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Math Help - Calc One Maximization Problem

  1. #1
    Forum Admin topsquark's Avatar
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    Calc One Maximization Problem

    I recently completed a Calculus proficiency exam on tutor.com. Here is a question they asked me:

    An open box is to be made from a 20 inch square piece of cardboard by cutting four equal square pieces from each corner and turning up the sides. What is the side length of the pieces that should be cut out so that the box will have maximum volume?

    This question can be done, using partial derivatives, but this solution (that the cardboard is a square with sides \sqrt{20} inches long) was not listed as one of the possible answers. Not to mention the exam was supposed to use Calc I techniques, of which I doubt partial derivatives are included.

    Is there any way to generate a solution here without resorting to partial derivatives? (I finally answered the question by process of elimination: only one answer listed was physically possible.)

    -Dan
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  2. #2
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    Hey TP:

    If you let the side of one of the little squares be x, then the volume of the box would be:

    V=x(20-2x)(20-2x)=4x^{3}-80x^{2}+400x

    Differentiate: 12x^{2}-160x+400=3x^{2}-40x+100=(x-10)(3x-10)



    The solutons are 10/3 and 10

    x=10/3 is the one to use.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    An open box is to be made from a 20 inch square piece of cardboard by cutting four equal square pieces from each corner and turning up the sides. What is the side length of the pieces that should be cut out so that the box will have maximum volume?
    Quote Originally Posted by galactus View Post
    Hey TP:

    If you let the side of one of the little squares be x, then the volume of the box would be:

    V=x(20-2x)(20-2x)=4x^{3}-80x^{2}+400x

    Differentiate: 12x^{2}-160x+400=3x^{2}-40x+100=(x-10)(3x-10)



    The solutons are 10/3 and 10

    x=10/3 is the one to use.
    Ah, I see what happened. There are two possible ways to read the phrase "20 inch square piece of cardboard." You took it to be a 20 in x 20 in piece of material (which generates the only possible physically correct answer) and I took it to mean that the area of the cardboard is 20 in^2.

    Hmmm... Anyone else think the wording here is ambiguous? Or am I just having a perception problem today?

    -Dan
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  4. #4
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    20 inch square is not 20 square inches.

    Umm, the in^2.
    20 in^2
    No, it's not that here.
    Otherwise it is written in in^2.

    20 inch square is more like 20-inch square.
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  5. #5
    Eater of Worlds
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    I recently completed a Calculus proficiency exam on tutor.com.
    I done that a few months back. I passed all the tests with flying colors, but am unable to participate because I am relegated to a damn, dial-up.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by galactus View Post
    I done that a few months back. I passed all the tests with flying colors, but am unable to participate because I am relegated to a damn, dial-up.
    Sorry to hear that. My woes include moving (for the second time in a month) and from what I've been told I'll lose my DSL for about two weeks. Ugh.

    -Dan
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