Arectangular 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 1536 incubed, what were the original dimensions of the piece of metal?

Step-by-step explanation:

This is good!  The length of the metal is 20 inches longer than the width.  But we fold up 4-inch flaps on each side, so the height of the box is 4 inches, and the length and width will each decrease by 8 inches (4 inches on each side).  Since the length and width decrease by the same amount, the length will still be 20 inches longer than the width.  So the equations would be:

l = w + 20

vol = l x w x h = 1716 in3

1716 = (w + 20)(w)(4)

1716 = 4w2 + 80w

4w2 + 80w - 1716 = 0

4(w2 + 20w - 429) = 0

4(w + 33)(w - 13) = 0

w + 33 = 0, so w = -33

w - 13 = 0, so w = 13

Since the width can't be negative, the width of the box is 13 and the length is 33 (13 + 20).

But remember, the length and width of the piece of metal are each 8 inches longer than the box, so it was 41 inches by 21 inches.

answer:

step-by-step explanation: