if a rectangular piece of metal is 5 inches longer than it is wide squares with sides 1 inch longer or cut from the four corners and the flaps are folded upward to form and open box if the volume of the box is 546 inches to the third power what were the original dimensions of the metal?

1. Let the width of the rectangular piece of metal be \( w \) inches. Then, the length will be \( w + 5 \) inches.

2. After cutting squares with sides 1 inch from each corner and folding the flaps upward, the resulting dimensions of the open box will be:

- Width: \( w - 2 \) inches

- Length: \( w + 3 \) inches

- Height: 1 inch

3. The volume \( V \) of the box is given as 546 cubic inches. Using the volume formula \( V = \text{length} \times \text{width} \times \text{height} \):

546 = (w + 3)(w - 2) \cdot 1

4. Simplify and solve for \( w \):

546 = (w + 3)(w - 2)

546 = w^2 + 3w - 2w - 6

546 = w^2 + w - 6

w^2 + w - 552 = 0

5. Solve the quadratic equation using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

a = 1, b = 1, c = -552

w = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-552)}}{2 \cdot 1}

w = \frac{-1 \pm \sqrt{1 + 2208}}{2}

w = \frac{-1 \pm \sqrt{2209}}{2}

w = \frac{-1 \pm 47}{2}

6. We have two possible solutions for \( w \):

w=\frac{46}{2}=23\quad

w = \frac{-48}{2} = -24 \quad \text{(not possible, width and length must be positive)}

7. Therefore, the original dimensions of the piece of metal are:

\boxed{\text{Width }=23\text{ inches, Length }=28\text{ inches}}