# Thread: word problem involving box

1. ## word problem involving box

Constructing a Box
An open box is to be constructed from a rectangular piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the length of the rectangle is twice its width and if the box is to hold 4 cubit feet, what should be the dimensions of the sheet metal?

The first thing I did was draw a picture and label the dimensions.
If you were looking directly down on the box from above, the right side is 3w, the left inside of the box between the one foot squares is w-2, the longer side of the back is 2w, as well as the long side of the front 2w, the inside of the front between the two one foot squares is 2w-2. I wish I could put my picture on here, so you could get a better idea of what I am trying to say.
Then I did the following:
l= 2w -2 w= w-2 h=1
v= (2w-2)(w-2)(1)
v= 2w^2-8w +4
4= 2w^2 -6w +4
0= 2w^2-6w
0= 2w(w-3)
w= 3
Would someone look this over and let me know if this is the right way to go about this problem or if there is another approach.
Thank you,
Keith

2. Originally Posted by keith
Constructing a Box
An open box is to be constructed from a rectangular piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the length of the rectangle is twice its width and if the box is to hold 4 cubit feet, what should be the dimensions of the sheet metal?

The first thing I did was draw a picture and label the dimensions.
If you were looking directly down on the box from above, the right side is 3w, the left inside of the box between the one foot squares is w-2, the longer side of the back is 2w, as well as the long side of the front 2w, the inside of the front between the two one foot squares is 2w-2. I wish I could put my picture on here, so you could get a better idea of what I am trying to say.
Then I did the following:
l= 2w -2 w= w-2 h=1
v= (2w-2)(w-2)(1)
v= 2w^2-8w +4
4= 2w^2 -6w +4
0= 2w^2-6w
0= 2w(w-3)
w= 3
Would someone look this over and let me know if this is the right way to go about this problem or if there is another approach.
Thank you,
Keith
I have pretty much the same method, however, your work is incomplete, you haven't told us the dimensions:

See diagram below

For the box:

Length = 2x - 2 ft
Width = x - 2 ft
height = 1 ft

Now volume = l*w*h
=> V = (2x - 2)(x - 2)(1)

we want V = 4

so V = (2x - 2)(x - 2) = 4
=> 2x^2 - 6x + 4 = 4
=> 2x^2 - 6x = 0
=> 2x(x - 3) = 0
=> 2x = 0 or x - 3 = 0
=> x = 0 ........impossible
or x = 3

Take x = 3

So the dimensions of the box are:
Length = 2(3) - 2 ft = 4 ft
Width = (3) - 2 ft = 1 ft
height = 1 ft

EDIT (Thanks Soroban): The question asked for the dimensions of the original metal sheet, not the box. The length of the metal sheet is 2x = 6, and the width is x = 3

3. Hello, Keith!

You did a great job!

Constructing a Box
An open box is to be constructed from a rectangular piece of sheet metal
by removing a square of side 1 foot from each corner and turning up the edges.
If the length of the rectangle is twice its width and if the box is to hold 4 ft³.
what should be the dimensions of the sheet metal?

The first thing I did was draw a picture and label the dimensions. .Good!
Code:
      : - - - -  2W  - - - - :
- *---*--------------*---* -
: |   |              |   | 1
: * - *              * - * -
: |                      |
W |                      | W-2
: |                      |
: * - *              * - * -
: |   |              |   | 1
- *---*--------------*---* -
: 1 : - - 2W-2 - - : 1 :
Then I did the following:

. . L = 2W -2, .W = W-2, .H = 1

. . V .= .(2W - 2)(W - 2)(1) .= .2W² - 8w + 4

. . 2W² - 6W + 4 .= .4 . . 2W² - 6W .= .0 . . 2W(W - 3) .= .0

Hence: .W = 3
Punchline . The width of the sheet metal is 3 feet; its length is 6 feet.

. .
Good work!