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Math Help - [SOLVED] Maximum error involved

  1. #1
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    [SOLVED] Maximum error involved

    A closed box to be found to have length 2ft, witdh 4ft and hight 3ft, where the measurement of each dimension is made with a maximum possible error of +-0.02ft. The top of the box is made from material that costs only 2 Dollars/ ft^2. The material for the sides and bottom costs only 1.50 Dollars/ ft^2. What is the maximum error involved in the computation of the cost of the box?

    1) In that question, gonna find top,bottom and side surface areas with errors and calculate the cost?

    2) Couldn't understand what it wants in question as answer. The difference between max and min value? ,, Only max value? or Only min value? ...

    3) Just noticed that i dont know how to make multiplying with errors. So how can i calculate areas with errors?

    p.s.--> I asked that question before but i just noticed that i didn't write the information about cost of sides and bottom. Now the question is complete.
    Last edited by Lafexlos; March 30th 2010 at 04:00 AM.
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  2. #2
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    Hello, Lafexlos!

    A closed box to be found to have length 2ft, width 4ft and height 3ft.
    The measurement of each dimension has a maximum possible error of ±0.02ft.

    The top of the box is made from material that costs $2/ft².
    The material for the sides and bottom costs $1.50/ft².
    What is the maximum error involved in the computation of the cost of the box?
    Let L = length, W = width, H = height.

    We are given: . \begin{Bmatrix}L \:=\: 2 && dL \:=\: \pm0.02 \\ W \:=\: 4 && dW \:=\: \pm0.02 \\ H \:=\: 3 && dH \:=\: \pm0.02 \end{Bmatrix}


    The area of the top is LW ft².
    . . Its cost is: . 2LW dollars.

    The rest of the box has a total area of: LW + 2LH + 2WH ft².
    . . Its cost is: . \tfrac{3}{2}(LW + 2LH + 2WH) dollars.

    The total cost is: . C \;=\;2LW + \tfrac{3}{2}(LW + 2LH + 2WH)

    . . We have: . C \;=\;\tfrac{7}{2}LW + 3LH + 3WH


    Take differentials: . dC \;=\;\tfrac{7}{2}(L\,dW + W\,dL) + 3(L\,dH + H\,dL) + 3(W\,dH + H\,dW)

    Substitute the given data:
    . . dC \;=\;\tfrac{7}{2}\bigg[2(0.02) + 4(0.02)\bigg] + 3\bigg[2(0.02) + 3(0.02)\bigg] + 3\bigg[4(0.02) + 3(0.02)\bigg]  \;=\;1.14


    Therefore, the maximum error of the cost is: . dC \:=\:\$1.14

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  3. #3
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    Arggh. Again that "error = differential" thing.
    Ok. Thanks.
    I think i won't forget from now on.
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