Solving for a Problem Involving Volume

[holmes1172]

Okay, I have a rectangular piece of metal that is 10 inches longer than it is wide. Squares with 2in. long sides are cut from the four corners and the flaps are folded up to form an open box. The volume of this box is 832<sup>3 in. , what were the original dimensions of the piece of metal?

Okay, so:</sup>

width=x
length=x+10
volume=832³
four corner squares=2

832²=(x+10)(x)(2)
832=x²+20x(2x)
I know I'm supposed to subtract 832 from both sides giving me 0=x²+20x(2x) but somehow this doesn't feel right with what I've already come up with. Could somebody steer me in how I need to correct or go about this problem? Don't solve it for me, just explain if you could. Thanks!

 

[Quaid]

width of the metal sheet = x

length of the metal sheet = x+10

832²=(x+10)(x)(2)

There are different widths and different lengths, in this exercise, so I added clarifications above.

That volume equation needs to use the dimensions of the box. (2 is the correct height; the other two dimensions are incorrect.)

You have confused the width of the box with the width of the metal sheet.

Same with the lengths.

The metal sheet is x+10 by x. The dimensions of the box's base are different.

In the diagram below, the pinkish rectangle is the sheet of metal.

The red line shows the length of the box, and the blue line shows the width of the box.

Do you understand your mistake, when you wrote your volume equation?

 

[Quaid]

Hi holmes1172:

Did you finish this exercise?

:cool:

 

[holmes1172]

Hi holmes1172:

Did you finish this exercise?

:cool:

Yes. Sorry it took me so long to respond here. The answer was 20 inches by 30 inches...