To determine the dimensions of the open box, we start by understanding how the dimensions relate to the original square sheet of metal.
Let \( x \) be the length of one side of the original square sheet. After cutting out squares of size \( 3 \) inches from each corner and folding up the sides, the resulting dimensions of the open box will be:
- Length: \( x - 2 \times 3 = x - 6 \) inches (subtracting 3 inches from each side)
- Width: \( x - 2 \times 3 = x - 6 \) inches
- Height: \( 3 \) inches (the height is simply the size of the square cut from the corners)
The volume \( V \) of the open box can be calculated using the formula:
\[ V = \text{Length} \times \text{Width} \times \text{Height} \]
Substituting the expressions we found for the dimensions of the box, we have:
\[ V = (x - 6)(x - 6)(3) \]
This volume must equal \( 147 \) cubic inches. Therefore, we set up the equation:
\[ 3(x - 6)^2 = 147 \]
Now, we can simplify this equation. First, divide both sides by \( 3 \):
\[ (x - 6)^2 = 49 \]
Next, we take the square root of both sides:
\[ x - 6 = 7 \quad \text{or} \quad x - 6 = -7 \]
Solving the first equation:
\[ x - 6 = 7 \implies x = 13 \]
Solving the second equation:
\[ x - 6 = -7 \implies x = -1 \quad (\text{This is not a valid length.}) \]
So, we have \( x = 13 \) inches for the side length of the original square sheet.
Now, we can find the dimensions of the open box:
- Length: \( x - 6 = 13 - 6 = 7 \) inches
- Width: \( x - 6 = 13 - 6 = 7 \) inches
- Height: \( 3 \) inches
Thus, the dimensions of the open box are:
- Length: \( 7 \) inches
- Width: \( 7 \) inches
- Height: \( 3 \) inches
To confirm, we calculate the volume:
\[ V = 7 \times 7 \times 3 = 147 \text{ cubic inches} \]
This is consistent with the given volume requirement. Therefore, the dimensions of the open box are:
\[ \boxed{7 \text{ inches} \text{ (length)} , \text{ and } , 7 \text{ inches} \text{ (width)}} \]