College Algebra

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On a rectangular piece of cardboard with perimeter 11 inches, three parallel and equally spaced creases are made. The cardboard is then folded along the creases to make a rectangular box with open ends. Letting x represent the distance (in inches) between the creases, use a graphing calculator to find the value of that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places.

values of x that maximizes volume = in
Maximum volume= in^3

  • College Algebra -

    I would start with a drawing as I do for most problems. Draw a reactangle 4x one side and b the other.

    The perimeter is then 4x+b+4x+b=11

    8x+2b=11

    if this is folded to a tube then the volume of the tube is bx^2, i.e. a tube with cross sectional area x^2 and length b.

    so V=bx^2

    rearrangen and substitute for b into the equation above gives

    8x+2V/(x^2) = 11

    or

    8x^3+2V=11x^2

    or

    V=5.5x^2-4x^3

    which you can plot to find max V

    I got 1.54 in^3 as the max volume

    but check the maths!

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