5]: . 15x\left(\tfrac{2000}{x^2}\right)^2 \:=\:115,\!000 \quad\Rightarrow\quad \frac{60,\!000,\!000}{x^3} \:=\:115,\!000<br />

    . . x^3 \:=\:\frac{60,\!000,\!000}{115,\!000} \:=\:\frac{12,\!000}{23} \quad\Rightarrow\quad x \:=\:\sqrt[3]{\frac{12,000}{23}} \quad\Rightarrow\quad\boxed{ x \;=\;\frac{10}{23}\sqrt[3]{6348}}


    Substitute into [4]: . y \;=\;\frac{200}{\left(\frac{10}{23}\sqrt[3]{6348}\right)^2} \quad\Rightarrow\quad\boxed{ y \;=\;\frac{5}{3}\sqrt[3]{6348}}


    Substitute into [1]: . z \;=\;\frac{1000}{\left(\frac{10}{23}\sqrt[3]{6348}\right)\left(\frac{5}{3}\sqrt[3]{6348}\right)} \quad\Rightarrow\quad\boxed{ z \;=\;\frac{5}{23}\sqrt[3]{6348}}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. optimization help!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 12th 2009, 12:54 AM
  2. Optimization
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 8th 2009, 02:09 PM
  3. Optimization
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 29th 2009, 10:56 AM
  4. optimization
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 12th 2008, 10:47 AM
  5. Optimization
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: October 13th 2008, 06:44 PM

Search Tags


/mathhelpforum @mathhelpforum