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Thread: inequality word problem

  1. #1
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    inequality word problem

    Question: Susie is designing a stand for an exhibition. It is to have display screens round three sides and to be open on the other side. 40 m of screens are available for the stand and the owners want a rectangular floor space of at least 150 square metres.
    What limits must Susie work within for the lengths of the sides of the stand?


    total length of screen = 40 + 40 + 40 = 120
    perimeter = 2l + w

    2l +w = 120
    w = 120 -2l
    or  l =60 - \frac{1}{2}w

    w(60 - \frac{1}{2}w)\geq 150
     60w-\frac{1}{2}w^2\geq 150


    i need up with this but  w^2-120w+300\leq 0
    can some one tell me what i am doing wrong




    thank you for helping me
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  2. #2
    Junior Member Ahri's Avatar
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    Re: inequality word problem

    Doesn't $40m$ of screens available mean that the sum of three screens $S_1 + S_2 + S_3 = 40m$?
    Last edited by Ahri; Jun 11th 2017 at 01:36 PM.
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  3. #3
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    Re: inequality word problem

    Quote Originally Posted by Ahri View Post
    Doesn't $40m$ of screens available mean that the sum of three screens $S_1 + S_2 + S_3 = 40m$?
    total length of screen = 40
    perimeter = 2l + w (because two of the length sides and 1 width will be used to cover the screen)

    2l +w = 40
    w = 40 -2l
    or  l =20 - \frac{1}{2}w

    w(20 - \frac{1}{2}w)\geq 150

     20w-\frac{1}{2}w^2\geq 150

     w^2-40w+300\leq 0

    my answer is  10\leq w\leq 30 but the book says  5\leq w\leq 15 where did i go wrong thank you for any help
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  4. #4
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    Re: inequality word problem

    $x(40-2x) \ge 150$

    $40x - 2x^2 \ge 150$

    $0 \ge 2x^2 - 40x + 150$

    $0 \ge x^2 - 20x + 75$

    $0 \ge (x-5)(x-15) \implies 5 < x < 15$
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