Help!!!

A rectangle piece of cardboard twice as long as wide is to be made into an open box by cutting 2 in. squares from each corner and bending up the sides. (a) Express the volume V of the box as a function of the width W of the piece of cardboard (b) find the domain of the function.

Thanks!

Length = 2W (inches)

Width = W (inches)
Height = 2 inches
V(W)=Length*Width*Height
=2W*W*2
=4W²
The minimum width of the box is 4 inches, which results in a box of zero volume.
Thus the domain of V(W) is [4,∞].

Thanks MathMate!

well i would find a factor of something by dividing numbers from 1 to 10 and figure it out by dividing it up

To find the volume V of the box, we need to first determine the length and height of the box.

(a) Let's start by expressing the length of the box in terms of the width of the cardboard. We are given that the length is twice as long as the width. So, the length L can be written as L = 2W.

Next, let's consider the height of the box. When we cut squares of 2 inches from each corner and fold up the sides, the height will be 2 inches. Therefore, the height H is fixed at H = 2 inches.

Finally, we can calculate the volume V using the formula for the volume of a rectangular box: V = Length × Width × Height. Substituting the values for length, width, and height, we have:

V = (2W) × W × 2 = 4W^2

Therefore, the volume V of the box is given as a function of the width W, V(W) = 4W^2.

(b) To find the domain of the function, we need to consider any restrictions or limitations on the width W. Since the problem states that the width is twice the length, we can assume that the width should have a positive value.

Therefore, the domain of the function is all positive real numbers, or W > 0.

In summary:
(a) The volume function V(W) = 4W^2.
(b) The domain of the function is W > 0.