Answer
$20\ in$ by $30\ in$
Work Step by Step
Step 1. Assume the original width is $x$, the dimensions of the box would be $x-4$ by $x+10-4$ by $2$
Step 2. The volume of the box would be $V=(x-4)(x+6)(2)$
Step 3. Let $V=832$, we have $(x-4)(x+6)(2)=832$ or $x^2+2x-24=416$ which gives $x^2+2x-440=0$
Step 4. Solve the above equation and discard the negative solution to get $x=20\ in$
Step 5. The original dimensions of the metal are $20\ in$ by $30\ in$