Question

in exercise 23, suppose that the piece of metal has length twice the width and 4-in squares are cut from the corners. if the volume of the box is 1536in^3 what were the original dimensions of the piece of metal?

Answer 1

To find the original dimensions of the piece of metal, we calculate the width and then the length, knowing the volume of the resultant box and the fact that length is twice the width with 4-inch squares removed from each corner.

Step-by-step explanation:

The student is being asked to find the original dimensions of a piece of metal from which a box was made by cutting 4-inch squares from each corner and folding up the sides. To solve this, let's denote the original width of the metal as w inches, and therefore, the length will be 2w inches since it's stated to be twice the width.

After cutting out the 4-inch squares from each corner, the new dimensions of the metal would be (w - 8) inches in width and (2w - 8) inches in length because we subtract 4 inches from each side of the width and length. The resulting height of the box, after folding it, would be 4 inches.

The volume of the folded box is given to be 1536 in3. We use the volume formula for a box (Volume = length × width × height) to set up the equation:

(2w - 8) × (w - 8) × 4 = 1536

Solving this equation will give us the value of w, which can then be used to find the original dimensions of the metal piece by substituting back into 2w for the length and w for the width.

Answer 2

l=2w
l x w=4 in²
V=lx wx H=1536in^3, so H= 1536 /lxw=384
and l x (l/2) xH =1536
l²/2 x384 =1536, implies I=sqrt(1536/192) so I= 2.82
and when l = 2.82=2w we can get w=I/2=2.82/2=1.41
finally the original dimensions of the piece of metal were
I=2.82, w=1.41 and H= 384 (the uniti is inch)