Answer:
15 in × 35 in
Explanation:
You want to know the original dimensions of a rectangular piece of metal from which 4" squares are cut from the corners and flaps folded to form an open box with volume 756 cubic inches. The metal is 20 inches longer than it is wide.
Box dimensions
When 4" squares are cut from the corners, the "flaps" left are 4" wide on each side of the original piece of metal. When those are folded up, the depth of the box is 4" and the dimensions of the bottom of the box are 8" smaller than the dimensions of the original metal piece.
The box dimensions are 4" × (w -8)" × ((w+20) -8)", so the volume is ...
V = LWH = 4(w-8)(w+12) = 4(w² +4w -96)
We want to find w when the volume is 756 cubic inches.
Solution
4(w² +4w -96) = 756
w² +4w -96 = 189 . . . . . divide by 4
w² +4w +4 = 289 . . . . . . add 100 to "complete the square"
(w+2)² = 17² . . . . . . . . . . write as squares
w +2 = 17 . . . . . . . . . . . . take the positive square root
w = 15 . . . . . . . . subtract 2
The original piece of metal was 15 inches wide and 35 inches long.
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Additional comment
The box dimensions are 4" × 7" × 27".