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Thread: worded inequalities

  1. #1
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    worded inequalities

    Q: The sketch below shows a plan of a living room. The length of the room is (2x-1) and the width of the room is (x+1) where x is measured in meters
    worded  inequalities-bos-1.jpg


    Given that the area of the room must be at least 135 square meters and the total length of the walls cannot exceed 54 meters.
    (a) Find the set of values of x that satisfy both constraints.
    (b) Hence find the maximum and minimum values of the area and perimeter of the room.


    My attempt:
    for part a
     p = (2x-1)+(2x-1)+(x+1)+(x+1)
     p=6x
    but p<54
     6x<54
    x<9
    therefore x+1<9+1 = x+1<10
    and  2(x)<18
     2x-1<17

    thus x+1<10 ; 2x-1<17

    please point out if i did anything wrong


    for part b

    i dont know how to answer part b

    please help
    thank you
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  2. #2
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    Re: worded inequalities

    in (a) you didn't address the area constraint.

    You correctly found the upper bound on $x$, but you need to find a lower bound as well.

    Spoiler:
    $area = length \times width = 2 x^2+x-1 \geq 135$

    $2x^2 + x - 136 \geq 0$

    $(x-8) (2 x+17)$

    $x =8,~-\dfrac {17}{2}$

    the parabola is upwards facing so it must be that $x \geq 8$ in order to satisfy $area \geq 135$

    Thus the bounds on $x$ are

    $8 \leq x \leq 9$

    b) within the bounds perimeter and area increase with increasing $x$

    use this to determine the max and min perimeter and area.
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  3. #3
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    Re: worded inequalities

    Quote Originally Posted by romsek View Post
    in (a) you didn't address the area constraint.

    You correctly found the upper bound on $x$, but you need to find a lower bound as well.

    Spoiler:
    $area = length \times width = 2 x^2+x-1 \geq 135$

    $2x^2 + x - 136 \geq 0$

    $(x-8) (2 x+17)$

    $x =8,~-\dfrac {17}{2}$

    the parabola is upwards facing so it must be that $x \geq 8$ in order to satisfy $area \geq 135$

    Thus the bounds on $x$ are

    $8 \leq x \leq 9$

    b) within the bounds perimeter and area increase with increasing $x$

    use this to determine the max and min perimeter and area.
    for part b
    WHEN X = 8

    length = (2x-1) = (2(8)-1) = 16-1=15 width = (x+1) = (8+1) = 9
    area = 15 x 9 = 135
    perimeter = 2(15) + 2(9) = 30 + 18 = 48


    WHEN X = 9
    length = (2x-1) = (2(9)-1) = 17
    width = (x+1) = (9+1) = 10
    area = 17 X 10 = 170
    perimeter = 2(17)+2(10) = 34+20 = 54

    minimum perimeter = 48, maximum area = 135
    maximum perimeter = 54, maximum area = 170

    Am I right
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  4. #4
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    Re: worded inequalities

    minimum perimeter = 48, maximum area = 135
    maximum perimeter = 54, maximum area = 170

    Am I right
    yes
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