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Math Help - Mins and maxes

  1. #1
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    Mins and maxes

    A man has 340 yards of fencing for enclosing two seprate fields, one of which is to be a rectangle twice as long as it is wide and the other is a square. The square field must contain at least 100 square yards and the rectanglar one must contain at least 800 square yards.
    a) if x is the width of the rectangular field, what are the maximum and minimum possible values of x?
    b)What is the greatest number of square yards that can be enclosed in the two fields? Explain.
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by calc_help123 View Post
    A man has 340 yards of fencing for enclosing two seprate fields, one of which is to be a rectangle twice as long as it is wide and the other is a square. The square field must contain at least 100 square yards and the rectanglar one must contain at least 800 square yards.
    a) if x is the width of the rectangular field, what are the maximum and minimum possible values of x?
    b)What is the greatest number of square yards that can be enclosed in the two fields? Explain.

    Let x = width of rectangle, so length of rectangle = 2x

    Let y = length of one side of square

    340 yards of fencing is available to enclose the two fields, so:

    perimeter of rectangle + perimeter of square = 340 yds

    <br />
6x + 4y = 340

    Then find the areas of the two fields:

    Area of rectangle = 2x^2 = 800
    Area of square = y^2 = 100

    Now use these equations to answer the questions. Give it a try and come back if you are still stuck!
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    I too had reached up to this point. For x of the rectangle, it is greater than or at least 20 yards and greater than or at least 10 yards for the square. but that is where i get stuck. could u or someone please explain the rest of the problem.
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  4. #4
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    Hello, calc_help123!

    A man has 340 yards of fencing for enclosing two separate fields,
    one of which is to be a rectangle twice as long as it is wide and the other is a square.

    The square field must contain at least 100 square yards
    and the rectanglar one must contain at least 800 square yards.

    a) If x is the width of the rectangular field, what are the maximum
    and minimum possible values of x?

    The rectangular field has width x and length 2x.
    The area of the rectangular field is: . A_1 \:=\:2x\cdot x \:=\:2x^2 yd˛.

    Its perimeter is 6x yards, leaving 340-6x yards for the square field.

    The side of the square field is: . \frac{340-6x}{4}\,=\,\frac{170-3x}{2} yards.

    The area of the square field is: . A_2 \:=\:\left(\frac{170-3x}{2}\right)^2 yd˛.


    We are told that: . A_1 \geq 800
    . . 2x^2 \:\geq \:800 \quad\Rightarrow\quad x^2 \:\geq \:400 \quad\Rightarrow\quad x \:\geq \:20

    We are told that: . A_2 \geq 100
    . . \left(\frac{170-3x}{2}\right)^2 \:\geq\:100 \quad\Rightarrow\quad \frac{170-3x}{2} \:\geq \:10 \quad\Rightarrow \quad -3x \:\geq\:-150 \quad\Rightarrow\quad x \:\leq\:50


    Therefore: . 20 \:\leq \:x \:\leq\:50




    b) What is the greatest number of square yards that can be enclosed
    in the two fields? Explain.

    The total area is: . A \:=\:2x^2 + \left(\frac{170-3x}{2}\right)^2  \quad\Rightarrow\quad A \;=\;\tfrac{1}{4}\left(17x^2 - 1020x + 28,\!900\right)

    The graph is an up-opening parabola.
    Its vertex (lowest point) is: . (30,3400)
    Code:
           |
           |                       *
           |
           |                      *:
           |                     * :
           |     *             *   :
           |     :  *       *      :
           |     :      *          :
           |     :      :          :
           |     :      :          :
         --+-----+------+----------+---
           |    20     30         50
           |
    Hence, maximum area occurs at the endpoints of the interval.

    Looking at the graph, we suspect it happens at x = 50,
    . . but we can verify this.

    At x=20\!:\;A \;=\;\tfrac{1}{4}\left(17\!\cdot\!20^2 - 1020\!\cdot\!20 + 28,\!900\right) \:=\:3825\text{ yd}^2

    At x=50\!:\;A \;=\;\tfrac{1}{4}\left(17\!\cdot\!50^2 - 1020\!\cdot\!50 + 28,\!900\right) \:=\:5100\text{ yd}^2 . . . There!


    The rectangle will be 50 × 100 yards.
    The square has sides of 10 yards.


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