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Math Help - strange small story problem-closed box area

  1. #1
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    strange small story problem-closed box area

    If a closed rectangular box should have volume 27ccm, find the dimensions
    that will make the surface area of closed box minimum possible.

    Nothing else is said about box...
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  2. #2
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    Quote Originally Posted by fiksi View Post
    If a closed rectangular box should have volume 27ccm, find the dimensions
    that will make the surface area of closed box minimum possible.

    Nothing else is said about box...
    Hi

    Let a, b and c the length of the box

    The volume of the box is V_0 = 27 = a b c

    The surface area is A(a,b,c) = 2ab + 2bc + 2ac = 2\left(ab + \frac{V_0}{a} + \frac{V_0}{b}\right)

    The extremum of A is obtained when \frac{\partial A}{\partial a} = 0 and \frac{\partial A}{\partial b} = 0

    Spoiler:

    b - \frac{V_0}{a^2} = 0

    a - \frac{V_0}{b^2} = 0

    V_0 = a^2b = ab^2 = abc \implies a = b = c

    The box must be a cube
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    Let a, b and c the length of the box

    The volume of the box is V_0 = 27 = a b c

    The surface area is A(a,b,c) = 2ab + 2bc + 2ac = 2\left(ab + \frac{V_0}{a} + \frac{V_0}{b}\right)

    The extremum of A is obtained when \frac{\partial A}{\partial a} = 0 and \frac{\partial A}{\partial b} = 0

    Spoiler:

    b - \frac{V_0}{a^2} = 0

    a - \frac{V_0}{b^2} = 0

    V_0 = a^2b = ab^2 = abc \implies a = b = c

    The box must be a cube
    So i don't even need 27ccm data, aye? I thought in the start that it should be 3x3x3, for the area to be min possible.
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  4. #4
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    I have considered the generic case but you need the data to compute a,b,c which are 3 cm
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  5. #5
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    Quote Originally Posted by running-gag View Post
    I have considered the generic case but you need the data to compute a,b,c which are 3 cm
    Of course I need it to know a,b,c. Your generic solution was nice as it applies all-round. I meant, I don't need any data about the box itself(that's what i meant), as it must be a cube. From 27ccm is 3,3,3.

    thx again. I guess I can apply this to all similar problems
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  6. #6
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    I prefer to use V_0, perform all the calculations and only at the end substitute the value because :
    - it can be applied to any value of V_0
    - it allows to find possible mistakes : when I find b - \frac{V_0}{a^2} = 0, I know that this is a homogeneous equation since b is expressed in cm, V0 in cm^3 and aČ in cm. If I use the value of V0, I am losing this possibility.
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