A rectangular piece of metal is 20 in longer than it is wide. squares with sides 4 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 1904 incubed, what were the original dimensions of the piece of metal?

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Riia Riia · Answer 1 · 2018-08-31T09:34:18+02:00

Let the width be x inch. SO length = x+20.

And 4 inch cut out at each corner, so we get[tex] Volume = length* width * height \\ 1904 = (x+20-2*4)*(x-2*4)*4 \\ 1904 = (x+12)* (x-8)*4 [/tex]

[tex] 1904 = 4(x^2+4x-96) \\ x^2 +4x-96 = 476 [/tex]

[tex] x^2 +4x-96-476=0 \\ x^2+4x -572=0 \\ (x+26)(x-22)=0 \\ x=22 in [/tex]

So the width is 22 inch, length is [tex] 22+20=42 [/tex] inch more then width, and height is 4 inch .

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