# Thread: Box Construction

1. ## Box Construction

(1) An open box is to be constructed from a square piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the box is to hold 4 cubic feet, what should be the dimensions of the sheet metal?

(2) Using the question above, solve if the piece of sheet metal is a rectangle whose length is twice the width.

2. Originally Posted by symmetry
(1) An open box is to be constructed from a square piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the box is to hold 4 cubic feet, what should be the dimensions of the sheet metal?

(2) Using the question above, solve if the piece of sheet metal is a rectangle whose length is twice the width.
(1)The height of the box will be 1 ft, so the area of the base must be 4 sq ft
So if the width of the box is w ft, the length is 4/w ft, and the size of the
sheet it is floded from is w+2 by (4/w)+2 ft.

Opps, did not notice that we has a square sheet of metal, as its square w=2, and the
original sheet is as Earboth says 4'x4'

(2)Here 2(w+2)=(4/w)+2, which you should solve for w, you will have two
roots one of which will not satisfy the conditions to be a width, and the
other will, and it will be the solution.

RonL

3. Originally Posted by symmetry
(1) An open box is to be constructed from a square piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the box is to hold 4 cubic feet, what should be the dimensions of the sheet metal?

(2) Using the question above, solve if the piece of sheet metal is a rectangle whose length is twice the width.
Hello,

to (1):

the volume is $V=A_{base} \cdot h$

the height is 1'. thus the base area must be 4 sqft. That means the base area has the dimension $2'\times 2'$. Therefore the piece of metal must have the dimension $4'\times 4'$

to(2):

let the width of the piece be x
then the length of the piece is 2x
If you cut off squares of $1'\times 1'$ the base area must have the same value:

$(x-2)(2x-2)=4$. Expand the LHS and solve for x. I've got x = 0 or x = 3.

That means the piece of metal has the dimension $3'\times 6'$.

EB