SOLUTION: 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
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Question 1134803: 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 816 in^3, what were the original dimensions of the piece of metal Answer by greenestamps(13200) (Show Source):
Let the dimensions of the original piece of metal be x and x+30.
When squares of side length 6 are cut from each corner to form the open box, the dimensions of the base are x-12 and (x+30)-2 = x+18; the height of the box is 6.
We are told that the volume of the box (length times width times height) is 816:
or
Reject the negative answer; it makes no sense in the problem.
ANSWER: The dimensions of the original piece of metal were x=16 and x+30 = 46.
