+52 A rectangular piece of metal is 20 in longer than it is wide. Squares with sides 4 in long are cut from the four coners and the flaps are folded upward to form an open box. If the volume of the box is 1904 in^3, what were the original dimensions of the piece of metal?
What is the width? in
What is the original length? in
+129899
Let the original width be = W
And the original length = W + 20
When the squares are cut the dimensions of the box = (W - 8) in, (W + 20 - 8) in and 4 in
And we have that
(W - 8) ( W + 20 - 8) (4) = 1903
(W - 8) ( W + 12) (4) = 1903 simplify
4W^2 + 16W - 384 = 1903
4W^2 + 16W - 2287 = 0
Using the quadratic formula to solve this ( and taking the positive answer) the solution is that W ≈ 21.995 in
And the length ≈ 21.995 + 20 ≈ 41.995 in
