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Math Help - Optimization: Minimization

  1. #1
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    Optimization: Minimization

    Hey everyone, here is a minimization problem from the Optimization II chapter in my book. I have the maximization down, but this I can't seem to get. Can someone work through this one for me and show the steps so I can see the flow?


    Minimizing Packagin Costs.
    A rectangular box is to have a square base and a volume of 20 ft cubed. If the material for the base costs 30 cents/square foot, the material for the sides costs 10 cents/square foot, and the material for the top costs 20 cents/square foot, determine the dimensions of the box that can be constructed at minimum cost.
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  2. #2
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    Volume is given by:

    x^{2}y=20.........[1]

    Cost with regard to surface area is:

    0.30x^{2}+0.40xy+0.20x^{2}..........[2]

    Solve [1] for y and sub into [2]:

    0.30x^{2}+0.40x(\frac{20}{x^{2}})+0.20x^{2}=0.5x^{  2}+\frac{8}{x}

    Diferentiate, set to 0 and solve for x. y will follow.
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  3. #3
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    Hello, 2taall!

    We use the same procedure of Minimization as for Maximization.
    The tricky part (of course) is setting up the function to optimize.
    I'll walk you thorough the set-up.


    A rectangular box is to have a square base and a volume of 20 ft³.
    The material for the base costs 30˘/ft², for the sides 10˘/ft², and the top 20˘/ft².
    Determine the dimensions of the box that can be constructed at minimum cost.
    Code:
                * - - - - *
              /         / |
            /         /   |
          * - - - - *     |
          |         |     |
          |         |     |
        y |         |     *
          |         |   /
          |         | / x
          * - - - - *
               x
    The volume of the box is: . x^2y \:=\:20\quad\Rightarrow\quad y \:=\:\frac{20}{x^2} .[1]

    The base has area x^2 ft². .Its cost is: . 30x^2 cents.

    The top has area x^2 ft². .Its cost is: . 20x^2 cents.

    The sides have area 4xy ft². .Their cost is: . 40xy cents.

    Hence, the total cost is: . C \;=\;30x^2 + 20x^2 + 40xy\quad\Rightarrow\quad C \:=\:50x^2 + 40xy .[2]


    Substitute [1] into [2]: . C \:=\:50x^2 + 40x\left(\frac{20}{x}\right)

    Hence, we must minimize: . C \;=\;50x^2 + 800x^{-1}

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  4. #4
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    Ok guys,

    I differentiated .50x^2 + 8/x

    I got f'x= x - (8/ x^2 )=

    solving for x:
    -8/x^2 = -x =

    -8= -x^3 =

    8=x^3 =

    taking the cubed root of each side:
    x=2

    Can you guys check that math?




    Now we have(hopefully) x=2 and y= 20/x^2


    EDIT: is the x=2 plugged back into the (x^2)y=20?
    in which case [(2)^2]y=20 would make y=5
    so the box would be 5 x 2 x 2 ?

    Thanks everyone.
    Last edited by 2taall; November 10th 2007 at 06:32 AM.
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  5. #5
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    I appreciate all the help you guys are giving. I only wish that I had found this forum at the beginning of my Calculus course!

    Was everything OK as far as my work went?

    Thanks.
    Last edited by 2taall; November 11th 2007 at 09:54 AM.
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  6. #6
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    Hello,2taal!

    Your work looks great!

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  7. #7
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    Thanks for looking it over!

    Feel free to dabble in my:
    Very hard area and definite integral problem
    thread in the calculus forum too!
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