A rectangular box has sides with lengths x, x+1, and x+2. If the volume of the box is 504 cubic inches, find the dimensions of the box.

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- Jul 16th 2009, 01:56 PMmoomoo762Someone please help solve this problem
A rectangular box has sides with lengths x, x+1, and x+2. If the volume of the box is 504 cubic inches, find the dimensions of the box.

- Jul 16th 2009, 02:19 PMskeeter
- Jul 16th 2009, 02:19 PMyeongil
You should know that the formula for the volume of a rectangular prism is

V = LWH.

So plug in:

$\displaystyle 504 = x(x + 1)(x + 2)$

(let's switch sides)

$\displaystyle x(x + 1)(x + 2) = 504$

Multiply out the left side, and afterwards, subtract 504 from both sides. You'll end up with a cubic equation equaling zero. Solve for x. Don't forget to reject any solution that gives you a negative number for**any**of the lengths.

01

EDIT: beaten to it by skeeter! (Rofl) - Jul 16th 2009, 02:45 PMmoomoo762
I'm stuck

I subtract 504 from both sides to make it equal to zero and I get

x^3+5x^2+2x-504=0

What's the best way to solve for x now? - Jul 16th 2009, 03:12 PMskeeter
- Jul 16th 2009, 03:30 PMmoomoo762
Using the rational root test I got 7

I know the other two sides are 8 and 9 but how do I go about finding them? - Jul 16th 2009, 04:21 PMskeeter
first of all, correct your expansion ...

$\displaystyle x(x+1)(x+2) = 504$

$\displaystyle x^3 + 3x^2 + 2x - 504 = 0$

you know 7 is a root ... use synthetic division

Code:`7] 1.......3........2.......-504`

............7.......70........504

-------------------------------

....1......10.......72.........0

$\displaystyle x^2 + 10x + 72 = 0$

since $\displaystyle b^2-4ac < 0$ , the other two roots are imaginary.

$\displaystyle x = 7$ is the only real root

go back to the original equation ...

$\displaystyle x(x+1)(x+2) = 504$

$\displaystyle 7(7+1)(7+2) = 504$ - Jul 16th 2009, 04:24 PMStroodle
$\displaystyle x(x+1)(x+2) = 504$

$\displaystyle x(x^2+3x+2)=504$

$\displaystyle x^3+3x^2+2x-504=0$

$\displaystyle (x-7)(x^2+10x+72)=0$

Notice that there are no real solutions for the quadratic factor as $\displaystyle \triangle <0$

So the only real solution is $\displaystyle x=7$

The question stated the sides are $\displaystyle x,\ x+1,\ x+2$ so one side is $\displaystyle 7$ and the others are $\displaystyle 7+1=8$ and $\displaystyle 7+2=9$