A rectangular piece of metal is 30 in longer than it is wide. Squares with sides 6 in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 2706 in^3, what were the original dimensions of the piece of metal?
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Let x be the width of the rectangular piece of metal.
The length of the rectangular piece of metal is x + 30.
The height of the box is 6 in.
The width of the box is x - 2(6 in) = x - 12 in.
The length of the box is x + 30 - 2(6 in) = x + 30 - 12 in = x + 18 in.
The volume of the box is (x - 12 in)(x + 18 in)(6 in) = 2706 in^3.
Expanding the equation, we get 6x^2 + 72x - 216x - 2592 = 2706 in^3.
Combining like terms, we get 6x^2 - 144x - 528 = 0.
Dividing both sides by 6, we get x^2 - 24x - 88 = 0.
We can solve the quadratic equation by factoring:
(x - 11)(x - 8) = 0.
The possible values for x are 11 and 8.
If x = 11, the length of the rectangular piece of metal is 11 + 30 = 41 in.
If x = 8, the length of the rectangular piece of metal is 8 + 30 = 38 in.
Therefore, the original dimensions of the piece of metal are either 11 in by 41 in or 8 in by 38 in. Answer: \boxed{11 \textrm{ in} \times 41 \textrm{ in}}.
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