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Math Help - Quadratic Functions- Maximum and Minimum Problems

  1. #1
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    Quadratic Functions- Maximum and Minimum Problems

    This problem is from the Addison-Wesley Mathematics 11 Text Book (Western Canada Edition)

    Chapter 2 - Quadratic Functions
    Section 2.5 - Maximum and Minimum problems

    pg. 131 Problem #16

    A 30-cm piece of wire is cut in two. One piece is bent into the shape of a square. The other piece is bent into the shape of a rectangle with a length-to-width ratio of 2:1. The sum of the areas of the square and rectangle is a minimum. What are the lengths of the two pieces?

    I have tried to solve most of the problem already; I just don't know how to end it. The back of the text says that the answer is 14.1cm, 15.9 cm but it could always be wrong. I would really appreciate if someone could help me Here is my attempt at it: (Im sorry if its messy, I tried to do the best I could on the computer.)

    Piece #1 + Piece #2 = 30cm
    Piece #1 = x
    Piece #2 = 30 - x

    Area of the square = 1/4x2

    Area of the rectangle = 2w2

    6w = 30-x
    w = (30-x)
    6
    w = 5 - (x/6)

    Area of the rectangle = 2w2
    = 2(5 - (x/6))2
    = (10 (x/3))2

    Sum of the areas = 1/4x2 + (10 (x/3))2

    I tired this equation in my graphing calculator but came up with a minimum vertex of x=9.23 and y=69.23, which is impossible because y must be less than 30.

    Thank you!

    -Olivia
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  2. #2
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    Hello, Olivia!

    You made a vailiant effort, but your set-up is off . . .


    A 30-cm piece of wire is cut in two.
    One piece is bent into the shape of a square.
    The other piece is bent into the shape of a rectangle with a length-to-width ratio of 2:1.
    The sum of the areas of the square and rectangle is a minimum.
    What are the lengths of the two pieces?
    Let x = length of wire for the rectangle.
    Then 30-x = length of wire for the square.

    The rectangle looks like this:
    Code:
                 2k
          * - - - - - - - *
          |               |
        k |               | k
          |               |
          * - - - - - - - *
                 2k
    The perimeter of the rectangle is: . 2k + k + 2k + k \:=\:6k
    Its wire is x cm long: . 6k = x\quad\Rightarrow\quad k = \frac{x}{6} . [1]

    The area of the rectangle is: . A_r \:=\:(2k)(k) \:=\:2k^2 . [2]

    Substitute [1] into [2]: . A_r \;=\;2\left(\frac{x}{6}\right)^2\quad\Rightarrow\q  uad\boxed{A_r\:=\:\frac{x^2}{18}}


    The square looks like this:
    Code:
          * - - - - - - *
          |             |
          |             |
          |             | (30-x)
          |             |
          |             |
          * - - - - - - *
              (30-x)
    The perimeter of the square is: 30 - x cm,
    . . so its side is: \frac{1}{4}(30 - x)

    The area of the square is: . A_s \;=\;\left(\frac{30-x}{4}\right)^2\quad\Rightarrow\quad\boxed{A_s\;=\;  \frac{x^2 - 60x + 900}{16}}


    Hence, the total area is: . A \;=\;\frac{x^2}{18} + \frac{x^2 - 60x + 900}{16} \;=\;\frac{1}{144}(17x^2 - 540x + 8100)

    And that is the function we must minimize . . .

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  3. #3
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    Thank you

    Thank you very much, I really do appreciate it
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