Chapter 1 - Equations and Inequalities - 1.5 Applications and Modeling with Quadratic Equations - 1.5 Exercises - Page 130: 28

$20\ in$ by $40\ in$

Step 1. Assume the original width is $x$, the dimensions of the box would be $x-8$ by $2x-8$ by $4$ Step 2. The volume of the box would be $V=(x-8)(2x-8)(4)$ Step 3. Let $V=1536$, we have $(x-8)(2x-8)(4)=1536$ or $(x-8)(x-4)=192$ which gives $x^2-12x-160=0$ 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 $40\ in$