Chapter 1 - Equations and Inequalities - 1.5 Applications and Modeling with Quadratic Equations - 1.5 Exercises - Page 130: 27

$20\ in$ by $30\ in$

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 $30\ in$