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