# check my work w/ this word problem

• Dec 16th 2009, 02:57 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 $\displaystyle 588cm^3$. 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
$\displaystyle 4x^3-80x^2+400x-588=0$
$\displaystyle 4(x^3-20x^2+100x-147)=0$

where is got a factor of (x-3)

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

wich gave me x=15.44 or x=1.55

so the length can be 3cm, 15.44cm, or 1.55cm?
• Dec 16th 2009, 03:40 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 $\displaystyle 588cm^3$. 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
$\displaystyle 4x^3-80x^2+400x-588=0$
$\displaystyle 4(x^3-20x^2+100x-147)=0$

where is got a factor of (x-3)

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

wich gave me x=15.44 or x=1.55

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

$\displaystyle x = 3$

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

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

$\displaystyle x= 13.322$

$\displaystyle x = 3.678$

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