Question

A rectangular piece of metal is 30 in longer than it is wide. Squares with sides 6 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 2400 incubed​, what were the original dimensions of the piece of​ metal?

Answer 1

22 inch by 52 inch

Explanation:

If the Width of the metal=x

The length is 30inch longer than its width= x+30

If Squares with sides 6 in long are cut from the four corners and the flaps are folded.

The length of the pre-folded box =x+30-12=(x+18)inch

The width of the pre-folded box = (x-6-6)inch = (x-12) inch

The height of the flap= 6 inch

Volume of the box =2400 inch cubed

Volume of a Cuboid=LWH

Therefore:

6(x-12)(x+18)=2400

6x²+36x-1296=2400

6x²+36x-1296-2400=0

6x²+36x-3696=0

6(x-22)(28+x)=0

x-22=0 or x+28=0

x=22 inch (Since length cannot be negative)

The original dimensions of the metal sheet are 22 inch by 52 inch