Arectangular piece of metal is 10 in. longer than it is wide. squares with sides 2 in. long are cut from the four corners

Volume of the box = length  * width * height

length of original peice = x + 10 , width = x

length of box  (x + 10) - 4  = x + 6
width of box = x - 4
height of box =  2

so we have 

2 * (x -4) * ( x + 6) = 832

2x^2 + 4x - 48 = 832

2x^2 + 4x - 880 = 0
x^2  + 2x - 440 = 0

(x + 22)(x - 20) = 0

x = 20

so width of original piece of metal = 20 ins , length = 30 inches.

The dimensions of the piece of metal can be represented by x and x+10.

Now, 2 in squares are being cut out of each corner.

So the new dimensions are x-4 (2in from each side) and x+10-4 or x+6.

When you fold it up, the height becomes 2 and the base has dimensions x-4 and x+6. Now plug this into the volume formula.

V=l*w*h

1302 = (x+6)(x-4)(2)

651   = x2+2x-24

0      = x2 + 2x-675

0      = (x+27)(x-25)

x=-27 reject since lengths cannot be negative or x=25.

So your original dimension for the piece of metal are 25 by 35.  

Hope this helped.

Step-by-step explanation:

The dimensions of the piece of metal can be represented by x and x+10.

Now, 2 in squares are being cut out of each corner.

So the new dimensions are x-4 (2in from each side) and x+10-4 or x+6.

When you fold it up, the height becomes 2 and the base has dimensions x-4 and x+6. Now plug this into the volume formula.

V=l*w*h

1302 = (x+6)(x-4)(2)

651   = x2+2x-24

0      = x2 + 2x-675

0      = (x+27)(x-25)

x=-27 reject since lengths cannot be negative or x=25.

So your original dimension for the piece of metal are 25 by 35.

Step-by-step explanation: