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Math Help - Maximum and minimum applications

  1. #1
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    Maximum and minimum applications

    Hey guys, need a hand with a problem.

    A cuboid parcel is to be tied with a length of string both across and longways, forming a cross on top of it. The parcel as a length of x, width of x and height of y. The fixed volume is 8000cm cubed.

    Find the length, L, of string in terms of x and y.

    What is the volume of the box in terms of x and y, and hence find y in terms of x.

    Using the previous equation, find the length of string, L, in terms of x only.

    What is the minimum length of L?

    Thanks.
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  2. #2
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    Re: Maximum and minimum applications

    I get L=4x+4y V=x^2y So x^2y=8000 y=8000/x^2 S0 L=4x+32000/x^2 Then differentiate and put =0
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  3. #3
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    Re: Maximum and minimum applications

    Ok so i did that, it came out x=25.2. That would be a side length yeah? So then i sub that into the L=4x+32000/x^2 equation to get L, then back into the L=4x+4y equation to get y?
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  4. #4
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    Re: Maximum and minimum applications

    Yes 25.2 is correct. You weren't asked to find L and x and y.
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