# Box Dimensions

• Mar 9th 2009, 07:32 AM
orangepower
Box Dimensions
The dimensions of a gift box are consecutive positive integers such that the height is the least integer and the length is the greatest integer. If the height is increased by 1cm, the width is increased by 2cm, and the length is increased by 3cm, then a larger box is constructed such that the volume is increased by 456cm3. determine the dimensions of each box.

and so far I got to
LWH=V
(L+3)(W+2)(H+1)= V+456

.___. then I got stuck >< I don't even know if I'm on the right track.
According to the answers in the back, the dimensions of the first box are
9cm by 8cm by 7cm
the second box would be
12cm by 10cm by 8cm

• Mar 9th 2009, 08:05 AM
earboth
Quote:

Originally Posted by orangepower
The dimensions of a gift box are consecutive positive integers such that the height is the least integer and the length is the greatest integer.
Therefore w = h+1 and l = h+2
If the height is increased by 1cm, the width is increased by 2cm, and the length is increased by 3cm, then a larger box is constructed such that the volume is increased by 456cm3. determine the dimensions of each box.

and so far I got to
LWH= h(h+1)(h+2) = h^3+3h^2+2h = V
((h+2)+3)((h+1)+2)(H+1)= V+456

.___. then I got stuck >< I don't even know if I'm on the right track.
According to the answers in the back, the dimensions of the first box are
9cm by 8cm by 7cm
the second box would be
12cm by 10cm by 8cm

I've added some necessary information :D
• Mar 9th 2009, 08:25 AM
Soroban
Hello, orangepower!

You overlooked the most powerful condition in the problem.

Quote:

The dimensions of a gift box are consecutive positive integers
such that the height is the least integer and the length is the greatest integer.
If the height is increased by 1 cm, the width is increased by 2 cm,
and the length is increased by 3 cm,
then a larger box is constructed such that the volume is increased by 456 cm³.
Determine the dimensions of the box.

We have: . $\begin{array}{ccc}\text{Height} &=& n \\ \text{Width} &=& n+1 \\ \text{Length} &=& n+2 \end{array}$

The original volume is: . $V_o \:=\:n(n+1)(n+2) \:=\:n^3 + 3n^2 + 2n$

The larger box has dimensions: . $\begin{array}{ccc}\text{Height} &=& n+1 \\ \text{Width} &=& n+3 \\ \text{Length} &=& n+5\end{array}$

The larger volume is: . $V_1 \:=\:(n+1)(n+3)(n+5) \:=\:n^3 + 9n^2 + 23n + 15$

The larger volume is 456 cm³ more than the original volume.
. . $n^3 + 9n^2 + 23n + 15 \;=\;n^3 + 3n^2 + 2n + 456$

This simplifies to: . $6n^2 + 21n - 441 \:=\:0 \quad\Rightarrow\quad 2n^2 + 7n - 147 \:=\:0$

. . which factors: . $(n-7)(2n+21) \:=\:0$

. . and has roots: . $n \:=\:7,\:\text{-}\tfrac{21}{2}$

Therefore, the dimensions of the box are: . $7\text{ by }8\text{ by }9$

• Mar 9th 2009, 09:12 AM
orangepower
^ I know how you found 7 but how did you find 8 and 9? D:

ahhh >< I finally got it THANKS!!!