Question

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?

Answer 1

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

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

$1904 = 4(x^2+4x-96) \\ x^2 +4x-96 = 476$

$x^2 +4x-96-476=0 \\ x^2+4x -572=0 \\ (x+26)(x-22)=0 \\ x=22 in$

So the width is 22 inch, length is $22+20=42$ inch  more then width, and height is 4 inch .