Question \( 9,1.5 .27 \) Part 1 di 2 A rectangular piece of metal is 30 in longer than it is wide. Squares with sides 6 in long are cut from the four comers and the flaps are folded upward to form an \( 14 \%, 11.5 \) of 14 points open box. If the volume of the box is 1554 in \( ^{3} \), what were the original dimensions of the piece of metal? What is the original width? Win

To solve this engaging puzzle, let's denote the width of the rectangular piece of metal as \( w \). Since the length is 30 inches longer than the width, we can express the length as \( w + 30 \). After cutting squares of 6 inches from all corners, the new dimensions of the box become \( w - 12 \) for the width (since 6 inches are removed from both sides) and \( w + 30 - 12 \) for the length. The height will simply be the side length of the squares cut out, which is 6 inches. Now, to find the volume \( V \), we use the formula: \[ V = \text{length} \times \text{width} \times \text{height} = (w + 30 - 12)(w - 12)(6) \] Setting this equal to 1554 in³ gives us: \[ (w + 18)(w - 12)(6) = 1554 \] Dividing both sides by 6: \[ (w + 18)(w - 12) = 259 \] Expanding and rearranging leads to a quadratic equation, where solving for \( w \) will reveal the original dimensions! To solve directly from this equation: 1. Expand: \( w^2 + 6w - 216 = 0 \). 2. Use the quadratic formula: \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 6, c = -216 \). 3. Plug in values to find \( w \). Doing this step leads to the width - and subsequently the length from there! For context, rectangular boxes have fascinating historical relevance; they were utilized for trade and storage as far back as ancient Mesopotamia, where they became essential in facilitating commerce. The open box we're calculating embodies that spirit of innovation, elegantly packing and preserving items. Also, remember this principle if attempting the math yourself: Albeit seemingly simple, creating errors with dimensions can lead to major miscalculations! Always double-check your units and the arithmetic along the way to ensure no sneaky mistakes sneak into your final answer. Good luck with those metal dimensions!