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Math Help - Functions and Their Graphs

  1. #1
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    Functions and Their Graphs

    Constructing an Open Box: An open box with a square base is requried to have a volume of 10 cubic feet.

    a) Express the amount A of material used to make such a box as a function of the length x of a side of the square base. MY answer: S=x^2 + 4x(10/x^2)
    b)How much material is required for a base 1 foot by 1 foot?
    c)How much material is required for a base 2 feet by 2 feet?
    d) Graph A=A(x). For what value of x is A smallest?

    I cannot figure out b,c,d...thank you
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  2. #2
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    Quote Originally Posted by spinka View Post
    Constructing an Open Box: An open box with a square base is requried to have a volume of 10 cubic feet.

    a) Express the amount A of material used to make such a box as a function of the length x of a side of the square base. MY answer: S=x^2 + 4x(10/x^2)
    b)How much material is required for a base 1 foot by 1 foot?
    c)How much material is required for a base 2 feet by 2 feet?
    d) Graph A=A(x). For what value of x is A smallest?

    I cannot figure out b,c,d...thank you
    So if x = length of one side of the square base, and the volume is 10 cu.ft., then,

    Volume = (x^2)*y = 10
    and y = 10/(x^2) ft.

    a) A = Area of base + area of 4 sides
    A = x^2 +4(x)(y)
    A = x^2 +4x[10/(x^2)] ----you got this far.
    A = x^2 +40/x
    In function form,
    A(x) = x^2 +40/x ----------answer.

    b) If x=1, what is A(1)?
    A(1) = 1^2 +40/1 = 1 +40 = 41 sq.ft. --------answer.

    c) If x=2, what is A(2)?
    A(2) = 2^2 +40/2 = 4 +20 = 24 sq.ft. --------answer.

    d) Graph A(x) = x^2 +40/x, and find the minimum A.

    I don't know how to graph this on calculators or in computers. I can do it by getting mamy points or ordered pairs.
    It looks like it is an skewed hyperbola, with branches in the 1rst and 3rd quadrants.

    In my rough sketch, minimum A appears to be between x=2 and x=3.

    That's as far as I can go.
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  3. #3
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    Well, ticbol pretty much summed it up nicely. Although, rather than checking it manually, you can find out the minimum yourself using the first derivative test.

    You got a) right, and it was S=x^2 + \frac{40}{x}. Now use the first derivative test to find out it's minimum.

    S'(x) = 2x - \frac{40}{x^2}

    Now find the points at which the first derivative is zero.

    S'(x) = 0

    0 = 2x - \frac{40}{x^2}

    \frac{40}{x^2} = 2x

    40 = 2x^3

    x = \sqrt[3]{20} \approx 2.714417617

    Which is in the interval Ticbol gave. If you plug in values before \sqrt[3]{20}, you'll find it negative. If you plug in values after \sqrt[3]{20}, you find it positive. Thus, it's a minimum. :-)
    Last edited by Chop Suey; July 17th 2008 at 12:49 AM.
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