# check my work w/ this word problem

• Dec 16th 2009, 03:57 PM
extraordinarymachine
check my work w/ this word problem
ok so i'm not sure if my answers are right

An open box is to be formed by cutting four identical squares from the coners of a sheet of metal 20cm by 20cm, and folding up the metal to form sides. the capasity of the box must be $588cm^3$. what is the side length of the squares?

this is what i got

let x be the length of the missing sides

x(20-2x)(20-2x)=588
$4x^3-80x^2+400x-588=0$
$4(x^3-20x^2+100x-147)=0$

where is got a factor of (x-3)

and was left with $x^2-17x+49$

wich gave me x=15.44 or x=1.55

so the length can be 3cm, 15.44cm, or 1.55cm?
• Dec 16th 2009, 04:40 PM
skeeter
Quote:

Originally Posted by extraordinarymachine
ok so i'm not sure if my answers are right

An open box is to be formed by cutting four identical squares from the coners of a sheet of metal 20cm by 20cm, and folding up the metal to form sides. the capasity of the box must be $588cm^3$. what is the side length of the squares?

this is what i got

let x be the length of the missing sides

x(20-2x)(20-2x)=588
$4x^3-80x^2+400x-588=0$
$4(x^3-20x^2+100x-147)=0$

where is got a factor of (x-3)

and was left with $x^2-17x+49$

wich gave me x=15.44 or x=1.55

so the length can be 3cm, 15.44cm, or 1.55cm?

$x = 3$

$x^2-17x+49 = 0$

$x = \frac{17 \pm \sqrt{93}}{2}$

$x= 13.322$

$x = 3.678$

is it physically possible for $x = 13.322$ ???