# Thread: worded inequalities

1. ## 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

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

thank you

2. ## 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.

3. ## Re: worded inequalities

Originally Posted by romsek
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

4. ## Re: worded inequalities

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

Am I right
yes