Results 1 to 2 of 2

Math Help - Optimizing a rectangle

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    19

    Optimizing a rectangle

    Hi, Im having a problem finishing off a problem. It is a calc ab problem

    A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and 5$ for the sides, what is the cost of the least expensive tank?

    So the information i gathered was:
    SA=LW+2(LH+WH)
    W=4
    V=36

    S0 since V=LWH=36, 4LH=36, L=9/H
    Then, SA= 36/h+18+8h
    At first i thought I would just need to optimize that and go -36h^-2 + 8 and found the min was at 2.121 but then I thought you might need to factor in the price using another equation. Could someone help me with this problem?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,715
    Thanks
    633
    Hello, calc_help123!

    A tank with a rectangular base and rectangular sides is to be open at the top.
    It is to be constructed so that its width is 4 m and its volume is 36 m.
    If building the tank costs $10/m for the base and $5/m for the sides,
    what is the cost of the least expensive tank?
    We can not separate the surface area and the cost of materials.
    . . We must come up with a single Cost Function ... which we will minimize.
    Code:
             * - - - - *
            /|        /|   
           / |       / | y
          * - - - - *  |
          |         |  *
        y |         | /
          |         |/ 4
          * - - - - *
               x

    The width of the box is 4 m.
    Let x = length.
    Let y = height.

    The volume will be 36 m.
    So we have: . 4xy \:=\:36 \quad\Rightarrow\quad y \:=\:\frac{9}{x} .[1]


    Now we will create a Cost Function.


    The front and back has area: . 2 \times xy \:=\:2xy m.

    The left and right has area: . 2 \times (4y) \:=\:8y m.

    The total area of the sides of the box is: . 2xy + 8y m.

    . . At $5/m, the sides will cost: . 5(2xy + 8y) \:=\:10xy + 40y dollars.


    The base has area: . 4x m.

    . . At $10/m, its cost is: . 10(4x) \:=\:40x dollars.


    Hence, the total cost is: . C \;=\;10xy + 40y + 40x .[2]


    Substitute [1] into [2]: . C \;=\;10x\left(\frac{9}{x}\right) + 40\left(\frac{9}{x}\right) + 40x

    . . and we have: . C \;=\;90 + 360x^{-1} + 40x


    And that is the function we must minimize . . .

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: June 2nd 2011, 04:56 PM
  2. Replies: 3
    Last Post: March 24th 2010, 06:32 PM
  3. Replies: 1
    Last Post: November 23rd 2009, 06:41 PM
  4. Optimizing
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 1st 2009, 01:50 PM
  5. Optimizing a rectangle
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 18th 2009, 05:55 PM

Search Tags


/mathhelpforum @mathhelpforum