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Math Help - Function Applications Lab

  1. #1
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    Function Applications Lab

    Hey there guys. My teacher really loves throwing wickedly difficult (or at least they are to us) problems at us, and he just recently gave us a new one that has all of us stumped yet again. As always, I appreciate any and all help given to me. Alas, onto the questions:

    1) The manufacturing industry invests significant effort in designing efficient packaging for its products. Suppose that a box is made by folding up the flaps after squares are cut from a 20-inch by 25-inch piece of cardboard. What size squares should be cut out to make a box with the greatest volume?

    2) A gardener has 120 feet of deer-resistant fence. She wants to enclose a rectangular vegetable garden in her backyard, and wants the enclosed area to be at least 800 square feet. What range of values is possible for the length of her garden. Write your answer in interval notation.

    My group members and I have taken a look at both problems, and honestly have no idea how to begin, so I came here in the hopes that somebody could help us out. Thanks again in advance. It is appreciated.
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  2. #2
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    Re: Function Applications Lab

    Hello, Seshiru!

    2) A gardener has 120 feet of deer-resistant fence.
    She wants to enclose a rectangular vegetable garden in her backyard,
    and wants the enclosed area to be at least 800 square feet.
    What range of values is possible for the length of her garden?
    Write your answer in interval notation.

    You should be able to visualize the rectangular garden,
    . . but here is a diagram anyway.
    Let x = length of the garden.
    Code:
                x
          *-----------*
          |           |
        y |           | y
          |           |
          *-----------*
                x
    The amount of fencing is: . 2x + 2y \:=\:120 \quad\Rightarrow\quad y \:=\:60-x .[1]

    The area of the garden is: . A \:=\:xy .[2]

    Substitute [1] into [2]: . A \;=\;x(60-x)

    The area is to be at least 800 square feet: . x(60-x) \;\ge\;800

    We have: . x^2 - 60x + 800 \;\le\;0

    This is a parabola: . y \;=\;x^2-60x + 800
    When is the graph negative (below the x-axis)?
    Answer: between its x-intercepts.

    We have: . x^2 -60x+800 \:=\:0 \quad\Rightarrow\quad (x-20)(x-40) \:=\:0

    Hence, the x-intercepts are 20 and 40.


    The range of x is: . [20,\,40]
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