1. ## 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

2. ## Re: inequality word problem

Doesn't $40m$ of screens available mean that the sum of three screens $S_1 + S_2 + S_3 = 40m$?

3. ## Re: inequality word problem

Originally Posted by Ahri
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

4. ## Re: inequality word problem

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

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

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

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

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